How to prove that rank is a similarity invariant?

[Raskolnikov]

1. Prove that the rank of a matrix is invariant under similarity.Notes so far:
Let A, B, P be nxn matrices, and let A and B be similar. That is, there exists an invertible matrix P such that B = P^-1AP. I know the following relations so far: rank(P)=rank(P^-1)=n ; rank(A) = rank(A^T); rank(A) + nullity(A) = n . However, I'm unable to write a full proof of the theorem. It makes sense intuitively, but I really would like a written proof.Thanks for your help!

 

[Office_Shredder]

Rank is the dimension of the image of the transformation. Can you compare the dimension of the image of B and A?

 

[Raskolnikov]

Yea, it sort of just hit me. Just to verify I've written it up soundly, how does this sound to you?

Let $$B = P^{-1}AP.$$ Consider $$T_A:V \rightarrow V$$ and $$T_B: V \rightarrow V.$$ Then $$(P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow A\textbf{x}=\textbf{0}$$, simply multiply by $$P$$ on left and $$P^{-1}$$ on right. Therefore, $$N(T_A) = N(T_B).$$ Equivalently, $$rank(A) = rank(B)$$.

 

[vela]

Yea, it sort of just hit me. Just to verify I've written it up soundly, how does this sound to you?

Let $$B = P^{-1}AP.$$ Consider $$T_A:V \rightarrow V$$ and $$T_B: V \rightarrow V.$$ Then $$(P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow A\textbf{x}=\textbf{0}$$, simply multiply by $$P$$ on left and $$P^{-1}$$ on right. Therefore, $$N(T_A) = N(T_B).$$ Equivalently, $$rank(A) = rank(B)$$.

That doesn't work because the x gets in the way when you try to multiply by P^-1 on the right.

 

[Raskolnikov]

That doesn't work because the x gets in the way when you try to multiply by P^-1 on the right.

Attempt #2:
hmmm...well, since P is invertible, $$R(T_P) = R^n.$$
So $$(P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow AP\textbf{x}=\textbf{0} <br /> \Leftrightarrow T_A(T_P(\textbf{x})) = \textbf{0}.$$ But, since $$R(T_P) = R^n ,$$ this is equivalent to $$T_A(\textbf{x}) = A\textbf{x}= \textbf{0}.$$

How is this?

 

[vela]

That doesn't work either. For example, working in R², suppose you have

$$A=\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$

and

$$P=\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$$

Then

$$P^{-1}AP\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$

but

$$A\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix}$$

So P^-1APx=0 doesn't imply Ax=0.

 

[Raskolnikov]

Alright, I think I'm over-complicating things. Since P and P^-1 are invertible, they are equivalent to multiplication of some elementary matrices. Thus, multiplying A on the left by P^-1 is solely elementary row operations, and multiplying by P on the right is solely elementary column operations. Neither of these affect the dimensions of the rowspace or columnspace of A. Thus, rank(A) remains the same.

How does this sound?

 

[HallsofIvy]

That sounds more complicated than necessary to me. Go back to your original idea but be more careful- If $B= P^{-1}AP$ and Bx= 0, then (P^{-1}AP)x= P^{-1}(AP)x= 0 so, multiplying on both sides by P, APx= 0. Now let $y= P^{-1}x$ so that $x= Py$. APx= AP(P^{-1}P)x= A(Px)= Ay= 0.

 

[Raskolnikov]

Now let $y= P^{-1}x$ so that $x= Py$. APx= AP(P^{-1}P)x= A(Px)= Ay= 0.

I don't see how that logically follows. From your definition of y, we should get $$APx = AP(Py) = AP^2y=0.$$

Regardless, I think I stumbled on a somewhat more elegant proof. It uses the general fact that $$rank(AB) \ \leq \ min\{rank(A),rank(B)\}.$$

Let $B= P^{-1}AP$. Then $$rank(B) = rank(P^{-1}AP) \leq rank(P^{-1})rank(A)rank(P) \leq rank(A),$$ since $$rank(P) = rank(P^{-1}) = n \geq rank(A).$$ Similarly, since $$A=PBP^{-1}$$, we easily find that $$rank(A) \leq rank(B).$$ Therefore, $$rank(A)=rank(B).$$