# Similar Matrices have the same rank

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 ?

• 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...
– 5xum
Commented Apr 2, 2019 at 8:37
• Isn't the rank also the number of non zero eigen values? And similar matrices have the same spectrum Commented Apr 2, 2019 at 8:39
• @FareedAF The rank is only the number of non zero eigen values if the matrices are square.
– 5xum
Commented Apr 2, 2019 at 8:47
• @5xum The OP is clearly talking about square matrices, otherwise the product $MBM^{-1}$ doesn’t make sense.
– amd
Commented Apr 2, 2019 at 17:28

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