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Let $F$ be a field and $V$ be a finite dimensional vector space. Suppose $T,S:V \longrightarrow V$ are two linear maps that are conjugate. Now consider some $f(x) \in F[x]$. I want to show that the rank of the maps $f(T),f(S): V \longrightarrow V$ are the same. First, we note that by the Rank-Nullity theorem,

$dim(Ker(f(T))) + dim(Im(f(T))) = dim(Ker(f(S))) + dim(Im(f(S)))$

and thus, $rnk(f(T)) = rnk(f(S))$ if and only if $null(f(T)) = null(f(S))$. Therefore, I am trying to show that the kernel of $f(T)$ is equal to the kernel of $f(S)$.

I begin with the fact that $Ker(f(T)) = \{v \in V | f(T)(v) = 0\}$. Similarly for $Ker(f(S))$. Not sure how to proceed.

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Hint: You know that $T$ and $S$ are conjugate. What about $f(T)$ and $f(S)$? – froggie Dec 13 '12 at 17:57
When you write conjugate, do you mean $\langle Ax, y \rangle = \langle x, A^\ast y \rangle$ like this? – Rudy the Reindeer Dec 13 '12 at 18:20
@MattN. what the heck is $A$? – afedder Dec 13 '12 at 19:54
@froggie $f(T)$ and $f(S)$ would be conjugate as well...but how would this help with rank? – afedder Dec 13 '12 at 19:55
@MattN. when I say conjugate, I mean there exists an invertible linear map $R$ such that $T=RSR^{-1}$ – afedder Dec 13 '12 at 19:58
up vote 1 down vote accepted

As you said in the comments, $f(S)$ and $f(T)$ are conjugate linear transformations. So the problem boils down to proving the following statement:

Suppose $A$ and $B$ are linear maps $V\to V$ that are conjugate, say $B = RAR^{-1}$. Then the rank of $A$ and $B$ are the same.

To prove this, we must show that the rank of $A$ and $RAR^{-1}$ are the same. Since $R^{-1}$ is invertible, it is surjective. Therefore the image of $AR^{-1}$ is the same as the image of $A$. We can then conclude that the image of $RAR^{-1}$ is $R(Im(A))$. But since $R$ is invertible, the dimension of $R(Im(A))$ is the same as the dimension of $Im(A)$. This proves that the rank of $RAR^{-1}$ is the same as the rank of $A$.

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