We show that \(-1\) is a root of the characteristic polynomial \(\det(A - \lambda I)\). In other words, show that \(\det(A - (-1)I) = 0\).

Now, \(A - (-1)I = \begin{bmatrix} 0 & 2 \\ 1 & 1\end{bmatrix} - \begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & 2\end{bmatrix}\). The determinant of \(\begin{bmatrix} 1 & 2 \\ 1 & 2\end{bmatrix} = 1\cdot 2 - 2 \cdot 1 = 0.\) Hence, \(-1\) is an eigenvalue of \(A\).

We need to show that \(Au = \lambda u\) for some scalar \(\lambda\).

Note that \[A u = \begin{bmatrix} 2 & 2 \\ 6 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \end{bmatrix} = \begin{bmatrix} -2 \\ 4 \end{bmatrix} = -2\begin{bmatrix} 1 \\ -2 \end{bmatrix} = -2u. \] Hence, \(u\) is an eigenvector of \(A\) with eigenvalue value \(-2\).

One can certainly first obtain all the eigenvalues of \(A\). However, we will work directly with the definition of an eigenvector instead. In particular, we want to obtain all values of \(a\) such that \(Au = \lambda u\) for some scalar \(\lambda\).

Note that \[A u = \begin{bmatrix} 2a - 2 \\ a + 1 \end{bmatrix}.\] Hence, we want \[\begin{bmatrix} 2a - 2 \\ a + 1 \end{bmatrix} = \begin{bmatrix} \lambda a \\ -\lambda \end{bmatrix},\] or equivalently, \begin{align*} 2a - 2 & = \lambda a \\ a + 1 & = -\lambda \end{align*} From the second equation, \(\lambda = -a -1\). Substituting for \(\lambda\) in the first equation gives \[2a - 2 = (-a-1)a,\] or equivalently, \[a^2 + 3a - 2 = 0.\] Using the quadratic formula, we obtain \(a = \frac{-3\pm \sqrt{17}}{2}\).

One should now check that both values work. The details are omitted.

We first determine the geometric multiplicity of the eigenvalue \(2\). Note that \(A - 2I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & -2 \\ 0 & 1 & -2\end{bmatrix}\). Its RREF is \(\begin{bmatrix} 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\). Since there is only one pivot column (the second column), the dimension of \(N(A-2I)\) is \(3-1= 2\). Hence, the geometric multiplicy is \(2\).

To determine the algebraic multiplicity, note that the characteristic polynomial is \(p_A = \left | \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & -2 \\ 0 & 1 & -\lambda \end{array} \right |= (2-\lambda)\left| \begin{array}{ccc} 3-\lambda & -2 \\ 1 & -\lambda \end{array}\right | = (2-\lambda)(-3\lambda + \lambda^2 +2) = (2-\lambda)(1-\lambda)(2-\lambda).\)

Hence, \(2\) appears as a root of \(p_A\) twice. So the algebraic multiplicy is \(2\).

Let \(A\) be an \(n\times n\) triangular matrix. Let \(a_i\) denote the \(i\)th entry on the diagonal of \(A\) for \(i = 1,\ldots,n\).

The eigenvalues are precisely the roots of the characteristic polynomial \(p_A = \det(A-\lambda I_n)\). Since \(A\) is triangular, \(A-\lambda I_n\) is also triangular and therefore has determinant \((a_1-\lambda)(a_2-\lambda)\cdots(a_n-\lambda)\). Thus, the roots of \(p_A\) are precisely \(a_1,\ldots,a_n\). Hence, the eigenvalues of \(A\) are the entries on the diagonal.

We need to find a basis for the nullspace of \(A - I = \begin{bmatrix} 0 & 1 & -1 \\ 0 & 2 & -2 \\ 0 & 1 & -1\end{bmatrix}\).

Clearly, the RREF of \(A-I\) is \(\begin{bmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\). Note that the first and third columns are nonpivot columns. Hence, a basis for the nullspace is given by \(\left\{\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix}\right\}\).

Let \(\lambda = 1+2i\). Then \begin{align*} \det(A-\lambda I) &= \begin{vmatrix} 1-\lambda & 4 \\ -1 & 1-\lambda\end{vmatrix} \\ &= (1-\lambda)^2-(-1)4 \\ &= (-2i)^2 +4 = 4i^2 + 4 = -4 + 4 = 0. \end{align*} Hence, \(\lambda\) is an eigenvalue of \(A\).

To obtain a basis for the eigenspace, we find a basis for the nullspace of \(A - (1+2i)I = \begin{bmatrix} -2i & 4 \\ -1 & -2i\end{bmatrix}\) whose RREF is \(\begin{bmatrix} 1 & 2i \\ 0 & 0 \end{bmatrix}\). Hence, a basis for the nullspace is given by \(\left\{\begin{bmatrix} -2i \\ 1 \end{bmatrix}\right\}\).

We first find the eigenvalues of \(A\) by solving for \(\lambda\) in \(\det(A - \lambda I) = 0.\) That is \[\left| \begin{array}{rrr} 1-\lambda & 1 & 0 \\ 0 & 2-\lambda & 0 \\ 0 & 2 & 1-\lambda \end{array}\right| = 0,\] or equivalently, \[(1-\lambda)^2(2-\lambda) = 0.\] Hence, the eigenvalues are \(1\) and \(2\).

Note that \(A - I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 0 \end{bmatrix}\). Its RREF is \(\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\). Hence, a basis for the nullspace of \(A-I\) is given by \(\left \{ \begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} \right\}\).

Note that \(A - 2I = \begin{bmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 2 & -1 \end{bmatrix}\). Its RREF is \(\begin{bmatrix} 1 & 0 & -\frac{1}{2} \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0 \end{bmatrix}\). Its RREF is A basis for the nullspace of this matrix is given by \(\left \{ \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ 1\end{bmatrix} \right\}.\)

Hence, setting \(D = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{bmatrix}\) and \(P = \begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2}\\ 0 & 1 & 1\end{bmatrix}\), we have \(A = PDP^{-1}\).

Let \(\lambda\) be an eigenvalue of \(A^\mathsf{T}A\). Let \(u\) be an eigenvector of \(A^\mathsf{T}A\) with eigenvalue \(\lambda\).

Then \(0 \leq \| A u\|^2 = (Au)^\mathsf{T} (Au) = u^\mathsf{T} A^\mathsf{T} A u = u^\mathsf{T} (\lambda u) = \lambda \|u\|^2.\)

Since \(u \neq 0\), \(\|u\|^2 \gt 0\). Hence, \(\lambda \geq 0\).

As \(A\) is \(2\times 2\), if \(A\) has two distinct eigenvalues, then \(A\) will be diagonalizable. So we must first determine all values of \(a\) for which \(A\) has a unique eigenvalue.

The characteristic polynomial of \(A\) is \begin{eqnarray*} \left | \begin{array}{ccc} 10 - 5a -\lambda & -16 + 10a \\ 5 - 3a & -8 + 6a -\lambda \end{array} \right | & = & (10-5a -\lambda)(-8+6a-\lambda) - (-16+10a)(5-3a) \\ & = & -80 + 100a - 30a^2 - (2+a)\lambda + \lambda^2 + 80 - 98a + 30 a^2 \\ & = & 2a - (2+a)\lambda + \lambda^2\\ & = & (2-\lambda)(a-\lambda). \end{eqnarray*} Thus, for \(A\) to have a unique eigenvalue, we must have \(a = 2\). Note that when \(a = 2\), \(A = \begin{bmatrix} 0 & 4 \\ -1 & 4 \end{bmatrix}\) and \(2\) is the unique eigenvalue. Now, \(A - 2I = \begin{bmatrix} -2 & 4 \\ -1 & 2 \end{bmatrix}\). It is clear that this matrix has rank 1 and therefore its nullity is less than 2. Hence, the geometric multiplicity of the only eigenvalue 2 is less than 2 and so \(A\) is not diagonalizable. Hence, the answer is “yes”.

Since \(A\) has \(n\) distinct eigenvalues, \(A\) is diagonalizable. Hence, there exist an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(A = PDP^{-1}\).

Thus \(\det(A) = \det(PDP^{-1}) = \det(P)\det(D)\det(P^{-1}) =\det(D)\). But \(D\) is a diagonal matrix. Hence, \(\det(D)\) is the product of the diagonal entries which are precisely the eigenvalues of \(A\).

By the inverse formula, \(P^{-1} = \frac{1}{\det(P)}\begin{bmatrix} a & -1 \\ -3 & 1 \end{bmatrix}.\) (One can also compute \(P^{-1}\) directly.) Multiplying out the matrices gives \begin{eqnarray*} PDP^{-1} & = & \frac{1}{a-3} \begin{bmatrix} a+3 & -2 \\ 6a & -3-a\end{bmatrix}\\ \end{eqnarray*} Since all the entries of \(A\) must be integers and \(a\) is an integer, \(a-3\) must divide \(-2\), the top-right entry in the matrix above. Thus, the only possibilities for \(a\) are \(1,2,4,5\) because the only integers that divide \(-2\) are \(-1,-2,1,2\). One can check that all 4 values lead to integer matrices. Hence, \(a = 2\) is a desired value.

As discusses previously, it suffices to show that \({\cal S}_{n\times n}\) is closed under vector addition and scalar multiplication as defined for \(\mathbb{R}^{n\times n}\).

Let \(A,B \in {\cal S}_{n\times n}\). Let \(\lambda \in \mathbb{R}\). Then \(A^\mathsf{T} = A\) and \(B^\mathsf{T} = B\).

Let \(C = A + B\). Then \(C^\mathsf{T} = (A+B)^\mathsf{T} = A^\mathsf{T} + B^\mathsf{T} = A + B = C\). Hence, \(C \in {\cal S}_{n \times n}\).

Let \(D = \lambda A\). Then \(D^\mathsf{T} = (\lambda A)^\mathsf{T} = \lambda A^\mathsf{T} = \lambda A = D\). Hence, \(D \in {\cal S}_{n\times n}\).