1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to prove that rank is a similarity invariant?

  1. Jul 2, 2010 #1
    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!
     
  2. jcsd
  3. Jul 2, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Rank is the dimension of the image of the transformation. Can you compare the dimension of the image of B and A?
     
  4. Jul 10, 2010 #3
    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].
     
  5. Jul 10, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    That doesn't work because the x gets in the way when you try to multiply by P-1 on the right.
     
  6. Jul 10, 2010 #5
    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?
     
    Last edited: Jul 10, 2010
  7. Jul 10, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  8. Jul 10, 2010 #7
    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?
     
  9. Jul 11, 2010 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  10. Jul 12, 2010 #9
    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to prove that rank is a similarity invariant?
Loading...