Prove that $M$ and $TMT^{-1}$ have the same determinant and eigenvalues.

Question

Prove that $M$ and $TMT^{-1}$ have the same determinant and eigenvalues.

So I'm confused as to what T is supposed to be, are they referring to the Transpose?

I know that $T(T^{-1})= 1 $ but I know that I can't move stuff around because it may not be commutative.

Edit: Just realized that $det(T)det(M)det(T^{-1})$ is commutative since the determinant is not a matrix but actually a number value.

Answer 1

$\det(TMT^{-1})=\det(T)\det(M)\det(T^{-1})=\det(T)\det(M)\frac{1}{\det(T)}=\det(M)$.
Because a matrix $A$ has an eigenvalue $\lambda$ iff $(A-\lambda I)\vec{x}=\vec{0}$ for some non-zero $\vec{x}$, assume $(M-\lambda I)\vec{v}=\vec{0}$ for some non-zero $\vec{v}$. Note that $T\vec{v}≠\vec{0}$ because $T$ is invertible, and
$(TMT^{-1}-\lambda I)(T\vec{v})=TMT^{-1}T\vec{v}-\lambda T\vec{v}=TM\vec{v}-\lambda T\vec{v}=T(M-\lambda I)\vec{v}=\vec{0}$. Repeat this argument going the other way and you have shown the eigenvalues are identical.

This result shows the eigenvalues of different bases of the same vector space are identical, and the eigenvectors differ only by the relative coordinates they are written in.

Answer 2

T is probably an invertible matrix so;

$det(M)=det(T)\frac{1}{det(T)}det(M)=det(T)det(M)det(T^{-1})=det(TMT^{-1})$

And

$$det(TMT^{-1}-\lambda I)=det(T(M-\lambda I)T^{-1})=det(T)det(T^{-1})det(M-\lambda I)=det(M-\lambda I)$$ So they have same char.polynomial. so same eigenvalues