A \(\displaystyle 4 \times 4\) matrix has four not necessarily distinct eigenvalues, \(\displaystyle \lambda_{1} , \lambda_{2} , \lambda_{3} , \lambda_{4}\), which are the solutions of its characteristic polynomial equation 

\(\displaystyle (\lambda - \lambda_{1}) (\lambda - \lambda_{2}) (\lambda - \lambda_{3} ) (\lambda - \lambda_{4})=0\)

If there are only two distinct eigenvalues, 1 and \(\displaystyle -1\), then one of the following three situations happens:

1 has multiplicity 3 and \(\displaystyle -1\) has multiplicity 1, in which case \(\displaystyle \lambda_{1} = \lambda_{2} = \lambda_{3}=1 , \lambda_{4} = -1\),

and the characteristic equation is 

\(\displaystyle (\lambda - 1) (\lambda - 1) (\lambda - 1 ) (\lambda +1)=0\)

or

\(\displaystyle \lambda^{4}-2 \lambda^{3}+2 \lambda - 1 = 0\)

1 has multiplicity 2 and \(\displaystyle -1\) has multiplicity 2, in which case \(\displaystyle \lambda_{1} = \lambda_{2} = 1 , \lambda_{3}= \lambda_{4} = -1\),

and the characteristic equation is 

\(\displaystyle (\lambda - 1) (\lambda - 1) (\lambda + 1 ) (\lambda +1)=0\)

or 

\(\displaystyle \lambda ^{4} - 2 \lambda^{2} + 1 = 0\)

1 has multiplicity 1 and \(\displaystyle -1\) has multiplicity 3, in which case \(\displaystyle \lambda_{1} = 1 , \lambda_{2} = \lambda_{3}= \lambda_{4} = -1\)

and the characteristic equation is 

\(\displaystyle (\lambda - 1) (\lambda + 1) (\lambda + 1 ) (\lambda +1)=0\)

or

\(\displaystyle \lambda^{4}+2 \lambda^{3}-2 \lambda - 1 = 0\)

Of the four equations given as choices, only \(\displaystyle \lambda ^{4}+ 2 \lambda^{2} + 1 = 0\) cannot be a characteristic equation of the matrix.

By the Cayley-Hamilton Theorem, a matrix is a solution of its own characteristic polynomial equation. None of the choices addresses the characteristic equation of \(\displaystyle A\).

Assume that 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

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

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

Changing to reduced row echelon form:

\(\displaystyle R1 \leftrightarrow R3\)

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

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

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

\(\displaystyle \frac{1}{2}R2 \rightarrow R2\)

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

\(\displaystyle -3R2+ R3 \rightarrow R3\)

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

The matrix is in reduced row echelon form. There is one column which does not contain leading 1s, so 2 is indeed an eigenvalue. The eigenspace is the set of all vectors \(\displaystyle (x, y, z)\) such that \(\displaystyle x= -z , y = -2z\) - that is, the set of vectors \(\displaystyle t(-1,- 2, 1)\). The eigenspace has dimension 1.

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

\(\displaystyle A = \begin{bmatrix} {\color{Red}\textbf{a} } & 4 & -2 \\ 1 & {\color{Red}\textbf{3}} & 0 \\ - 6 & 4 & {\color{Red}\textbf{a}} \end{bmatrix}\)

Set the trace equal to 10 and solve for \(\displaystyle a\):

\(\displaystyle \textup{Tr} (A ) = 10\)

\(\displaystyle a+ a + 3 = 10\)

\(\displaystyle 2 a+ 3 = 10\)

\(\displaystyle 2a = 7\)

\(\displaystyle a = 3.5\)

Examine the columns of \(\displaystyle A\):

\(\displaystyle A= \begin{bmatrix} {\color{Red} 0.1} & {\color{DarkGreen} 0.4} & {\color{Blue} 0. 7} \\ {\color{Red} 0. 5} & {\color{DarkGreen} 0. 2} &{\color{Blue} 0.3} \\ {\color{Red} 0.4 }&{\color{DarkGreen} 0.4} & {\color{Blue} 0} \end{bmatrix}\)

The entries of each column add up to 1:

Column 1: \(\displaystyle 0.1 + 0.5 + 0.4 = 1\)

Column 2: \(\displaystyle 0.4+ 0.2 + 0.4 = 1\)

Column 3: \(\displaystyle 0.7+0.3 + 0 = 1\)

It follows that \(\displaystyle A\) could be a stochastic matrix for a state system; one of the properties of such a matrix is that one of its eigenvalues must be 1.

A \(\displaystyle 4 \times 4\) matrix will have four eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since \(\displaystyle 2-3i\) is a zero, so it \(\displaystyle 2+3i\). 

We now know three of the zeroes, and since only one eigenvalue is unknown, it must be a real value. However, there is no restriction on this zero. Therefore, we have no way of determining the fourth zero - and, consequently, no way of figuring out the characteristic equation with any certainty. 

The sum of the eigenvalues of a matrix is equal to the trace of a matrix - the sum of its diagonal elements; the product of a matrix is equal to its determinant. 

Therefore, to find \(\displaystyle a\), it is necessary to first find \(\displaystyle b\). The trace of \(\displaystyle A\) is equal to diagonal sum \(\displaystyle b+7\), so set that equal to the sum of the given eigenvalues and solve:

\(\displaystyle b+7 = (4+ \sqrt{3}) + (4- \sqrt{3})\)

\(\displaystyle b+7 = 8\)

\(\displaystyle b =1\)

The matrix is therefore

\(\displaystyle A = \begin{bmatrix} 7 & a \\ -3 & 1 \end{bmatrix}\)

\(\displaystyle a\) can be found by setting the determinant - the upper-left to lower-right product minus the upper-right to lower-left product - equal to the product of the eigenvalues:

\(\displaystyle 7(1)-a(-3) = (4+ \sqrt{3}) (4- \sqrt{3})\)

\(\displaystyle 3a+ 7 = 13\)

\(\displaystyle 3a = 6\)

\(\displaystyle a= 2\),

the correct choice.

\(\displaystyle A\) is a lower triangular matrix - all elements above its main diagonal are equal to 0. Such a matrix has as its eigenvalues its diagonal elements. All three diagonal elements in \(\displaystyle A\) are 1, so 1 is the only eigenvalue of \(\displaystyle A\).

To find the eigenvector(s) of \(\displaystyle A\) corresponding to \(\displaystyle \lambda = 1\), first, find the matrix

\(\displaystyle A - \lambda I = A - I\). 

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

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

Perform the Gauss-Jordan elimination process to get this matrix in reduced row-echelon form:

\(\displaystyle R1 \leftrightarrow R2\)

\(\displaystyle R2 \leftrightarrow R3\)

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

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

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

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

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

\(\displaystyle \frac{1}{2} R2 \rightarrow R2\)

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

This matrix is interpreted as

\(\displaystyle x = 0\), \(\displaystyle y = 0\), \(\displaystyle z\) arbitrary.

The eigenvectors must all take the form

\(\displaystyle \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}\)

for some scalar \(\displaystyle z\) - equivalently, all eigenvectors of \(\displaystyle A\) are scalar multiples of 

\(\displaystyle \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\).

A \(\displaystyle 3 \times 3\) matrix will have three (not necessarily distinct) eigenvalues, which are the zeroes of its characteristic polynomial equation. Since all of the entries of the matrix are real, all of the coefficients will be real as well. It follows that any imaginary zeroes must occur in conjugate pairs, so, since \(\displaystyle 1 +5 i\) is a zero of this equation, so is its complex conjugate, \(\displaystyle 1 -5 i\). 

The characteristic equation of a \(\displaystyle 3 \times 3\) equation sets a degree-3 polynomial equal to 0. Since 2, \(\displaystyle 1 +5 i\), and \(\displaystyle 1 -5 i\) are its zeroes, this polynomial is

\(\displaystyle (\lambda -2) [\lambda -(1+5i)][\lambda -(1-5i)]\)

\(\displaystyle =(\lambda -2) [(\lambda -1)-5i] [(\lambda -1)+5i]\)

\(\displaystyle =(\lambda -2) [(\lambda -1)^{2}- (5i)^{2}]\)

\(\displaystyle =(\lambda -2) (\lambda ^{2} -2\lambda +1 +25)\)

\(\displaystyle =(\lambda -2) (\lambda ^{2} -2\lambda +26)\)

\(\displaystyle = \lambda ^{3} -4\lambda ^{2} +30\lambda -52\),

The characteristic equation is 

\(\displaystyle \lambda ^{3} -4\lambda ^{2} +30\lambda -52 = 0\).

The two eigenvectors given, \(\displaystyle \textbf{v}_{1}\) and \(\displaystyle \textbf{v}_{2}\), are linearly independent, since they are not scalar multiples of each other; therefore, they form a basis of the 2-dimensional eigenspace of \(\displaystyle \lambda\). \(\displaystyle \textbf{u}\) is an eigenvector corresponding to \(\displaystyle \lambda\) if and only if it is a linear combination of the two given basis vectors, or, equivalently, if there exist \(\displaystyle a,b\) so that 

\(\displaystyle a \textbf{v}_{1} +b \textbf{v}_{2} = \textbf{u}\)

Rewrite this as follows:

\(\displaystyle a \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} +b \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ x \end{bmatrix}\)

This is equivalent to a system of three linear equations in two variables:

\(\displaystyle a-b = 1\)

\(\displaystyle -2a+b = 0\)

\(\displaystyle a-2b = x\)

Solve the system by first, rewriting the second equation as 

\(\displaystyle b= 2a\)

The first equation becomes

\(\displaystyle a-2a = 1\)

\(\displaystyle -a = 1\)

\(\displaystyle a = -1\)

Substitute in the first equation:

\(\displaystyle -1 -b = 1\)

\(\displaystyle -b = 2\)

\(\displaystyle b = -2\)

Now substitute for \(\displaystyle a\) and \(\displaystyle b\) in the third equation:

\(\displaystyle -1-2 (-2) = x\)

\(\displaystyle x= 3\), the correct response.