# How to prove that rank is a similarity invariant?

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(AT); 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.

Related Calculus and Beyond Homework Help News on Phys.org
Office_Shredder
Staff Emeritus
Gold Member
Rank is the dimension of the image of the transformation. Can you compare the dimension of the image of B and A?

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
Staff Emeritus
Homework Helper
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.

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 $$(P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow AP\textbf{x}=\textbf{0} \Leftrightarrow T_A(T_P(\textbf{x})) = \textbf{0}.$$ But, since $$R(T_P) = R^n ,$$ this is equivalent to $$T_A(\textbf{x}) = A\textbf{x}= \textbf{0}.$$

How is this?

Last edited:
vela
Staff Emeritus
Homework Helper
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.

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
Homework Helper
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.

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, $$rank(A)=rank(B).$$