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Prove that :- If $2$ matrices $A$ and $B$ are similar then they will have the same rank.

Proof is given here but I can't understand both answers which are related to image and kernel. I have seen all the video lectures of Prof. Gilbert Strang but I have not seen these things in those lectures. I only know that $A$ and $B$ are similar iff $A$ = $MBM^{-1}$ for some invertible square matrix $M$ but I can't proceed further. Is there any simple proof of it ?
Please help.

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    $\begingroup$ What definition of rank are you working with? Usually, rank is defined as the dimension of the image, so it's hard to talk about rank without talking about images... $\endgroup$
    – 5xum
    Commented Apr 2, 2019 at 8:37
  • $\begingroup$ Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum $\endgroup$ Commented Apr 2, 2019 at 8:39
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    $\begingroup$ @FareedAF The rank is only the number of non zero eigen values if the matrices are square. $\endgroup$
    – 5xum
    Commented Apr 2, 2019 at 8:47
  • $\begingroup$ @5xum The OP is clearly talking about square matrices, otherwise the product $MBM^{-1}$ doesn’t make sense. $\endgroup$
    – amd
    Commented Apr 2, 2019 at 17:28

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My favourite proof goes along the lines of

$rk(B)\geq rk(MBM^{-1}) = rk(A)$, as multiplying can only reduce rank (or keep it unchanged), never increase it. Now note that $B = M^{-1}AM$, so we similarily get $rk(A)\geq rk(M^{-1}AM) = rk(B)$

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  • $\begingroup$ Thank you! @Arthur $\endgroup$
    – ankit
    Commented Apr 2, 2019 at 8:57

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