Let $V=Mat_n(\mathbb{R})$ and $T:V\rightarrow V$ the linear operator $T(A)=A^t$. Find the characteristic polynomial of $T$

Question

Let $V=Mat_n(\mathbb{R})$ and $T:V\rightarrow V$ the linear operator $T(A)=A^t$. Find the characteristic polynomial of $T$.

My work:

$(T\circ T)(A)=T(T(A))=T(A^t)=(A^t)^t=A$

$P_{T\circ T}(x)=P_A(x)=det(xI-A)$

then

$P_T(x)=P_{A^t}(x)=det(xI-A^t)$

Is correct my reasoning? Can someone help me? thanks!

Answer 1

In fact, $T$ is a symmetry ($T\circ T=Id$), then its characteristic polynomial is $$ (X-1)^{d_{i}}(X+1)^{d_{ai}} $$ where $d_{i}$ (resp. $d_{ai}$) is the dimension of the invariant (resp. antiinvariant space). Using the matrix units $(E_{ij})_{1\leq i,j,\leq n}$, you can build bases of these spaces.

For the invariant space (symmetric matrices), one has $$ (E_{ij}+E_{ji})_{1\leq i\leq j,\leq n} $$ hence $\frac{n(n+1)}{2}$ matrices and, similarly, for the anti-invariant space (antisymmetric matrices), one has $$ (E_{ij}-E_{ji})_{1\leq i< j,\leq n} $$ hence $\frac{n(n-1)}{2}$ matrices. the desired polynoomial is then $$ (X-1)^{\frac{n(n+1)}{2}}(X+1)^{\frac{n(n-1)}{2}} $$

Hope this helps.