Here's the question,
Let $T$ be the transformation of 2 by 2 real symmetric matrices defined by:
\begin{bmatrix}a&b\\b&c\end{bmatrix}>>>>\begin{bmatrix}c&-b\\-b&a\end{bmatrix}
then which of the following statements is NOT true?
- $\det(T)=-1$
- $T^{-1}=T$
- $T$ is linear
- the space of 2 by 2 real symmetric matricies with only zeros in the main diagonal is an eigenspace of $T$.
- $\lambda=2$ is an eigen value of $T$
To solve this I used:
\begin{bmatrix}1&5\\5&1\end{bmatrix}>>>>\begin{bmatrix}1&-5\\-5&1\end{bmatrix}
I found the $\det(T)=-24$, so 1 is not correct. And I also found that the eigenvalues of this matrix I chose were, $-4$ and $-6$. Therefore is 5 is also NOT correct.
Am I correct in assuming this matrix? Also am I correct in my answers or am I missing something, if so please explain.
Thank You
Best Answer
A basis of the vector space of symmetric $2\times2$ matrices is $$ \mathscr{B}= \left\{ A_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}; A_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}; A_3=\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\} $$
If $T$ is linear, then its matrix with respect to $\mathscr{B}$ is given by $$ X= \begin{bmatrix} C_{\mathscr{B}}(T(A_1)) & C_{\mathscr{B}}(T(A_2)) & C_{\mathscr{B}}(T(A_3)) \end{bmatrix} $$ where $C_{\mathscr{B}}(A)$ is the coordinate vector of $A$ with respect to $\mathscr{B}$. Then, since $T(A_1)=A_3$, $T(A_2)=-A_2$ and $T(A_3)=A_1$, $$ X= \begin{bmatrix} 0&0&1\\ 0&-1&0\\ 1&0&0 \end{bmatrix} $$ Now, for $A=\left[\begin{smallmatrix}a&b\\b&c\end{smallmatrix}\right]$, we have $$ C_{\mathscr{B}}(A)=\begin{bmatrix}a\\b\\c\end{bmatrix} $$ and $$ X\begin{bmatrix}a\\b\\c\end{bmatrix}= \begin{bmatrix} 0&0&1\\ 0&-1&0\\ 1&0&0 \end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}= \begin{bmatrix}c\\-b\\a\end{bmatrix}=C_{\mathscr{B}}(T(A)) $$ Therefore $T$ is linear. Note that this could be proved directly, by simple computations, but $T$ is useful for the other questions.
The determinant of $T$ is the same as the determinant of any associated matrix, so $\det T=\det X=1$.
The eigenvalues of $T$ are the same as the eigenvalues of $X$. Since $$ \det(X-\lambda I)=\det \begin{bmatrix} -\lambda&0&1\\ 0&-1-\lambda&0\\ 1&0&-\lambda \end{bmatrix}= (-1-\lambda)^2(1-\lambda) $$ the eigenvalues are $1$ and $-1$.
Is the subspace of symmetric matrices with $0$ on the diagonal an eigenspace? This is easier without $X$: the eigenspace relative to $-1$ is the set of matrices $A=\left[\begin{smallmatrix}a&b\\b&c\end{smallmatrix}\right]$ such that $T(A)=-A$ or $$ \begin{bmatrix} c & -b \\ -b & a \end{bmatrix}= \begin{bmatrix} -a & -b\\ -b & -c \end{bmatrix} $$ So the condition is $a=c$ and this is not having $0$ on the diagonal.
The eigenspace relative to $1$ is the space of matrices $A$ such that $T(A)=A$ and the condition reads $$ \begin{bmatrix} c & -b \\ -b & a \end{bmatrix}= \begin{bmatrix} a & b\\ b & c \end{bmatrix} $$ which gives $a=c$ and $b=0$.
The fact that $T=T^{-1}$ is obvious: just observe that $T(T(A))=A$ for any (symmetric) matrix $A$. This already tells you that $2$ is not an eigenvalue. Why?