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

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

Now the next step to take the determinant.

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

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

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

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

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

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

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

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

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

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

So our eigen values are

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

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

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.

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

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

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

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

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

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

This can be factored to

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

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

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

For $\displaystyle \lambda=-1$.

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

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

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

$\displaystyle R_1-2R_2\rightarrow R_2$

$\displaystyle R_1-R_3\rightarrow R_3$

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

This can be reduced to 

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

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

We need to take the dot product and set it equal to zero, and pick a value for $\displaystyle x$, and $\displaystyle z$.

Let $\displaystyle x=1$, and $\displaystyle z=0$.

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

$\displaystyle x+4x+4z+0=0$

$\displaystyle 5x+4z=0$

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

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

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

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

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

After row reducing, the matrix looks like

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

So our equations are then

$\displaystyle x-z=0$, and $\displaystyle 2y-z=0$, which can be rewritten as $\displaystyle x=z$, $\displaystyle z=2y$.

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

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

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

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

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

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

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

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

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