Linear Algebra - Eigenvalues and Eigenvectors of Symmetric Matrices

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

$\lambda_{1}=-26.37;\lambda_{2}=0.91;\lambda_{3}=29.46$

$\lambda_{1}=-24.37;\lambda_{2}=2.91;\lambda_{3}=31.46$

$\lambda_{1}=-30.37;\lambda_{2}=-3.09;\lambda_{3}=25.46$

$\lambda_{1}=-18.37;\lambda_{2}=8.91;\lambda_{3}=37.46$

Explanation

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

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

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

The eigenvalues are 0 and .

The eigenvalues are and .

The only eigenvalue is .

The only eigenvalue is 0.

Explanation

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 .

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

where is a real number.

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

can be any real number.

$-4 \sqrt{2} < a < 4 \sqrt{2}$

$a < - 4 \sqrt{2}$ or $a > 4 \sqrt{2}$

Explanation

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 .

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

$\lambda_{1}=-34.75;\lambda_{2}=-3.90;\lambda_{3}=-0.35$

$\lambda_{1}=-36.75;\lambda_{2}=-5.90;\lambda_{3}=-2.35$

$\lambda_{1}=-29.75;\lambda_{2}=1.10;\lambda_{3}=4.65$

$\lambda_{1}=-42.75;\lambda_{2}=-11.90;\lambda_{3}=-8.35$

Explanation

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

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

$\lambda_{1}=-25.16;\lambda_{2}=0.26;\lambda_{3}=19.90$

$\lambda_{1}=-27.16;\lambda_{2}=-1.74;\lambda_{3}=17.90$

$\lambda_{1}=-30.16;\lambda_{2}=-4.74;\lambda_{3}=14.90$

$\lambda_{1}=-34.16;\lambda_{2}=-8.74;\lambda_{3}=10.90$

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

Eigenvalues and Eigenvectors of Symmetric Matrices

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