By definition, if \(\displaystyle \textbf{v}\) is an eigenvector of a matrix \(\displaystyle A\), there is a scalar \(\displaystyle \lambda\) such that

\(\displaystyle A\textbf{v} = \lambda \textbf{v}\)

Let \(\displaystyle \textbf{v}=\begin{bmatrix} 4\\ 2\\ -3 \end{bmatrix}\), the given eigenvector of \(\displaystyle G\). Then 

\(\displaystyle G\textbf{v} = \lambda \textbf{v}\) for some real \(\displaystyle \lambda\). From this equation, it follows that any scalar multiple of eigenvector \(\displaystyle c\bold{v}\) of a matrix is also an eigenvector of the matrix, since:

\(\displaystyle c \cdot G\textbf{v} =c \cdot \lambda \textbf{v}\)

\(\displaystyle G \cdot c \cdot \textbf{v} = \lambda \cdot c \cdot \textbf{v}\)

\(\displaystyle G (c \textbf{v} )= \lambda (c \textbf{v})\),

making \(\displaystyle c \textbf{v}\) an eigenvector as well.

Let \(\displaystyle \textbf{u} =\begin{bmatrix}16\\ 8\\ -12 \end{bmatrix}\). Each element in \(\displaystyle \textbf{u}\) is four times the corresponding element in \(\displaystyle \textbf{v}\), so \(\displaystyle \textbf{u}= 4\textbf{v}\) - that is, it is a scalar multiple of a known eigenvector of \(\displaystyle G\). That makes \(\displaystyle \textbf{u}\) itself an eigenvector of \(\displaystyle G\).

To find the eigenvectors, we first need to find the eigenvalues. We can do by calculating

\(\displaystyle det(A-\lambda I) = \begin{vmatrix} -\lambda & 2\\ 1& -\lambda \end{vmatrix} = (-\lambda)^2 - (2) = \lambda^2-2\).

Setting this equal to \(\displaystyle 0\), we have \(\displaystyle 0= \lambda^2 -2\).

Hence \(\displaystyle \lambda = \pm\sqrt{2}\), are our eigenvalues.

Now we find the eigenvectors associated with each eigenvalue. Let's start with \(\displaystyle \lambda = \sqrt{2}\).

First plug the eigenvalue into the equation \(\displaystyle A-\lambda I = 0\), and solve the resulting system of equations.

\(\displaystyle \begin{align*} -\sqrt{2}x +2y&=0 \\ x-\sqrt{2}y&=0 \end{align*}\)

Solving this system gives \(\displaystyle \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}-\sqrt{2} & \\1 & \end{bmatrix}t\) once it is parametrized .

Hence \(\displaystyle \begin{bmatrix}-\sqrt{2} & \\1 & \end{bmatrix}\) is one of our eignvectors.

Repeating the above process with \(\displaystyle \lambda = -\sqrt{2}\) gives the equations

\(\displaystyle \begin{align*} \sqrt{2}x +2y&=0 \\ x+\sqrt{2}y&=0 \end{align*}\), with the solution \(\displaystyle \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}\sqrt{2} & \\1 & \end{bmatrix}t\). Hence \(\displaystyle \begin{bmatrix} \sqrt{2}\\1 \end{bmatrix}\) is the other eigenvector.

To find the eigenvalues, we first need to characteristic polynomial of our matrix

\(\displaystyle \begin{vmatrix} 1-\lambda& 5& 3& 2& 9\\ 0& 2-\lambda& 3& 5& 8\\ 0& 0& 3-\lambda& 3& 3\\ 0& 0& 0& 9-\lambda& 6\\ 0& 0& 0& 0& 3-\lambda \end{vmatrix}\)

This determinant is easy to evaluate since it is of an upper triangular matrix; we simply multiply the diagonal entries together.

\(\displaystyle =(1-\lambda)(2-\lambda)(3-\lambda)^2(9-\lambda)\).

The roots of this polynomial are \(\displaystyle 1,2,3,9\), but since the root \(\displaystyle 3\) has multiplicity \(\displaystyle 2\), we only have \(\displaystyle 4\) distinct eigenvalues, not \(\displaystyle 5\).

Suppose \(\displaystyle A\) is an invertible matrix with nonzero eigenvalue \(\displaystyle \lambda\). Suppose \(\displaystyle x\) is the associated eigenvector. It can be shown that \(\displaystyle \frac{1}{\lambda}\) is an eigenvalue of \(\displaystyle A^{-1}\). We have

\(\displaystyle Ax=\lambda x\)

We can multiply both sides by \(\displaystyle A^{-1}\) to get

\(\displaystyle A^{-1}Ax=A^{-1}\lambda x\)

\(\displaystyle Ix=A^{-1}\lambda x\)

\(\displaystyle x=\lambda A^{-1}x\)

Divide both sides by \(\displaystyle \lambda\) to get

\(\displaystyle A^{-1}x=\frac{1}{\lambda}x\)

So we get that \(\displaystyle \frac{1}{\lambda}\) is an eigenvalue of \(\displaystyle A^{-1}\)

It it is certainly the case that with a square matrix \(\displaystyle A\) with eigenvector \(\displaystyle x\), \(\displaystyle A\) has an infinite number of eigenvectors. Take a nonzero real number \(\displaystyle b\). Then \(\displaystyle bx\) is also an eigenvector because

\(\displaystyle Abx=bAx=b\lambda x=\lambda(bx)\)

So the infinite list of vectors \(\displaystyle x,2x,3x,4x,...\) are all eigenvectors - therefore, \(\displaystyle A\) will have an infinite number of eigenvectors. They are all from the same eigenspace, though, so nothing impressive was done. 

This is false. Suppose \(\displaystyle x\) is an eigenvector of \(\displaystyle A\). \(\displaystyle 2x\) is also an eigenvector:

\(\displaystyle A(2x)=2Ax=2\lambda x=\lambda(2x)\)

But \(\displaystyle x\) and \(\displaystyle 2x\) are obviously not linearly independent: \(\displaystyle 2x=2\cdot x\iff 2\cdot x+(-1)\cdot 2x=0\)

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

\(\displaystyle \lambda I - G\)

\(\displaystyle = \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} b & b \\ b & 0 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} - \begin{bmatrix} b & b \\ b & 0 \end{bmatrix}\)

Subtracting elementwise:

\(\displaystyle = \begin{bmatrix} \lambda - b & -b \\ -b & \lambda - 0 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} \lambda - b & -b \\ -b & \lambda \end{bmatrix}\)

The determinant can be found by taking the upper-left-to-lower-right product and subtracting the upper-right-to-lower-left product; set this to 0 to get the characteristic equation.

\(\displaystyle \begin{vmatrix} \lambda - b & -b \\ -b & \lambda \end{vmatrix} = 0\)

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

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

For 4 to be an eigenvalue, \(\displaystyle \lambda = 4\) must be a solution of this equation, so substitute and solve for \(\displaystyle b\):

\(\displaystyle 4^{2} - 4b - b^{2}=0\)

\(\displaystyle b^{2}+ 4b - 16=0\)

Applying the quadratic formula:

\(\displaystyle b = \frac{-4 \pm \sqrt{4^{2}- 4(1)(-16)}}{2(1)}\)

\(\displaystyle b = \frac{-4 \pm \sqrt{16 - (-64) }}{2}\)

\(\displaystyle b = \frac{-4 \pm \sqrt{80 }}{2}\)

\(\displaystyle b = \frac{-4 \pm 4\sqrt{5}}{2}\)

\(\displaystyle b = -2 \pm 2\sqrt{5}\)

\(\displaystyle H\) is an upper triangular matrix - that is, the only element above its main (upper left to lower right) diagonal is a zero. As a consequence, its eigenvalue(s) are the elements in its main diagonal, both of which are \(\displaystyle b\). It therefore follows that for the matrix to have 4 as an eigenvalue, \(\displaystyle b = 4\).

Find the characteristic equation of \(\displaystyle A\) by obtaining the determinant of \(\displaystyle \lambda I - A\).

\(\displaystyle \lambda I - A\)

\(\displaystyle = \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} -\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ -2 & -4 & 1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} -\begin{bmatrix} 1 & 2 & 3 \\ 1 & -1 & 4 \\ -2 & -4 & 1 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} \lambda -1 &- 2 & -3 \\ -1 & \lambda +1 &- 4 \\ 2 & 4 & \lambda -1 \end{bmatrix}\)

The determinant of this \(\displaystyle 3 \times 3\) matrix can be found by adding the upper-left to lower-right products:

\(\displaystyle (\lambda - 1)(\lambda + 1) (\lambda - 1) + (-2)(-4)(2) + (-3 )(-1)(4)\)

\(\displaystyle =( \lambda^{3} - \lambda^{2} - \lambda + 1 )+ 16 +12\)

\(\displaystyle = \lambda^{3} - \lambda^{2} - \lambda + 29\)

Adding the upper-right to lower-left products:

\(\displaystyle (-3)(\lambda + 1)(2) + (-4)(4)(\lambda-1) + (\lambda - 1) (-2)(-1)\)

\(\displaystyle = (-6 \lambda -6) + (-16 \lambda+16) + (2\lambda - 2)\)

\(\displaystyle = -20 \lambda +8\)

And subtracting the latter from the former:

\(\displaystyle = \lambda^{3} - \lambda^{2} - \lambda + 29 - ( -20 \lambda +8)\)

\(\displaystyle = \lambda^{3} - \lambda^{2} +19 \lambda + 21\)

The characteristic equation is 

\(\displaystyle \lambda^{3} - \lambda^{2} +19 \lambda + 21 = 0\)

Factor the polynomial completely

\(\displaystyle (\lambda + 1) (\lambda ^{2} - 2 \lambda + 21) = 0\)

The solution set of this equation is comprised of the zeroes of both polynomial factors: 

\(\displaystyle \left \{ -1, 1 - i \sqrt{20}, 1 + i \sqrt{20} \right \}\)

Only \(\displaystyle -1\) is a choice.

Assume that \(\displaystyle -2\) is an eigenvalue of \(\displaystyle A\). Then, if \(\displaystyle b = \begin{bmatrix} b_{1} \\ b_{2} \\b_{3} \end{bmatrix}\) is one of its eigenvectors, it follows that

\(\displaystyle Ab = -2 b\), or, equivalently,

\(\displaystyle (A + 2I) b =O\),

where \(\displaystyle I,O\) are the \(\displaystyle 3 \times 3\) identity and zero matrices, respectively.

\(\displaystyle A= \begin{bmatrix} 2 & 3 & 6 \\ 0 & 4 & 4 \\ 2 & 0 & 4 \end{bmatrix}\), so

\(\displaystyle A + 2 I = \begin{bmatrix} 2+2 & 3 & 6 \\ 0 & 4+2 & 4 \\ 2 & 0 & 4+2 \end{bmatrix}\)

\(\displaystyle = \begin{bmatrix} 4& 3 & 6 \\ 0 & 6 & 4 \\ 2 &6& 2\end{bmatrix}\)

Changing to reduced row echelon form:

\(\displaystyle R1 \leftrightarrow R3\)

\(\displaystyle \begin{bmatrix} 2 &6& 2 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}\)

\(\displaystyle \frac{1}{2} R1 \rightarrow R1\)

\(\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 4& 3 & 6 \end{bmatrix}\)

\(\displaystyle -4R1+ R3 \rightarrow R3\)

\(\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 6 & 4 \\ 0& -9 & 2 \end{bmatrix}\)

\(\displaystyle \frac{1}{6} R2 \rightarrow R2\)

\(\displaystyle \begin{bmatrix} 1 &3&1 \\ 0 & 1 & \frac{2}{3} \\ 0& -9 & 2 \end{bmatrix}\)

\(\displaystyle -3R2 + R1 \rightarrow R1\)

\(\displaystyle 9 R2 + R3 \rightarrow R3\)

\(\displaystyle \begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 8 \end{bmatrix}\)

\(\displaystyle \frac{1}{8} R3 \rightarrow R3\)

\(\displaystyle \begin{bmatrix} 1 &0 &-1 \\ 0 & 1 & \frac{2}{3} \\ 0& 0& 1 \end{bmatrix}\)

We do not need to go further to see that this matrix will not have a row of zeroes. This means the rank of the matrix is 3, and the nullity is 0. If this happens, the tested value, in this case \(\displaystyle -2\), is not an eigenvalue.