An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all four eigenvalues are as follows:

\(\displaystyle |6-\sqrt{5 }| = 6-\sqrt{5 }\approx 3.76\)

\(\displaystyle |6+\sqrt{5 }| = 6+\sqrt{5 }\approx 8.24\)

\(\displaystyle |6-i\sqrt{5 }| =\sqrt{ 6^{2}+(-\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40\)

\(\displaystyle |6-i\sqrt{5 }| =\sqrt{ 6^{2}+(\sqrt{5 }) ^{2}}= \sqrt{36+5} = \sqrt{41} \approx 6.40\)

\(\displaystyle |6+\sqrt{5 }| > |6-i\sqrt{5 }| = |6+i\sqrt{5 }| >|6-\sqrt{5 }|\);

therefore, the dominant eigenvalue is \(\displaystyle 6+ \sqrt{5}\).

The eigenvalues of a matrix are the zeroes of its characteristic equation.

\(\displaystyle \lambda ^{3}- 9 \lambda ^{2}+ 9 \lambda - 81 = 0\)

can be solved by factoring out the polynomial. First, factor by grouping:

\(\displaystyle (\lambda ^{3}- 9 \lambda ^{2})+ (9 \lambda - 81 )= 0\)

\(\displaystyle \lambda ^{2} (\lambda - 9)+9 ( \lambda - 9)= 0\)

\(\displaystyle (\lambda ^{2} +9) ( \lambda - 9)= 0\)

The zeroes of the polynomial are the zeroes of the individual factors, which can be calculated to be

\(\displaystyle \lambda_{1} = -3i , \lambda_{2} = 3i, \lambda_{3} = 9\)

An eigenvalue is a dominant eigenvalue if its absolute value is strictly greater than those of all other eigenvalues. The absolute values of all three eigenvalues are:

\(\displaystyle |\lambda_{1}| = |-3i| = 3\)

\(\displaystyle |\lambda_{2}| = |3i| = 3\)

\(\displaystyle |\lambda_{3}| = |9|=9\)

\(\displaystyle |\lambda_{3}| > |\lambda_{1}| =|\lambda_{2}|\),

 so 9 is the dominant eigenvalue of \(\displaystyle A\).

A necessary and sufficient condition for a matrix to have 0 as an eigenvalue is for the matrix to have determinant 0. \(\displaystyle A\) has an all-zero column:

\(\displaystyle A = \begin{bmatrix} \textbf{{\color{Red} 0}} & 4 & -7 & 1 \\ \textbf{{\color{Red} 0}} & 1 & -2 & 3 \\ \textbf{{\color{Red} 0}} & 3 & 1 & -5 \\\textbf{{\color{Red} 0}} & -4 & 2 & 5 \end{bmatrix}\)

This is a sufficient condition for a matrix to have determinant 0. It follows that 0 is an eigenvalue of \(\displaystyle A\).

One way to identify the matrix with these eigenvalues is to note that the sum of the eigenvalues of a matrix is equal to the trace of the matrix, and the product of the eigenvalues is equal to its determinant. Therefore, for a matrix to have 4 and 7 as eigenvalues, its trace must be 11 and its determinant must be 28.

The trace of a matrix is the sum of the elements in its main diagonal; the determinant is the product of these two elements minus the product of the other two:

\(\displaystyle \begin{bmatrix} \textbf{{\color{Blue} -1}} & -10\\ 4& \textbf{{\color{Blue} 12}} \end{bmatrix}\)

Trace: \(\displaystyle -1+12 = 11\)

Determinant: \(\displaystyle -1(12)-(-10)(4)= 28\)

\(\displaystyle \begin{bmatrix} \textbf{{\color{Blue} 2}} &-2 \\5 & \textbf{{\color{Blue} 9}} \end{bmatrix}\)

Trace: \(\displaystyle 2+9= 11\)

Determinant: \(\displaystyle 2(9)- (-2)(5) = 28\)

\(\displaystyle \begin{bmatrix} \textbf{{\color{Blue} 3}} & -1\\4 & \textbf{{\color{Blue} 8}} \end{bmatrix}\)

Trace: \(\displaystyle 3+8= 11\)

Determinant: \(\displaystyle 3(8)- (-1)4 = 28\)

\(\displaystyle \begin{bmatrix} \textbf{{\color{Blue} 5}} & 1 \\2 & \textbf{{\color{Blue} 6}} \end{bmatrix}\)

Trace: \(\displaystyle 5+6 = 11\)

Determinant: \(\displaystyle 5(6)- 1(2) = 28\)

All four matrices have the requisite trace and determinant, so all four have the eigenvalues 4 and 7.

An eigenvalue of a matrix is the dominant eigenvalue if its absolute value is strictly greater than that of all of its other eigenvalues. The absolute values of the three eigenvalues are:

\(\displaystyle |-9| = 9\)

\(\displaystyle |8|=8\)

\(\displaystyle |10i| = 10\).

\(\displaystyle |10i| > |9i|> |8|\), so \(\displaystyle 10i\) is the dominant eigenvalue.

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\(\displaystyle < 2,1,2>\).

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

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

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