If $T:V\longrightarrow V$ is diagonalizable, then $T$ has a characteristic polynomial of the form $f(x)=(x-c_{1})^{n_{1}}\dotsc(x-c_{k})^{n_{k}}$.

Question

If $c_{1},\dotsc, c_{k}$ are the distinct eigenvalues of $T$ and $W_{i}$ is the eigenspace of $c_{i}$, then $T$ has characteristic polynomial of the form $f(x)=(x-c_{1})^{\dim W_{1}}\dotsc (x-c_{k})^{\dim W_{k}}$.

How do I prove this?

Answer 1

First there's a typo: should be eigenvalues $c_i $.

You have that $PAP^{-1}=\Delta $, where $\Delta $ is diagonal with $\rm {dim}(W_i) \,c_i $'s on the diagonal for each $i $.

But the characteristic polynomials of similar matrices are the same, because $\rm {det}(A-\lambda I)=\rm {det}(P (A-\lambda I)P^{-1})=\rm {det}(\Delta-\lambda I) $, by linearity and the fact that the determinant is a homomorphism.

Now, nothing is left but to compute the characteristic polynomial of $\Delta $.

Answer 2

I found an answer. If $B$ is an eigenbasis of $T$ for $V$, then $[T]_{B}$ is diagonal. Hence for each eigenvalue $c_{i}$, $[T-c_{i}I]_{B}$ is diagonal, and the nullity of a diagonal matrix is equal to the number of zeroes on the diagonal.