Eigenvalues and Eigenvectors of Symmetric Matrices - Linear Algebra

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Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}3&2&15\2&-7&-16\15&-16&-1\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}3&2&15\2&-7&-16\15&-16&-1\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}3 - \lambda&2&15\2&- \lambda - 7&-16\15&-16&- \lambda - 1\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{From that, we can find the eigenvalues:}\&(-1)\cdot (5\lambda ^{2} - 502\lambda + \lambda ^{3} + 128)\&\lambda_{1}=-25.16;\lambda_{2}=0.26;\lambda_{3}=19.90\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align}$

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Question

Find the Eigen Values for Matrix .

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

Answer

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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Question

Find the eigenvalues and set of mutually orthogonal

eigenvectors for the following matrix.

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

Answer

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 , which can be rewritten as . In vector form it looks like, .

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

Let , and .

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

Now we pick another value for , and so that the result is zero. The easiest ones to pick are , and .

So the orthogonal vectors for $\lambda=-1$ are , and .

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

, and , which can be rewritten as , .

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}-5&17&-20\17&-3&-2\-20&-2&12\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}-5&17&-20\17&-3&-2\-20&-2&12\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}- \lambda - 5&17&-20\17&- \lambda - 3&-2\-20&-2&12 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Knowing this, we can expand and solve:}\&(-1)\cdot (\lambda ^{3} - 4\lambda ^{2} - 774\lambda + 708)\&\lambda_{1}=-26.37;\lambda_{2}=0.91;\lambda_{3}=29.46\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align}$

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Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}-13&10&-11\10&-9&11\-11&11&-17\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}-13&10&-11\10&-9&11\-11&11&-17\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}- \lambda - 13&10&-11\10&- \lambda - 9&11\-11&11&- \lambda - 17\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{From that, we can find the eigenvalues:}\&(-1)\cdot (149\lambda + 39\lambda ^{2} + \lambda ^{3} + 47)\&\lambda_{1}=-34.75;\lambda_{2}=-3.90;\lambda_{3}=-0.35\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align}$

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Question

$\begin{align}&\text{Find the eigenvalues for the matrix }\&A=\begin{bmatrix}18&-12&9\-12&-16&4\9&4&7\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, usually denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I is the identity matrix. For our matrix }A=\begin{bmatrix}18&-12&9\-12&-16&4\9&4&7\end{bmatrix}\&\text{We can define a matrix of the form }\begin{bmatrix}18 - \lambda&-12&9\-12&- \lambda - 16&4\9&4&7 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }3\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&\text{Using this, we can solve for our eigenvalues:}\&(-1)\cdot (\lambda ^{3} - 9\lambda ^{2} - 515\lambda + 2880)\&\lambda_{1}=-21.38;\lambda_{2}=5.39;\lambda_{3}=24.99\&\text{Note that since our matrix was symmetric, our eigenvalues are real.}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$C= \begin{bmatrix} 4 & a \ a & 8 \end{bmatrix}$ ,

where is a real number.

For to have two real eigenvalues, what must be true for ?

Answer

Any real value of makes a symmetric matrix with real entries. It holds that any eigenvalues of must be real regardless of the value of .

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Question

$G = \begin{bmatrix} b & b \ b & b \end{bmatrix}$

Give the set of eigenvalues of in terms of , if applicable.

Answer

An eigenvalue of is a zero of the characteristic equation formed from the determinant of $\lambda I - G$ , so find this determinant as follows:

$\lambda I - G$

$= \lambda \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} - \begin{bmatrix} b & b \ b & b \end{bmatrix}$

$= \begin{bmatrix} \lambda & 0 \ 0 & \lambda \end{bmatrix} - \begin{bmatrix} b & b \ b & b \end{bmatrix}$

Subtracting elementwise:

$= \begin{bmatrix}\lambda - b & - b \- b & \lambda -b \end{bmatrix}$

Set the determinant to 0 and solve for $\lambda$ :

$\begin{vmatrix}\lambda - b & - b \- b & \lambda -b \end{vmatrix} = 0$

The determinant can be found by taking the upper-left-to-lower-right product and subtracting the upper-right-to-lower-left product:

$(\lambda - b)(\lambda - b) - (-b) (-b) = 0$

$\lambda ^{2} - 2 \lambda b + b^{2} - b^{2} = 0$

$\lambda ^{2} - 2 \lambda b = 0$

$\lambda \left (\lambda - 2b \right ) = 0$ ,

so the eigenvalues of this matrix are 0 and .

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Question

Find the Eigen Values for Matrix .

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

Answer

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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Question

Find the eigenvalues and set of mutually orthogonal

eigenvectors for the following matrix.

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

Answer

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 , which can be rewritten as . In vector form it looks like, .

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

Let , and .

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

Now we pick another value for , and so that the result is zero. The easiest ones to pick are , and .

So the orthogonal vectors for $\lambda=-1$ are , and .

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

, and , which can be rewritten as , .

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

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Question

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

Answer

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

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Question

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

Answer

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

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