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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−k3m3k3+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_{3}} & \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

    Put λ=ω2\lambda=\omega^2. Substituting the given masses and stiffnesses gives

    A=13(100−700−70140−700−70100).A=\frac{1}{3}\begin{pmatrix}100&-70&0\\-70&140&-70\\0&-70&100\end{pmatrix}.

    A nonzero eigenvector requires det⁡(A−λI)=0\det(A-\lambda I)=0. Expanding the determinant,

    det⁡(A−λI)=(1003−λ)[(1003−λ)(1403−λ)−2(703)2]=(1003−λ)(λ2−80λ+14003).\begin{aligned} \det(A-\lambda I) &=\left(\frac{100}{3}-\lambda\right) \left[\left(\frac{100}{3}-\lambda\right) \left(\frac{140}{3}-\lambda\right)-2\left(\frac{70}{3}\right)^2\right]\\ &=\left(\frac{100}{3}-\lambda\right) \left(\lambda^2-80\lambda+\frac{1400}{3}\right). \end{aligned}

    Thus, in ascending order,

    λ1=40−101023≈6.3350,λ2=1003≈33.3333,λ3=40+101023≈73.6650.\lambda_1=40-\frac{10\sqrt{102}}{3}\approx6.3350, \quad \lambda_2=\frac{100}{3}\approx33.3333, \quad \lambda_3=40+\frac{10\sqrt{102}}{3}\approx73.6650.

    For λ1\lambda_1, the first and third rows of (A−λ1I)x=0(A-\lambda_1 I)x=0 give

    x2=100−3λ170x1=102−27x1,x3=x1.x_2=\frac{100-3\lambda_1}{70}x_1 =\frac{\sqrt{102}-2}{7}x_1, \qquad x_3=x_1.

    Substitution also satisfies the second row. Taking x1=1x_1=1 gives the smallest mode:

    v1=(1(102−2)/71)≈(11.15711).v_1=\begin{pmatrix}1\\(\sqrt{102}-2)/7\\1\end{pmatrix} \approx\begin{pmatrix}1\\1.1571\\1\end{pmatrix}.

    Its eigenspace is {tv1:t∈R}\{t v_1:t\in\mathbb R\}; eigenvectors require t≠0t\ne0. All three masses move in phase, with the middle mass having the largest amplitude.

    Smallest mode: amplitudes proportional to 1, 1.1571, 1

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

    Solution

    From Q1, the intermediate eigenvalue is λ2=100/3\lambda_2=100/3. Hence

    A−λ2I=13(0−700−7040−700−700).A-\lambda_2 I=\frac{1}{3} \begin{pmatrix}0&-70&0\\-70&40&-70\\0&-70&0\end{pmatrix}.

    The first row gives x2=0x_2=0; the second then gives x1+x3=0x_1+x_3=0. Taking x3=1x_3=1 gives v2=(−1,0,1)Tv_2=(-1,0,1)^T. Its length is 2\sqrt2, so

    v^2=12(−101)≈(−0.707100.7071).\widehat v_2=\frac{1}{\sqrt2}\begin{pmatrix}-1\\0\\1\end{pmatrix} \approx\begin{pmatrix}-0.7071\\0\\0.7071\end{pmatrix}.

    Either sign is a valid normalized eigenvector. The two outer masses move in opposite directions, while the middle mass is stationary in this mode.

    Intermediate mode: opposite outer-mass amplitudes and zero middle amplitude

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

    Solution

    The largest eigenvalue is λ3=40+10102/3≈73.6650\lambda_3=40+10\sqrt{102}/3\approx73.6650. The first and third rows of (A−λ3I)x=0(A-\lambda_3 I)x=0 give

    x2=100−3λ370x1=−102+27x1,x3=x1.x_2=\frac{100-3\lambda_3}{70}x_1 =-\frac{\sqrt{102}+2}{7}x_1, \qquad x_3=x_1.

    These values satisfy the second row as well. An unscaled eigenvector is

    v3=(1−(102+2)/71)≈(1−1.72851).v_3=\begin{pmatrix}1\\-(\sqrt{102}+2)/7\\1\end{pmatrix} \approx\begin{pmatrix}1\\-1.7285\\1\end{pmatrix}.

    Every nonzero multiple is also an eigenvector. The outer masses move together, in the opposite direction to the middle mass.

    Largest mode: outer amplitudes 1, middle amplitude -1.7285

  4. Combining the results obtained from Q1–Q3, 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−k3m3k3+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_{3}} & \frac{k_{3}+k_{4}}{m_{3}} \end{array}\right]
    Solution

    Keep the eigenvectors exact before multiplying. Write

    a=102−27,b=−102+27,a−b=21027.a=\frac{\sqrt{102}-2}{7},\qquad b=-\frac{\sqrt{102}+2}{7},\qquad a-b=\frac{2\sqrt{102}}7.

    Using the eigenvectors from Q1–Q3 as columns,

    P=(1−1/21a0b11/21),D=diag⁡(40−101023,1003,40+101023).P=\begin{pmatrix}1&-1/\sqrt2&1\\a&0&b\\1&1/\sqrt2&1\end{pmatrix}, \qquad D=\operatorname{diag}\left(40-\frac{10\sqrt{102}}3,\frac{100}3, 40+\frac{10\sqrt{102}}3\right).

    To find P−1P^{-1}, solve P(c1,c2,c3)T=(x1,x2,x3)TP(c_1,c_2,c_3)^T=(x_1,x_2,x_3)^T. Adding and subtracting its first and third equations gives c1+c3=(x1+x3)/2c_1+c_3=(x_1+x_3)/2 and c2=(x3−x1)/2c_2=(x_3-x_1)/\sqrt2. The middle equation ac1+bc3=x2ac_1+bc_3=x_2 then gives

    P−1=(−b2(a−b)1a−b−b2(a−b)−12012a2(a−b)−1a−ba2(a−b)).P^{-1}=\begin{pmatrix} -\dfrac{b}{2(a-b)}&\dfrac{1}{a-b}&-\dfrac{b}{2(a-b)}\\[6pt] -\dfrac{1}{\sqrt2}&0&\dfrac{1}{\sqrt2}\\[6pt] \dfrac{a}{2(a-b)}&-\dfrac{1}{a-b}&\dfrac{a}{2(a-b)} \end{pmatrix}.

    Each column satisfies Avi=λiviAv_i=\lambda_i v_i, so AP=PDAP=PD and P−1AP=DP^{-1}AP=D. The diagonal entries of DD are the corresponding eigenvalues, in the same order as the columns of PP.

  5. Determine [k1+k2m1−k2m10−k2m2k2+k3m2−k3m20−k3m3k3+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_{3}} & \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

    From A=PDP−1A=PDP^{-1}, adjacent factors P−1PP^{-1}P cancel, giving

    A50=PD50P−1,D50=diag⁡(λ150,λ250,λ350).A^{50}=PD^{50}P^{-1},\qquad D^{50}=\operatorname{diag}(\lambda_1^{50},\lambda_2^{50},\lambda_3^{50}).

    Use the exact values from Q4 before raising them to the 50th power. The result is

    A50≈1093(0.46255450−0.799525780.46255450−0.799525781.38198087−0.799525780.46255450−0.799525780.46255450).A^{50}\approx10^{93} \begin{pmatrix} 0.46255450&-0.79952578&0.46255450\\ -0.79952578&1.38198087&-0.79952578\\ 0.46255450&-0.79952578&0.46255450 \end{pmatrix}.

    The matrix entries shown are rounded; PD50P−1PD^{50}P^{-1} is the exact result. Every eigenvector of AA remains an eigenvector of A50A^{50}, with eigenvalue λi50\lambda_i^{50}. Here the three eigenvalues are positive and distinct, so taking the 50th power preserves their distinct eigenspaces.

  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 \mathbf{P}^{-1}=\frac{1}{\operatorname{det}(\mathbf{P})} \operatorname{adjoint}(\mathbf{P})=\frac{\mathbf{1}}{0} \operatorname{adjoint}(\mathbf{P})=\text { undefined }

    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.