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

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.

• $T$ is, presumably, just some other (invertible) matrix, such as you might use when changing bases. – mweiss May 23 '17 at 2:23
• Are you allowed to use $det(AB)=det(A)det(B)$? – edm May 23 '17 at 2:32
• This is a property of similar matrices. Search for the same on this site to see similar questions. – StubbornAtom May 23 '17 at 7:03

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

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

• We should be showing eigenvalue of M, not T. Moreover, this does not show the multiplicity of eigenvalues are the same. – edm May 23 '17 at 2:41
• Which "they" are you talking about? – edm May 23 '17 at 2:50
• Now you are even assuming that $v$ is an eigenvector of both $T$ and $M$. – edm May 23 '17 at 2:52