How to prove that rank is a similarity invariant?

  • #1
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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.


Thanks for your help!
 

Answers and Replies

  • #2
Office_Shredder
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Rank is the dimension of the image of the transformation. Can you compare the dimension of the image of B and A?
 
  • #3
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Yea, it sort of just hit me. Just to verify I've written it up soundly, how does this sound to you?

Let [tex] B = P^{-1}AP.[/tex] Consider [tex] T_A:V \rightarrow V [/tex] and [tex] T_B: V \rightarrow V. [/tex] Then [tex] (P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow A\textbf{x}=\textbf{0} [/tex], simply multiply by [tex] P [/tex] on left and [tex] P^{-1} [/tex] on right. Therefore, [tex] N(T_A) = N(T_B). [/tex] Equivalently, [tex] rank(A) = rank(B) [/tex].
 
  • #4
vela
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Yea, it sort of just hit me. Just to verify I've written it up soundly, how does this sound to you?

Let [tex] B = P^{-1}AP.[/tex] Consider [tex] T_A:V \rightarrow V [/tex] and [tex] T_B: V \rightarrow V. [/tex] Then [tex] (P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow A\textbf{x}=\textbf{0} [/tex], simply multiply by [tex] P [/tex] on left and [tex] P^{-1} [/tex] on right. Therefore, [tex] N(T_A) = N(T_B). [/tex] Equivalently, [tex] rank(A) = rank(B) [/tex].
That doesn't work because the x gets in the way when you try to multiply by P-1 on the right.
 
  • #5
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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, [tex] R(T_P) = R^n. [/tex]
So [tex] (P^{-1}AP)\textbf{x} = \textbf{0} \Leftrightarrow AP\textbf{x}=\textbf{0}
\Leftrightarrow T_A(T_P(\textbf{x})) = \textbf{0}. [/tex] But, since [tex] R(T_P) = R^n , [/tex] this is equivalent to [tex] T_A(\textbf{x}) = A\textbf{x}= \textbf{0}. [/tex]

How is this?
 
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  • #6
vela
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That doesn't work either. For example, working in R2, suppose you have

[tex]A=\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}[/tex]

and

[tex]P=\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}[/tex]

Then

[tex]P^{-1}AP\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}[/tex]

but

[tex]A\begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}1 \\ 0\end{bmatrix}[/tex]

So P-1APx=0 doesn't imply Ax=0.
 
  • #7
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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?
 
  • #8
HallsofIvy
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That sounds more complicated than necessary to me. Go back to your original idea but be more careful- If [itex]B= P^{-1}AP[/itex] and Bx= 0, then (P^{-1}AP)x= P^{-1}(AP)x= 0 so, multiplying on both sides by P, APx= 0. Now let [itex]y= P^{-1}x[/itex] so that [itex]x= Py[/itex]. APx= AP(P^{-1}P)x= A(Px)= Ay= 0.
 
  • #9
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Now let [itex]y= P^{-1}x[/itex] so that [itex]x= Py[/itex]. 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 [tex] APx = AP(Py) = AP^2y=0. [/tex]

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

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

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