Skip to main content

Tutorial 7: Matrix Algebra for Homogeneous Linear Algebraic System

  1. The following linear algebraic equations represents the 3 dof mass spring systems below:

    Figure for Question 1
    [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m2k3+k4m3]{x1x2x3}−ω2[100010001]{x1x2x3}={000}\left[\begin{array}{ccc} \frac{k_{1}+k_{2}}{m_{1}} & -\frac{k_{2}}{m_{1}} & 0 \\ -\frac{k_{2}}{m_{2}} & \frac{k_{2}+k_{3}}{m_{2}} & -\frac{k_{3}}{m_{2}} \\ 0 & -\frac{k_{3}}{m_{2}} & \frac{k_{3}+k_{4}}{m_{3}} \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right\}-\omega^{2}\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\left\{\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right\}=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}

    Determine the smallest eigenvalue and the corresponding eigenvector for the eigenvalue/ eigenvector problem if k1=k4=15Nm,k2=k3=35Nmk_{1}=k_{4}=15 \frac{N}{m}, k_{2}=k_{3}=35 \frac{N}{m}, and m1=m2=m_{1}=m_{2}= m3=1.5 kgm_{3}=1.5 \mathrm{~kg}. Then, draw the eigenvector.

    Solution
    [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m2k3+k4m3]{x1x2x3}−ω2[100010001]{x1x2x3}={000}[15+351.5−351.50−351.535+351.5−351.50−351.535+151.5]{x1x2x3}−ω2[100010001]{x1x2x3}={000}[1003−ω2−7030−7031403−ω2−7030−7031003−ω2]{x1x2x3}={000}\begin{aligned} \left[\begin{array}{ccc}\frac{k_1+k_2}{m_1} & -\frac{k_2}{m_1} & 0 \\ -\frac{k_2}{m_2} & \frac{k_2+k_3}{m_2} & -\frac{k_3}{m_2} \\ 0 & -\frac{k_3}{m_2} & \frac{k_3+k_4}{m_3}\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}-\omega^2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \left[\begin{array}{ccc}\frac{15+35}{1.5} & -\frac{35}{1.5} & 0 \\ -\frac{35}{1.5} & \frac{35+35}{1.5} & -\frac{35}{1.5} \\ 0 & -\frac{35}{1.5} & \frac{35+15}{1.5}\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}-\omega^2\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \left[\begin{array}{ccc}\frac{100}{3}-\omega^2 & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\omega^2 & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\omega^2\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\} \end{aligned}

    For eigenvalue/eigenvector problem: ([A]−λi[I]){x}i={0}\left([A]-\lambda_i[I]\right)\{x\}_i=\{0\}

    Coefficient matrix, [A]=[1003−7030−7031403−7030−7031003][A]=\left[\begin{array}{ccc}\frac{100}{3} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}\end{array}\right]

    Eigenvalue, λ=ω2\lambda=\omega^2 in this case

    For nontrivial solution, {x1x2x3}≠{000}\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\} \neq\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\},

    ∣1003−ω2−7030−7031403−ω2−7030−7031003−ω2∣=0∣1003−λ−7030−7031403−λ−7030−7031003−λ∣=0\begin{aligned} \left|\begin{array}{ccc} \frac{100}{3}-\omega^2 & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\omega^2 & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\omega^2 \end{array}\right|&=0 \\ \left|\begin{array}{ccc} \frac{100}{3}-\lambda & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\lambda & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\lambda \end{array}\right|&=0 \end{aligned}0=(1003−λ)[(1403−λ)(1003−λ)−(−703)(−703)]−(−703)((−703)(1003−λ))0=(1003−λ)[λ2−80λ+91009]+[49009λ−49000027]0=[−λ3+3403λ2−331009λ+91000027]+[49009λ−49000027]0=−λ3+3403λ2−94003λ+1400009\begin{aligned} 0&=\left(\frac{100}{3}-\lambda\right)\left[\left(\frac{140}{3}-\lambda\right)\left(\frac{100}{3}-\lambda\right)-\left(-\frac{70}{3}\right)\left(-\frac{70}{3}\right)\right]-\left(-\frac{70}{3}\right)\left(\left(-\frac{70}{3}\right)\left(\frac{100}{3}-\lambda\right)\right) \\ 0&=\left(\frac{100}{3}-\lambda\right)\left[\lambda^2-80\lambda+\frac{9100}{9}\right]+\left[\frac{4900}{9}\lambda-\frac{490000}{27}\right] \\ 0&=\left[-\lambda^3+\frac{340}{3} \lambda^2-\frac{33100}{9}\lambda+\frac{910000}{27}\right]+\left[\frac{4900}{9}\lambda-\frac{490000}{27}\right] \\ 0&=-\lambda^3+\frac{340}{3} \lambda^2-\frac{9400}{3} \lambda+\frac{140000}{9} \end{aligned}

    Using calculator or software:

    >> polynomial = [-1 340/3 -9400/3 140000/9]
    >> root=roots(polynomial)
    λ1=3839606=6.3350λ2=1003=33.3333λ3=14954203=73.6650\begin{aligned} \lambda_1&=\frac{3839}{606}=6.3350 \\ \lambda_2&=\frac{100}{3}=33.3333 \\ \lambda_3&=\frac{14954}{203}=73.6650 \end{aligned}

    For case 1 (λ1=3839606=6.3350=ω12)\left(\lambda_1=\frac{3839}{606}=6.3350=\omega_1^2\right):

    [1003−3839606−7030−7031403−3839606−7030−7031003−3839606]{x1x2x3}1={000}[16361606−7030−7038147202−7030−70316361606]{x1x2x3}1={000}→R1→R1(60616361)[1−121614070−7038147202−7030−70316361606]{x1x2x3}1={000}→R2→R2−R1(−703)[1−12161407004134205−7030−70316361606]{x1x2x3}1={000}→R2→R2(2054134)[1−12161407001−184915980−70316361606]{x1x2x3}1={000}→R3→R3−R2(−703)[1−12161407001−18491598000]{x1x2x3}1={000}→R1→R1−R2(−12161407)[10−101−18491598000]{x1x2x3}1={000}\begin{aligned} \begin{bmatrix}\frac{100}{3}-\frac{3839}{606} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\frac{3839}{606} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\frac{3839}{606}\end{bmatrix}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \begin{bmatrix}\frac{16361}{606} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{8147}{202} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{16361}{606}\end{bmatrix}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1\left(\frac{606}{16361}\right)]{} \left[\begin{array}{ccc}1 & -\frac{1216}{1407} & 0 \\ -\frac{70}{3} & \frac{8147}{202} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{16361}{606}\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R2\rightarrow R2-R1\left(\frac{-70}{3}\right)]{} \left[\begin{array}{ccc}1 & -\frac{1216}{1407} & 0 \\ 0 & \frac{4134}{205} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{16361}{606}\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R2\rightarrow R2\left(\frac{205}{4134}\right)]{} \left[\begin{array}{ccc}1 & -\frac{1216}{1407} & 0 \\ 0 & 1 & -\frac{1849}{1598} \\ 0 & -\frac{70}{3} & \frac{16361}{606}\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R3\rightarrow R3-R2\left(\frac{-70}{3}\right)]{} \left[\begin{array}{ccc}1 & -\frac{1216}{1407} & 0 \\ 0 & 1 & -\frac{1849}{1598} \\ 0 & 0 & \mathbf{0} \end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1-R2\left(\frac{-1216}{1407}\right)]{} \left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & -\frac{1849}{1598} \\ 0 & 0 & \mathbf{0} \end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \end{aligned}

    Note: RREF shows rank 2 (i.e. 2 linearly independent vectors)

    x1−x3=0≫x1=x3x_1-x_3=0 \gg x_1=x_3

    x2−18491598x3=0≫x2=18491598x3x_2-\frac{1849}{1598} x_3=0 \gg x_2=\frac{1849}{1598} x_3

    Eigenspace for {x1x2x3}λ=6.3350={x318491598x3x3}=t{1184915981}∣x3=t\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_{\lambda=6.3350}=\left\{\begin{array}{c} x_3 \\ \frac{1849}{1598} x_3 \\ x_3 \end{array}\right\}=\left.t\left\{\begin{array}{c} 1 \\ \frac{1849}{1598} \\ 1 \end{array}\right\}\right|_{x_3=t}, where t∈Rt\in \mathbb{R}

    Eigenvector, {x1x2x3}λ=6.3350={11.15711}\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_{\lambda=6.3350}=\left\{\begin{array}{c} 1 \\ 1.1571 \\ 1 \end{array}\right\}

    Figure for Solution 1

  2. Continue from Q1, determine the second largest eigenvalue and the corresponding normalised eigenvector. Then, draw the eigenvector.

    Solution

    For case 2 (λ2=1003=33.3333=ω22)\left(\lambda_2=\frac{100}{3}=33.3333=\omega_2^2\right):

    [1003−1003−7030−7031403−1003−7030−7031003−1003]{x1x2x3}2={000}[0−7030−703403−7030−7030]{x1x2x3}2={000}→R1↔R2[−703403−7030−70300−7030]{x1x2x3}2={000}→R1→R1(−3/70)[1−4710−70300−7030]{x1x2x3}2={000}→R2→R2(−3/70)[1−4710100−7030]{x1x2x3}2={000}→R3→R3−R2(−70/3)[1−471010000]{x1x2x3}2={000}→R1→R1−R2(−4/7)[101010000]{x1x2x3}2={000}\begin{aligned} \left[\begin{array}{ccc} \frac{100}{3}-\frac{100}{3} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\frac{100}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\frac{100}{3} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \left[\begin{array}{ccc} 0 & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{40}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R 1 \leftrightarrow R 2]{}\left[\begin{array}{ccc} -\frac{70}{3} & \frac{40}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & 0 \\ 0 & -\frac{70}{3} & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R 1 \rightarrow R 1 (-3 / 70)]{} \left[\begin{array}{ccc} 1 & \frac{-4}{7} & 1 \\ 0 & -\frac{70}{3} & 0 \\ 0 & -\frac{70}{3} & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R 2 \rightarrow R 2 (-3 / 70)]{} \left[\begin{array}{ccc} 1 & \frac{-4}{7} & 1 \\ 0 & 1 & 0 \\ 0 & -\frac{70}{3} & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R 3 \rightarrow R 3 - R 2 (-70 / 3)]{} \left[\begin{array}{ccc} 1 & \frac{-4}{7} & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R 1 \rightarrow R 1 - R 2 (-4 / 7)]{} \left[\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_2&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \end{aligned}

    Note: RREF shows rank 2 (i.e 2 linearly independent vectors)

    x1+x3=0≫x1=−x3x_1+x_3=0\gg x_1=-x_3

    x2=0x_2=0

    Eigenspace for {x1x2x3}λ=33.3333={−x30x3}=t{−101}∣x3=t\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=33.3333}=\left\{\begin{array}{c}-x_3 \\ 0 \\ x_3\end{array}\right\}=\left.t\left\{\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right\}\right|_{x_3=t}, where t∈Rt \in \mathbb{R}

    Note: If it is not specific the type of eigenvector, normally unscaled eigenvector by let t=1t=1 is provided for manual calculation.

    Eigenvector, {x1x2x3}λ=33.3333={−101}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=33.3333}=\left\{\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right\}

    Normalised Eigenvector, {x1x2x3}λ=33.3333, normalised =12{−101}={−0.707100.7071}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=33.3333, \text { normalised }}=\frac{1}{\sqrt{2}}\left\{\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right\}=\left\{\begin{array}{c}-0.7071 \\ 0 \\ 0.7071\end{array}\right\}

    Figure for Solution 2

  3. Continue from Q1, determine the largest eigenvalue and the corresponding unscaled eigenvector. Then, draw the eigenvector.

    Solution
    [1003−14954203−7030−7031403−14954203−7030−7031003−14954203]{x1x2x3}3={000}[−8147202−7030−703−16442609−7030−703−8147202]{x1x2x3}={000}→R1→R1(−202/8147)[181414070−703−16442609−7030−703−8147202]{x1x2x3}3={000}→R2→R2−R1(−70/3)[1814140700−8302615−7030−703−8147202]{x1x2x3}3={000}→R2→R2(−615/8302)[1814140700110255930−703−8147202]{x1x2x3}3={000}→R3→R3−R2(−70/3)[181414070011025593000]{x1x2x3}3={000}→R1→R1−R2(814/1407)[10−1011025593000]{x1x2x3}3={000}\begin{aligned} \left[\begin{array}{ccc} \frac{100}{3}-\frac{14954}{203} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3}-\frac{14954}{203} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3}-\frac{14954}{203} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \left[\begin{array}{ccc} \frac{-8147}{202} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{-16442}{609} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{-8147}{202} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\}\\ \xrightarrow[R1\rightarrow R1(-202/8147)]{} \left[\begin{array}{ccc} 1 & \frac{814}{1407} & 0 \\ -\frac{70}{3} & \frac{-16442}{609} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{-8147}{202} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \xrightarrow[R2\rightarrow R2-R1(-70/3)]{} \left[\begin{array}{ccc} 1 & \frac{814}{1407} & 0 \\ 0 & \frac{-8302}{615} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{-8147}{202} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \xrightarrow[R2\rightarrow R2(-615/8302)]{} \left[\begin{array}{ccc} 1 & \frac{814}{1407} & 0 \\ 0 & 1 & \frac{1025}{593} \\ 0 & -\frac{70}{3} & \frac{-8147}{202} \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \xrightarrow[R3\rightarrow R3-R2(-70/3)]{} \left[\begin{array}{ccc} 1 & \frac{814}{1407} & 0 \\ 0 & 1 & \frac{1025}{593} \\ 0 & 0 & \mathbf{0} \\ \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \xrightarrow[R1\rightarrow R1-R2(814/1407)]{} \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & \frac{1025}{593} \\ 0 & 0 & \mathbf{0} \\ \end{array}\right]\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_3&=\left\{\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right\} \\ \end{aligned}

    Note: RREF shows rank 2 (i.e. 2 linearly independent vectors)

    x1−x3=0≫x1=x3x_1-x_3=0 \gg x_1=x_3

    x2+1025593x3=0≫x2=−1025593x3x_2+\frac{1025}{593} x_3=0 \gg x_2=-\frac{1025}{593} x_3

    Eigenspace for {x1x2x3}λ=73.6650={x3−1025593x3x3}=t{1−10255931}∣x3=t\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_{\lambda=73.6650}=\left\{\begin{array}{c} x_3 \\ -\frac{1025}{593} x_3 \\ x_3 \end{array}\right\}=\left.t\left\{\begin{array}{c} 1 \\ -\frac{1025}{593} \\ 1 \end{array}\right\}\right|_{x_3=t}, where t∈Rt\in\mathbb{R}

    Note: Unscaled eigenvectors can be any vector of eigenspace. Normally we just let t=1t=1

    Eigenvector, {x1x2x3}λ=73.6650={1−1.72851}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=73.6650}=\left\{\begin{array}{c}1 \\ -1.7285 \\ 1\end{array}\right\}

    Figure for Solution 3

  4. Combining the results obtained from Q1-Q4, obtain the eigenvector matrix, PP and diagonalize the following matrix. Comment the relationship between the diagonal matrix and the eigenvalue.

    [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m2k3+k4m3]\left[\begin{array}{ccc} \frac{k_{1}+k_{2}}{m_{1}} & -\frac{k_{2}}{m_{1}} & 0 \\ -\frac{k_{2}}{m_{2}} & \frac{k_{2}+k_{3}}{m_{2}} & -\frac{k_{3}}{m_{2}} \\ 0 & -\frac{k_{3}}{m_{2}} & \frac{k_{3}+k_{4}}{m_{3}} \end{array}\right]
    Solution
    A=[1003−7030−7031403−7030−7031003]P=[1−0.707111.15710−1.728510.70711]\mathbf{A}=\left[\begin{array}{ccc} \frac{100}{3} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3} \end{array}\right] \mathbf{P}=\left[\begin{array}{ccc} 1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1 \end{array}\right]

    Diagonal matrix, D=P−1AP\mathbf{D}=\mathbf{P}^{\mathbf{- 1}} \mathbf{A P}

    P−1=1det⁡(P)adjoint⁡(P)\mathbf{P}^{-\mathbf{1}}=\frac{\mathbf{1}}{\operatorname{det}(\mathbf{P})} \operatorname{adjoint}(\mathbf{P})cofactor⁡(P)=[+∣0−1.72850.70711∣−∣1.1571−1.728511∣+∣1.1571010.7071∣−∣−0.707110.70711∣+∣1111∣−∣1−0.707110.7071∣+∣−0.707110−1.7285∣−∣111.1571−1.7285∣+∣1−0.70711.15710∣]=[1.2222−2.88560.81821.41420−1.41421.22222.88560.8182]\begin{aligned} \operatorname{cofactor}(\mathbf{P})&=\begin{bmatrix} +\left|\begin{matrix} 0&-1.7285\\0.7071&1 \end{matrix}\right| & -\left|\begin{matrix} 1.1571&-1.7285\\1&1 \end{matrix}\right| & +\left|\begin{matrix} 1.1571&0\\1&0.7071 \end{matrix}\right|\\ -\left|\begin{matrix} -0.7071&1\\0.7071&1 \end{matrix}\right| & +\left|\begin{matrix} 1&1\\1&1 \end{matrix}\right| & -\left|\begin{matrix} 1&-0.7071\\1&0.7071 \end{matrix}\right|\\ +\left|\begin{matrix} -0.7071&1\\0&-1.7285 \end{matrix}\right| & -\left|\begin{matrix} 1&1\\1.1571&-1.7285 \end{matrix}\right| & +\left|\begin{matrix} 1&-0.7071\\1.1571&0 \end{matrix}\right| \end{bmatrix}\\ &=\begin{bmatrix} 1.2222&-2.8856&0.8182\\ 1.4142&0&-1.4142\\ 1.2222&2.8856&0.8182 \end{bmatrix} \end{aligned}

    adjoint⁡(P)=[cofactor⁡(P)]T=[1.22221.41421.2222−2.885602.88560.8182−1.41420.8182]\operatorname{adjoint}(\mathbf{P})=\left[\operatorname{cofactor}\left(\mathbf{P}\right)\right]^T=\left[\begin{array}{ccc}1.2222 & 1.4142 & 1.2222 \\ -2.8856 & 0 & 2.8856 \\ 0.8182 & -1.4142 & 0.8182\end{array}\right]

    det⁡(P)=∣1−0.707111.15710−1.728510.70711∣=−(−1.7285)(0.7071)−(−0.7071)(1.1571+1.7285)+(1.1571)(0.7071)=4.0808\begin{aligned} \operatorname{det}(\mathbf{P})&=\left|\begin{array}{ccc}1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1\end{array}\right|\\ &=-(-1.7285)(0.7071)-(-0.7071)(1.1571+1.7285)+(1.1571)(0.7071)\\ &=4.0808 \end{aligned}P−1=14.0808[1.22221.41421.2222−2.885602.88560.8182−1.41420.8182]=[0.29950.34650.2995−0.707100.70710.2005−0.34650.2005]\mathbf{P}^{-1}=\frac{1}{4.0808}\left[\begin{array}{ccc} 1.2222 & 1.4142 & 1.2222 \\ -2.8856 & 0 & 2.8856 \\ 0.8182 & -1.4142 & 0.8182 \end{array}\right]=\left[\begin{array}{ccc} 0.2995 & 0.3465 & 0.2995 \\ -0.7071 & 0 & 0.7071 \\ 0.2005 & -0.3465 & 0.2005 \end{array}\right]D=P−1AP=[0.29950.34650.2995−0.707100.70710.2005−0.34650.2005][1003−7030−7031403−7030−7031003][1−0.707111.15710−1.728510.70711]=[1.89742.19541.8974−23.5705023.570514.7693−25.528714.7693][1−0.707111.15710−1.728510.70711]=[6.33500033.333300073.665];D=[λ1000λ2000λ3]\begin{aligned} D&=P^{-1} A P\\ &=\left[\begin{array}{ccc} 0.2995 & 0.3465 & 0.2995 \\ -0.7071 & 0 & 0.7071 \\ 0.2005 & -0.3465 & 0.2005 \end{array}\right]\left[\begin{array}{ccc} \frac{100}{3} & -\frac{70}{3} & 0 \\ -\frac{70}{3} & \frac{140}{3} & -\frac{70}{3} \\ 0 & -\frac{70}{3} & \frac{100}{3} \end{array}\right]\left[\begin{array}{ccc} 1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1 \end{array}\right]\\ &=\left[\begin{array}{ccc} 1.8974 & 2.1954 & 1.8974 \\ -23.5705 & 0 & 23.5705 \\ 14.7693 & -25.5287 & 14.7693 \end{array}\right]\left[\begin{array}{ccc} 1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1 \end{array}\right]\\ &=\left[\begin{array}{ccc} 6.335 & 0 & 0 \\ 0 & 33.3333 & 0 \\ 0 & 0 & 73.665 \end{array}\right] \quad ; \quad \mathbf{D}=\left[\begin{array}{ccc} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{array}\right] \end{aligned}

  5. Determine [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m2k3+k4m3]50\left[\begin{array}{ccc}\frac{k_{1}+k_{2}}{m_{1}} & -\frac{k_{2}}{m_{1}} & 0 \\ -\frac{k_{2}}{m_{2}} & \frac{k_{2}+k_{3}}{m_{2}} & -\frac{k_{3}}{m_{2}} \\ 0 & -\frac{k_{3}}{m_{2}} & \frac{k_{3}+k_{4}}{m_{3}}\end{array}\right]^{50} and comment on the change of it's eigenvalue and eigenvector, as compared to Q4\mathrm{Q} 4.

    Solution
    [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m2k3+k4m3]50=A50=PD50P−1=[1−0.707111.15710−1.728510.70711][6.33500033.333300073.665]50[0.29950.34650.2995−0.707100.70710.2005−0.34650.2005]=[1−0.707111.15710−1.728510.70711][6.3355000033.33335000073.66550][0.29950.34650.2995−0.707100.70710.2005−0.34650.2005]=1093[0.46255450−0.799525780.46255450−0.799525781.38198087−0.799525780.46255450−0.799525780.46255450]\begin{aligned} &\left[\begin{array}{ccc}\frac{k_1+k_2}{m_1} & -\frac{k_2}{m_1} & 0 \\ -\frac{k_2}{m_2} & \frac{k_2+k_3}{m_2} & -\frac{k_3}{m_2} \\ 0 & -\frac{k_3}{m_2} & \frac{k_3+k_4}{m_3}\end{array}\right]^{50}=\mathbf{A}^{50}=\mathbf{P D}^{50} \mathbf{P}^{-1}\\ &=\left[\begin{array}{ccc}1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1\end{array}\right]\left[\begin{array}{ccc}6.335 & 0 & 0 \\ 0 & 33.3333 & 0 \\ 0 & 0 & 73.665\end{array}\right]^{50}\left[\begin{array}{ccc}0.2995 & 0.3465 & 0.2995 \\ -0.7071 & 0 & 0.7071 \\ 0.2005 & -0.3465 & 0.2005\end{array}\right]\\ &=\left[\begin{array}{ccc}1 & -0.7071 & 1 \\ 1.1571 & 0 & -1.7285 \\ 1 & 0.7071 & 1\end{array}\right]\left[\begin{array}{ccc}6.335^{50} & 0 & 0 \\ 0 & 33.3333^{50} & 0 \\ 0 & 0 & 73.665^{50}\end{array}\right]\left[\begin{array}{ccc}0.2995 & 0.3465 & 0.2995 \\ -0.7071 & 0 & 0.7071 \\ 0.2005 & -0.3465 & 0.2005\end{array}\right]\\ &=10^{93}\left[\begin{array}{ccc}0.46255450 & -0.79952578 & 0.46255450 \\ -0.79952578 & 1.38198087 & -0.79952578 \\ 0.46255450 & -0.79952578 & 0.46255450\end{array}\right] \end{aligned}

    The entries above are rounded to eight significant figures; the exact factored form A50=PD50P−1A^{50}=PD^{50}P^{-1} remains the reliable result.

    Eigenvalue (A50)=λi50(\mathbf{A}^{50})=\lambda_i^{50}, for i=1,2,3i=1,2,3. It means each eigenvalue is raised to the power of 50 relative to the original eigenvalue. The eigenvectors of A50\mathbf{A}^{50} and A\mathbf{A} remain unchanged.

  6. Given B=[123011002]\mathbf{B}=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2\end{array}\right] and ∣B∣=2|\mathbf{B}|=2 has an eigenvalue of 2 . Find the remaining eigenvalues and develop the characteristic equation without developing the eigenvalue problem and without performing the determinant.

    Solution

    B=[123011002]\mathbf{B}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]

    ∣B∣=2|\mathbf{B}|=2 has an eigenvalue of 2

    B\mathbf{B} has 3 eigenvalues since it has 3×33 \times 3 matrix.

    From the property,

    Trace⁡(B)=∑λi=λ1+λ2+λ31+1+2=λ1+λ2+(2)λ1+λ2=2(1)\begin{aligned} \operatorname{Trace}(\mathbf{B})&=\sum \lambda_i=\lambda_1+\lambda_2+\lambda_3\\ 1+1+2&=\lambda_1+\lambda_2+(2) \\ \lambda_1+\lambda_2&=2 \end{aligned} \tag{1}

    From the property,

    Determinant⁡(B)=∏λi=λ1λ2λ32=λ1λ2(2)λ1λ2=1(2)\begin{aligned} \operatorname{Determinant}(\mathbf{B})&=\prod \lambda_i=\lambda_1 \lambda_2 \lambda_3\\ 2&=\lambda_1 \lambda_2(2) \\ \lambda_1 \lambda_2&=1 \end{aligned} \tag{2}

    Subs (2) into (1)

    λ1+1λ1=2λ12−2λ1+1=0(λ1−1)2=0λ1=1λ2=1λ1=1\begin{aligned} \lambda_1+\frac{1}{\lambda_1}&=2 \\ \lambda_1^2-2 \lambda_1+1&=0 \\ \left(\lambda_1-1\right)^2&=0 \\ \lambda_1&=1 \\ \lambda_2&=\frac{1}{\lambda_1}=1 \end{aligned}

    ∴λ1=1,λ2=1,λ3=2\therefore \lambda_1=1, \lambda_2=1, \lambda_3=2

    Characteristic equation:

    (λ−1)(λ−1)(λ−2)=0(λ2−2λ+1)(λ−2)=0λ3+(−4)λ2+(5)λ−2=0\begin{aligned} (\lambda-1)(\lambda-1)(\lambda-2)&=0 \\ \left(\lambda^2-2 \lambda+1\right)(\lambda-2)&=0\\ \lambda^3+(-4) \lambda^2+(5) \lambda-2&=0 \end{aligned}

  7. Continue Q6, using Cayley-Hamilton theorem to verify that:

    B−1=12B2+(−2)B+(52)I\mathbf{B}^{-1}=\frac{1}{2} \mathbf{B}^{2}+(-2) \mathbf{B}+\left(\frac{5}{2}\right) \mathbf{I} and B6=[(57)B2+(−108)B+52I]\mathbf{B}^{6}=\left[(57) \mathbf{B}^{2}+(-108) \mathbf{B}+52 \mathbf{I}\right]. Then, compute the B5\mathbf{B}^{5} via the theorem.

    Solution
    B3+(−4)B2+(5)B−2I=0B2+(−4)B+(5)I−2B−1=0B−1=12B2+(−2)B+(52)I (verified) B4+(−4)B3+(5)B2−2B=0B4=(4)B3−(5)B2+2BB4=[(16)B2−(20)B+8I]−(5)B2+2BB4=[(11)B2−(18)B+8I]B5=(11)B3−(18)B2+8BB5=[(44)B2−(55)B+22I]−(18)B2+8BB5=(26)B2−(47)B+22IB6=(26)B3−(47)B2+22BB6=[(104)B2+(−130)B+52I]−(47)B2+22BB6=[(57)B2+(−108)B+52I] (verified) \begin{aligned} \mathbf{B}^3+(-4) \mathbf{B}^2+(5) \mathbf{B}-2 \mathbf{I}&=\mathbf{0}\\ \mathbf{B}^2+(-4) \mathbf{B}+(5) \mathbf{I}-2 \mathbf{B}^{-1}&=\mathbf{0}\\ \mathbf{B}^{-1}&=\frac{1}{2} \mathbf{B}^2+(-2) \mathbf{B}+\left(\frac{5}{2}\right) \mathbf{I} \quad\text { (verified) }\\\\ \mathbf{B}^4+(-4) \mathbf{B}^3+(5) \mathbf{B}^2-2 \mathbf{B}&=\mathbf{0}\\ \mathbf{B}^4&=(4) \mathbf{B}^3-(5) \mathbf{B}^2+2 \mathbf{B}\\ \mathbf{B}^4&=\left[(16) \mathbf{B}^2-(20) \mathbf{B}+8 \mathbf{I}\right]-(5) \mathbf{B}^2+2 \mathbf{B}\\ \mathbf{B}^4&=\left[(11) \mathbf{B}^2-(18) \mathbf{B}+8 \mathbf{I}\right]\\ \mathbf{B}^5&=(11) \mathbf{B}^3-(18) \mathbf{B}^2+8 \mathbf{B}\\ \mathbf{B}^5&=\left[(44) \mathbf{B}^2-(55) \mathbf{B}+22 \mathbf{I}\right]-(18) \mathbf{B}^2+8 \mathbf{B}\\ \mathbf{B}^5&=(26) \mathbf{B}^2-(47) \mathbf{B}+22 \mathbf{I}\\ \mathbf{B}^6&=(26) \mathbf{B}^3-(47) \mathbf{B}^2+22 \mathbf{B}\\ \mathbf{B}^6&=\left[(104) \mathbf{B}^2+(-130) \mathbf{B}+52 \mathbf{I}\right]-(47) \mathbf{B}^2+22 \mathbf{B}\\ \mathbf{B}^6&=\left[(57) \mathbf{B}^2+(-108) \mathbf{B}+52 \mathbf{I}\right] \quad\text { (verified) }\\ \end{aligned}B=[123011002]\mathbf{B}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]B5=(26)B2−(47)B+22I=(26)[123011002][123011002]−(47)[123011002]+22[100010001]=(26)[1411013004]−(47)[123011002]+22[100010001]=[11014501310032]\begin{aligned} \mathbf{B}^5&=(26) \mathbf{B}^2-(47) \mathbf{B}+22 \mathbf{I}\\ &=(26)\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]-(47)\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]+22\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &=(26)\left[\begin{array}{lcc} 1 & 4 & 11 \\ 0 & 1 & 3 \\ 0 & 0 & 4 \end{array}\right]-(47)\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]+22\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &=\left[\begin{array}{ccc} 1 & 10 & 145 \\ 0 & 1 & 31 \\ 0 & 0 & 32 \end{array}\right] \end{aligned}

  8. Explain why the following equation : B5=PD5P−1\mathbf{B}^{5}=\mathbf{P D}^{5} \mathbf{P}^{-1} is not working to solve the Q7Q 7 problem?

    Solution

    Previously we obtained,

    B=[123011002]\mathbf{B}=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right]

    λ1=1,λ2=1,λ3=2\lambda_1=1, \lambda_2=1, \lambda_3=2

    where D=[100010002]\mathbf{D}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{array}\right]

    The eigenvalue/eigenvector problem: (B−λI)x=0(\mathbf{B}-\lambda \mathbf{I}) x=\mathbf{0}

    For λ1=1\lambda_1=1

    [1−12301−11002−1]{x1x2x3}1={000}[023001001]{x1x2x3}1={000}→R1→R1(1/2)[013/2001001]{x1x2x3}1={000}→R1→R1−R2(3/2)R3→R3−R2(1)[010001000]{x1x2x3}1={000}\begin{aligned} \left[\begin{array}{ccc}1-1 & 2 & 3 \\ 0 & 1-1 & 1 \\ 0 & 0 & 2-1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \left[\begin{array}{lll}0 & 2 & 3 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1(1/2)]{} \left[\begin{array}{lll}0 & 1 & 3/2 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1-R2(3/2)]{R3\rightarrow R3-R2(1)} \left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_1&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \end{aligned}

    Note: RREF shows rank 2 (i.e. 2 linearly independent vectors)

    x2=0x_2=0 x3=0x_3=0

    Eigenspace for {x1x2x3}λ=1={x100}=t{100}∣x1=t\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=1}=\left\{\begin{array}{l}x_1 \\ 0 \\ 0\end{array}\right\}=\left.t\left\{\begin{array}{l}1 \\ 0 \\ 0\end{array}\right\}\right|_{x_1=t}, where t∈Rt \in \mathbb{R}

    Note: Unscaled eigenvectors can be any vector of eigenspace. Normally we just let t=1t=1

    Eigenvector, {x1x2x3}λ=1={100}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=1}=\left\{\begin{array}{l}1 \\ 0 \\ 0\end{array}\right\}

    For repeated eigenvalue, λ2=1\lambda_2=1

    Eigenvector, {x1x2x3}λ=1={100}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=1}=\left\{\begin{array}{l}1 \\ 0 \\ 0\end{array}\right\}

    For λ3=2\lambda_3=2

    [1−22301−21002−2]{x1x2x3}={000}[−1230−11000]{x1x2x3}3={000}\begin{aligned} \left[\begin{array}{ccc}1-2 & 2 & 3 \\ 0 & 1-2 & 1 \\ 0 & 0 & 2-2\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \left[\begin{array}{ccc}-1 & 2 & 3 \\ 0 & -1 & 1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\} \end{aligned} →R2→R2(−1)R1→R1(−1)[1−2−301−1000]{x1x2x3}3={000}→R1→R1−R2(−2)[10−501−1000]{x1x2x3}3={000}\begin{aligned} \xrightarrow[R2\rightarrow R2(-1)]{R1\rightarrow R1(-1)} \left[\begin{array}{ccc} 1 & -2 & -3 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1-R2(-2)]{} \left[\begin{array}{ccc} 1 & 0 & -5 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\} \end{aligned}

    Note: RREF shows rank 2 (i.e. 2 linearly independent vectors)

    x1−5x3=0≫x1=5x3x_1-5 x_3=0 \gg x_1=5 x_3 x2−x3=0≫x2=x3x_2-x_3=0 \gg x_2=x_3

    Eigenspace for {x1x2x3}λ=2={5x3x3x3}=t{511}∣x3=t\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_{\lambda=2}=\left\{\begin{array}{c} 5 x_3 \\ x_3 \\ x_3 \end{array}\right\}=\left.t\left\{\begin{array}{l} 5 \\ 1 \\ 1 \end{array}\right\}\right|_{x_3=t} , where t∈Rt\in\mathbb{R}

    Note: Unscaled eigenvectors can be any vector of eigenspace. Normally we just let t=1t=1

    Eigenvector, {x1x2x3}λ=2={511}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=2}=\left\{\begin{array}{l}5 \\ 1 \\ 1\end{array}\right\}

    Eigenvector or modal matrix, P=[115001001]\mathbf{P}=\left[\begin{array}{lll} 1 & 1 & 5 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array}\right]

    det⁡(P)=1(0)−1(0)+5(0)=0\operatorname{det}(\mathbf{P})=1(0)-1(0)+5(0)=0 P−1=1det⁡(P)adjoint⁡(P)=10adjoint⁡(P)= undefined or ∞\mathbf{P}^{-1}=\frac{1}{\operatorname{det}(\mathbf{P})} \operatorname{adjoint}(\mathbf{P})=\frac{\mathbf{1}}{0} \operatorname{adjoint}(\mathbf{P})=\text { undefined or } \infty

    Thus, B5=PD5P−1\mathbf{B}^5=\mathbf{P D}^5 \mathbf{P}^{-1} can't be computed.

  9. Based on Q7\mathrm{Q} 7 and Q8\mathrm{Q} 8, discuss the advantage and disadvantage of diagonalization formula versus the Cayley-Hamilton theorem in solving the power of a matrix.

    Solution
    Diagonalization formula for power of a matrix
    Bk=PDkP−1\mathbf{B}^k=\mathbf{PD}^k\mathbf{P}^{-1}    		Cayley-Hamilton theorem
    f(A)=p0I+p1A+p2A2+…+pnAn=0f(\mathbf{A})=p_0\mathbf{I}+p_1\mathbf{A}+p_2\mathbf{A}^2+\ldots+p_n\mathbf{A}^n=0
    AdvantageCan compute the power of a matrix much faster with single equation.The formulation can be developed using characteristic equation only without the extensive calculation of the eigenvector and eigenvalues.
    Disadvantage
    • Require the complete eigenvalues and eigenvectors data for the diagonal matrix as well as the eigenvector matrix. This procedure can be long if compute manually.
    • In some cases, it can’t be computed because eigenvector matrix might not have inversion especially for repeated root case.
    		Need derivation of the theorem formula for the computation of the higher power of the matrix.

  10. Find all the eigenvalue and normalized eigenvectors in terms of eigenvalue matrix and eigenvector matrix for the matrix C\mathbf{C}.

    C=[011101110]\mathbf{C}=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]

    Then, verify the eigenvalue matrix and eigenvector matrix if they satisfy the eigenvalue/eigenvector problem, i.e. (C−λI)x=0(\mathbf{C}-\lambda \mathbf{I}) \boldsymbol{x}=\mathbf{0}.

    Solution
    C−λI=[−λ111−λ111−λ]\mathbf{C}-\lambda \mathbf{I}=\left[\begin{array}{ccc} -\lambda & 1 & 1 \\ 1 & -\lambda & 1 \\ 1 & 1 & -\lambda \end{array}\right]

    For non-trivial solution, ∣C−λI∣=−λ3+3λ+2=0|\mathbf{C}-\lambda \mathbf{I}|=-\lambda^3+3 \lambda+2=0

    λ1=−1,λ2=−1,λ3=2\lambda_1=-1, \lambda_2=-1, \lambda_3=2

    Eigenvalues or spectral matrix, D=[−1000−10002]\mathbf{D}=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2\end{array}\right]

    For λ1=−1\lambda_1=-1

    [111111111]{x1x2x3}={000}→R3→R3−R1R2→R2−R1[111000000]{x1x2x3}={000}\begin{aligned} \left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R3\rightarrow R3-R1]{R2\rightarrow R2-R1} \left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\} \end{aligned}

    Note: RREF shows rank 1 (i.e. 1 linearly independent vectors) x1+x2+x3=0x_1+x_2+x_3=0 x1=−x2−x3x_1=-x_2-x_3

     Eigenspace for {x1x2x3}λ=−1={−x2−x3x2x3}={−x2x20}+{−x30x3}=t{−110}∣x2=t+s{−101}∣x3=s, where t&S∈R\begin{aligned} \text { Eigenspace for }\left\{\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right\}_{\lambda=-1} &=\left\{\begin{array}{c} -x_2-x_3 \\ x_2 \\ x_3 \end{array}\right\} \\ &=\left\{\begin{array}{c} -x_2 \\ x_2 \\ 0 \end{array}\right\}+\left\{\begin{array}{c} -x_3 \\ 0 \\ x_3 \end{array}\right\}=\left.t\left\{\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right\}\right|_{x_2=t}+\left.s\left\{\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right\}\right|_{x_3=s} \text {, where }t\text{\&}S\in \mathbb{R} \end{aligned}

    Note: Unscaled eigenvectors can be any vector of eigenspace. Normally we just let t=1t=1 or s=1s=1

    Eigenvectors, {x1x2x3}λ=−1={−110}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=-1}=\left\{\begin{array}{c}-1 \\ 1 \\ 0\end{array}\right\} & {−101}\left\{\begin{array}{c}-1 \\ 0 \\ 1\end{array}\right\} for repeated eigenvalues λ1=−1,λ2=−1\lambda_1=-1, \lambda_2=-1 respectively.

    Note: Normalized eigenvectors has magnitude =1=1. It can be obtained by dividing the unscaled eigenvectors with the magnitude.

    Normalized eigenvectors, {x1x2x3}λ=−1={−1/21/20}&{−1/201/2}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=-1}=\left\{\begin{array}{c}-1 / \sqrt{2} \\ 1 / \sqrt{2} \\ 0\end{array}\right\} \&\left\{\begin{array}{c}-1 / \sqrt{2} \\ 0 \\ 1 / \sqrt{2}\end{array}\right\}

    For λ2=2\lambda_2=2

    [−2111−2111−2]{x1x2x3}3={000}→R2→R2(−1/2)R3→R3(−1/2)R1→R1(−1/2)[1−1/2−1/2−1/21−1/2−1/2−1/21]{x1x2x3}3={000}→R3→R3−R1(−1/2)R2→R2−R1(−1/2)[1−1/2−1/203/4−3/40−3/43/4]{x1x2x3}3={000}→R2→R2(4/3)[1−1/2−1/201−10−3/43/4]{x1x2x3}3={000}→R3→R3−R2(−3/4)[1−1/2−1/201−1000]{x1x2x3}3={000}→R1→R1−R2(−1/2)[10−101−1000]{x1x2x3}3={000}\begin{aligned} \left[\begin{array}{ccc}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[ \substack{ R2\rightarrow R2(-1/2)\\ R3\rightarrow R3(-1/2)} ]{R1\rightarrow R1(-1/2)} \left[\begin{array}{ccc}1 & -1/2 & -1/2 \\ -1/2 & 1 & -1/2 \\ -1/2 & -1/2 & 1\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R3\rightarrow R3-R1(-1/2)]{R2\rightarrow R2-R1(-1/2)} \left[\begin{array}{ccc}1 & -1/2 & -1/2 \\ 0 & 3/4 & -3/4 \\ 0 & -3/4 & 3/4\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R2\rightarrow R2(4/3)]{} \left[\begin{array}{ccc}1 & -1/2 & -1/2 \\ 0 & 1 & -1 \\ 0 & -3/4 & 3/4\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R3\rightarrow R3-R2(-3/4)]{} \left[\begin{array}{ccc}1 & -1/2 & -1/2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \xrightarrow[R1\rightarrow R1-R2(-1/2)]{} \left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right]\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_3&=\left\{\begin{array}{l}0 \\ 0 \\ 0\end{array}\right\}\\ \end{aligned}

    Note: RREF shows rank 2 (i.e. 2 linearly independent vectors)

    x1−x3=0≫x1=x3x_1-x_3=0 \quad \gg x_1=x_3 x2−x3=0≫x2=x3x_2-x_3=0 \quad \gg x_2=x_3

    Eigenspace for {x1x2x3}λ=2={x3x3x3}=s{111}∣x3=s\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=2}=\left\{\begin{array}{l}x_3 \\ x_3 \\ x_3\end{array}\right\}=\left.s\left\{\begin{array}{l}1 \\ 1 \\ 1\end{array}\right\}\right|_{x_3=s}, where t∈Rt \in \mathbb{R}

    Note: Unscaled eigenvectors can be any vector of eigenspace. Normally we just let s=1s=1

    Eigenvector, {x1x2x3}λ=2={111}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=2}=\left\{\begin{array}{l}1 \\ 1 \\ 1\end{array}\right\}

    Note: Normalized eigenvectors has magnitude =1=1. It can be obtained by dividing the unscaled eigenvectors with the magnitude.

    Normalized Eigenvector, {x1x2x3}λ=2={1/31/31/3}\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda=2}=\left\{\begin{array}{l}1 / \sqrt{3} \\ 1 / \sqrt{3} \\ 1 / \sqrt{3}\end{array}\right\}

    Eigenvector or modal matrix, P=[−1/2−1/21/31/201/301/21/3]\mathbf{P}=\left[\begin{array}{ccc}-1 / \sqrt{2} & -1 / \sqrt{2} & 1 / \sqrt{3} \\ 1 / \sqrt{2} & 0 & 1 / \sqrt{3} \\ 0 & 1 / \sqrt{2} & 1 / \sqrt{3}\end{array}\right]

    Eigenvalues or spectral matrix, D=[−1000−10002]\mathbf{D}=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2\end{array}\right]

    C=[011101110]\mathbf{C}=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right] (C−λI)x=0(\mathbf{C}-\lambda \mathbf{I}) \boldsymbol{x}=\mathbf{0} Cx=λx\mathbf{C} \boldsymbol{x}=\lambda \boldsymbol{x}

    C[{x1x2x3}λ1{x1x2x3}λ2{x1x2x3}λ3]=[λ1{x1x2x3}λ1λ2{x1x2x3}λ2λ3{x1x2x3}λ3]C[{x1x2x3}λ1{x1x2x3}λ2{x1x2x3}λ3]=[{x1x2x3}λ1{x1x2x3}λ2{x1x2x3}λ3][λ1000λ2000λ3]\begin{aligned} \mathbf{C}\left[\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_1} \quad\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_2} \quad\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_3}\right]&=\left[\lambda_1\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_1} \quad \lambda_2\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_2} \quad \lambda_3\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_3}\right]\\ \mathbf{C}\left[\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_1} \quad\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_2} \quad\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_3}\right]&=\left[\left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_1} \quad \left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_2} \quad \left\{\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right\}_{\lambda_3}\right] \begin{bmatrix} \lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3 \end{bmatrix} \end{aligned}

    CP=PD\mathbf{CP}=\mathbf{PD}

    LHS: [011101110][−12−12131201301213]=[0.70710.70711.1547−0.707101.15470−0.70711.1547]\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{ccc}-\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{3}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}}\end{array}\right]=\left[\begin{array}{ccc}0.7071 & 0.7071 & 1.1547 \\ -0.7071 & 0 & 1.1547 \\ 0 & -0.7071 & 1.1547\end{array}\right]

    RHS: [−1/2−1/21/31/201/301/21/3][−1000−10002]=[0.70710.70711.1547−0.707101.15470−0.70711.1547]\left[\begin{array}{ccc}-1 / \sqrt{2} & -1 / \sqrt{2} & 1 / \sqrt{3} \\ 1 / \sqrt{2} & 0 & 1 / \sqrt{3} \\ 0 & 1 / \sqrt{2} & 1 / \sqrt{3}\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2\end{array}\right]=\left[\begin{array}{ccc}0.7071 & 0.7071 & 1.1547 \\ -0.7071 & 0 & 1.1547 \\ 0 & -0.7071 & 1.1547\end{array}\right]

    Since LHS = RHS, the solutions of eigenvalue matrix, D\mathbf{D} & eigenvector matrix, P\mathbf{P} are verified.