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Linear Algebra : Eigenvalues and Eigenvectors of Symmetric Matrices

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Example Question #1 :Eigenvalues And Eigenvectors Of Symmetric Matrices

Find the Eigen Values for Matrix.

$A=\begin{bmatrix} 1 & -3 \\ 5& 4 \end{bmatrix}$

Possible Answers:

$\lambda=\frac{5+ \sqrt{51}}{2}$

$\lambda=\frac{5- i\sqrt{51}}{2}$

There are no eigen values

$\lambda=\frac{5+ i\sqrt{51}}{2}$

$\lambda=\frac{5- \sqrt{51}}{2}$

$\lambda=\frac{5+\sqrt{51}}{2}$

$\lambda=\frac{5- \sqrt{51}}{2}$

$\lambda=\frac{5+ i\sqrt{51}}{2}$

$\lambda=\frac{5- i\sqrt{51}}{2}$

Correct answer:

$\lambda=\frac{5+ i\sqrt{51}}{2}$

$\lambda=\frac{5- i\sqrt{51}}{2}$

解释:

The first step into solving for eigenvalues, is adding in a $-\lambda$ along the main diagonal.

$A=\begin{bmatrix} 1-\lambda & -3 \\ 5& 4-\lambda \end{bmatrix}$

Now the next step to take the determinant.

$\det(A)=(1-\lambda)(4-\lambda)-(-3)(5)$

$\det(A)=(1-\lambda)(4-\lambda)+15$

Now lets FOIL, and solve for $\lambda$ .

$(1-\lambda)(4-\lambda)+15=4-\lambda-4\lambda+\lambda^2+15$

$=\lambda^2-5\lambda+19$

Now lets use the quadratic equation to solve for $\lambda$ .

$\lambda=\frac{5\pm\sqrt{(-5)^2-4(19)}}{2(1)}$

$\lambda=\frac{5\pm\sqrt{25-76}}{2}$

$\lambda=\frac{5\pm\sqrt{-51}}{2}$

$\lambda=\frac{5\pm i\sqrt{51}}{2}$

So our eigen values are

$\lambda=\frac{5+ i\sqrt{51}}{2}$

$\lambda=\frac{5- i\sqrt{51}}{2}$

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Example Question #2 :Eigenvalues And Eigenvectors Of Symmetric Matrices

Find the eigenvalues and set of mutually orthogonal

eigenvectors for the following matrix.

$A=\begin{bmatrix} 3&2 &4 \\ 2& 0 &2 \\ 4&2&3\end{bmatrix}$

Possible Answers:

$\lambda=-1,8$

<1,2,0>, <4,2,-5>

No eigenvalues or eigenvectors exist

$\lambda=8$

<1,2,0>, <2,1,2>

$\lambda=-1,-1,8$

<1,2,0>, <4,2,-5>,<2,1,2>

$\lambda=1,8$

<4,2,-5>,<2,1,2>

Correct answer:

$\lambda=-1,-1,8$

<1,2,0>, <4,2,-5>,<2,1,2>

解释:

In this problem, we will get three eigen values and eigen vectors since it's a symmetric matrix.

To find the eigenvalues, we need to minus lambda along the main diagonal and then take the determinant, then solve for lambda.

$\\ |A-I\lambda|=\begin{vmatrix} 3-\lambda&2 &4 \\ 2& 0-\lambda &2\\ 4 & 2 & 3-\lambda \end{vmatrix}$

$=(3-\lambda)\begin{vmatrix} -\lambda& 2 \\ 2 & 3-\lambda \end{vmatrix} -2\begin{vmatrix} 2& 2\\ 4& 3-\lambda \end{vmatrix} +4 \begin{vmatrix} 2&-\lambda \\ 4&2 \end{vmatrix}$

$=(3-\lambda)((-\lambda)(3-\lambda)-(2)(2))-2((2)(3-\lambda)-(2)(4))+4((2)(2)-(-\lambda(4)))$

$=(3-\lambda)(\lambda^2-3\lambda-4)-2(-2-2\lambda)+4(4+4\lambda)$

$=-\lambda^3+3\lambda^2+4\lambda+3\lambda^2-9\lambda-12+4+4\lambda+16+16\lambda$

$=-\lambda^3+6\lambda^2+15\lambda+8$

This can be factored to

$-(\lambda+1)^2(\lambda-8)=0$

Thus our eigenvalues are at $\lambda=-1, 8$

Now we need to substitute $\lambda$ into or matrix in order to find the eigenvectors.

For $\lambda=-1$ .

$\\ (A+I)v=\begin{bmatrix} 3-(-1)&2 &4 \\ 2& 0-(-1) &2\\ 4 & 2 & 3-(-1)\end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

$\\ (A+I)v=\begin{bmatrix} 4 &2 &4 \\ 2& 1 &2\\ 4 & 2 & 4 \end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

Now we need to get the matrix into reduced echelon form.

$R_1-2R_2\rightarrow R_2$

$R_1-R_3\rightarrow R_3$

$\begin{bmatrix} 4 &2 &4 \\ 0& 0 &0\\ 0 & 0 & 0 \end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

This can be reduced to

$\begin{bmatrix} 2 &1 &2 \\ 0& 0 &0\\ 0 & 0 & 0 \end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

This is in equation form is 2x+y+2z=0 , which can be rewritten as y=-2x-2z . In vector form it looks like, <x, -2x-2z, z> .

We need to take the dot product and set it equal to zero, and pick a value for,.

Let x=1 , z=0 .

$<1,-2,0>\cdot<x,-2x-2z,z>=0$

x+4x+4z+0=0

5x+4z=0

Now we pick another value for,so that the result is zero. The easiest ones to pick are x=4 , z=-5 .

So the orthogonal vectors for $\lambda=-1$ are <1,-2,0> , <4,2,-5> .

Now we need to get the last eigenvector for $\lambda=8$ .

$\\ (A+I)v=\begin{bmatrix} 3-(8)&2 &4 \\ 2& 0-(8) &2\\ 4 & 2 & 3-(8)\end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

$\\ (A+I)v=\begin{bmatrix} -5 &2 &4 \\ 2& -8 &2\\ 4 & 2 & -5 \end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

After row reducing, the matrix looks like

$\begin{bmatrix} 1&0 &-1 \\ 0& 2 &-1\\ 0 & 0 & 0 \end{vmatrix}\left.\begin{matrix} 0\\ 0\\ 0 \end{bmatrix}\right|$

So our equations are then

x-z=0 , 2y-z=0 , which can be rewritten as x=z , z=2y .

Then eigenvectors take this form, <2y,y,2y> . This will be orthogonal to our other vectors, no matter what value of, we pick. For convenience, let's pick y=1 , then our eigenvector is <2,1,2> .

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Example Question #3 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-9&-2\\-2&-2\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-9.53;\lambda_{2}=-1.47$

$\lambda_{1}=-0.53;\lambda_{2}=7.53$

$\lambda_{1}=-15.53;\lambda_{2}=-7.47$

$\lambda_{1}=-12.53;\lambda_{2}=-4.47$

Correct answer:

$\lambda_{1}=-9.53;\lambda_{2}=-1.47$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-9&-2\\-2&-2\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 9&-2\\-2&- \lambda - 2\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&11\lambda + \lambda ^{2} + 14\\&\lambda_{1}=-9.53;\lambda_{2}=-1.47\end{align*}$

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Example Question #4 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Find the eigenvalues for the matrix }\\&A=\begin{bmatrix}7&8\\8&-10\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-14.17;\lambda_{2}=9.17$

$\lambda_{1}=-13.17;\lambda_{2}=10.17$

$\lambda_{1}=-22.17;\lambda_{2}=1.17$

$\lambda_{1}=-7.17;\lambda_{2}=16.17$

Correct answer:

$\lambda_{1}=-13.17;\lambda_{2}=10.17$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}7&8\\8&-10\end{bmatrix}\\&\text{We can define a matrix of the form }\begin{bmatrix}7 - \lambda&8\\8&- \lambda - 10\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Using this, we can solve for our eigenvalues:}\\&3\lambda + \lambda ^{2} - 134\\&\lambda_{1}=-13.17;\lambda_{2}=10.17\end{align*}$

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Example Question #5 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}4&-14&-12\\-14&10&13\\-12&13&1\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-1.64;\lambda_{2}=1.11;\lambda_{3}=39.54$

$\lambda_{1}=-8.64;\lambda_{2}=-5.89;\lambda_{3}=32.54$

$\lambda_{1}=-13.64;\lambda_{2}=-10.89;\lambda_{3}=27.54$

$\lambda_{1}=-9.64;\lambda_{2}=-6.89;\lambda_{3}=31.54$

Correct answer:

$\lambda_{1}=-9.64;\lambda_{2}=-6.89;\lambda_{3}=31.54$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}4&-14&-12\\-14&10&13\\-12&13&1\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}4 - \lambda&-14&-12\\-14&10 - \lambda&13\\-12&13&1 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (\lambda ^{3} - 15\lambda ^{2} - 455\lambda - 2096)\\&\lambda_{1}=-9.64;\lambda_{2}=-6.89;\lambda_{3}=31.54\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #6 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}12&18\\18&7\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-8.67;\lambda_{2}=27.67$

$\lambda_{1}=-2.67;\lambda_{2}=33.67$

$\lambda_{1}=-17.67;\lambda_{2}=18.67$

$\lambda_{1}=-10.67;\lambda_{2}=25.67$

Correct answer:

$\lambda_{1}=-8.67;\lambda_{2}=27.67$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}12&18\\18&7\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}12 - \lambda&18\\18&7 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&\lambda ^{2} - 19\lambda - 240\\&\lambda_{1}=-8.67;\lambda_{2}=27.67\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #7 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}8&-2&-20\\-2&-3&-9\\-20&-9&-3\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-31.00;\lambda_{2}=-8.84;\lambda_{3}=14.84$

$\lambda_{1}=-19.00;\lambda_{2}=3.16;\lambda_{3}=26.84$

$\lambda_{1}=-16.00;\lambda_{2}=6.16;\lambda_{3}=29.84$

$\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84$

Correct answer:

$\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}8&-2&-20\\-2&-3&-9\\-20&-9&-3\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}8 - \lambda&-2&-20\\-2&- \lambda - 3&-9\\-20&-9&- \lambda - 3\end{bmatrix}\\&\text{For a square matrix with dimensions of }3\\&det\begin{vmatrix} a&b&c \\ d&e&f\\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\\&\text{Knowing this, we can expand and solve:}\\&(-1)\cdot (\lambda ^{3} - 2\lambda ^{2} - 524\lambda + 84)\\&\lambda_{1}=-22.00;\lambda_{2}=0.16;\lambda_{3}=23.84\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #8 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-14&4\\4&-19\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-21.22;\lambda_{2}=-11.78$

$\lambda_{1}=-26.22;\lambda_{2}=-16.78$

$\lambda_{1}=-18.22;\lambda_{2}=-8.78$

$\lambda_{1}=-29.22;\lambda_{2}=-19.78$

Correct answer:

$\lambda_{1}=-21.22;\lambda_{2}=-11.78$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-14&4\\4&-19\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 14&4\\4&- \lambda - 19\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&33\lambda + \lambda ^{2} + 250\\&\lambda_{1}=-21.22;\lambda_{2}=-11.78\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #9 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Calculate the eigenvalues for the matrix }\\&\begin{bmatrix}-15&8\\8&1\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-20.31;\lambda_{2}=2.31$

$\lambda_{1}=-27.31;\lambda_{2}=-4.69$

$\lambda_{1}=-13.31;\lambda_{2}=9.31$

$\lambda_{1}=-18.31;\lambda_{2}=4.31$

Correct answer:

$\lambda_{1}=-18.31;\lambda_{2}=4.31$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-15&8\\8&1\end{bmatrix}\\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 15&8\\8&1 - \lambda\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{Knowing this, we can expand and solve:}\\&14\lambda + \lambda ^{2} - 79\\&\lambda_{1}=-18.31;\lambda_{2}=4.31\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Example Question #10 :Eigenvalues And Eigenvectors Of Symmetric Matrices

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\\&A=\begin{bmatrix}-16&-16\\-16&-4\end{bmatrix}\end{align*}$

Possible Answers:

$\lambda_{1}=-25.09;\lambda_{2}=9.09$

$\lambda_{1}=-36.09;\lambda_{2}=-1.91$

$\lambda_{1}=-33.09;\lambda_{2}=1.09$

$\lambda_{1}=-27.09;\lambda_{2}=7.09$

Correct answer:

$\lambda_{1}=-27.09;\lambda_{2}=7.09$

解释:

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-16&-16\\-16&-4\end{bmatrix}\\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 16&-16\\-16&- \lambda - 4\end{bmatrix}\\&\text{For a square matrix with dimensions of }2\\&det\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc\\&\text{From that, we can find the eigenvalues:}\\&20\lambda + \lambda ^{2} - 192\\&\lambda_{1}=-27.09;\lambda_{2}=7.09\\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align*}$

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Linear Algebra : Eigenvalues and Eigenvectors of Symmetric Matrices

Study concepts, example questions & explanations for Linear Algebra

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