Symmetric Matrix Transformation 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
 A: The options you've chosen to be incorrect are the right ones, but for the wrong reason in 5. You have to consider the eigenvalues of $T$, the matrix transformation, not the eigenvalues of the matrices it acts on.
To prove 6 wrong, you therefore have to prove that $T(A) = 2A$ has no non-zero solutions.
Edit: The reasoning for 1 happens to be wrong also, for similar motives. Again, you need the determinant of $T$, not the matrices it maps from or to.
A: This is a kind of question which makes us confused easily. 
First, what is $T?$ It is a transformation from a vector space to a vector space. Since the space is of finite dimension, $T$ can be written as a matrix and we can talk about its determinant and everything. Now we have two kinds of matrices here. One is 2x2 matrices on which $T$ maps, the other is the matrix for $T$. We will look at those 2x2 matrices as a vector space.
Let $V=\left\{ \begin{pmatrix}a & b\\ b& c \end{pmatrix} : a,b,c\in\mathbb{R} \right\}=\text{span}(v_1,v_2,v_2)$ where 
$$v_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},v_2=\begin{pmatrix}0&1\\1&0\end{pmatrix},v_3=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
So $\begin{pmatrix}a & b\\ b& c \end{pmatrix} = av_1+bv_2+cv_3.$
These are basis for $V.$ Now we write $T$ as a transformation matrix with respect to this basis. We have $ T(v_1)=v_3, T(v_2)=-v_2, T(v_3)=v_1.$ Thus
$$T=\begin{pmatrix}0&0&1\\0&-1&0\\1&0&0\end{pmatrix}.$$
Now we can check that:


*

*$\det(T)=1$

*$T^{-1}=T$

*$T$ is linear

*$T$ has eigenvalue $-1$ w.r.t. eigenvector $v_2.$

A: 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?
