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Now I know that one of the properties of similar matrices is that they have the same rank and also that Two n x n mat. A and B are similar if B=(P^-1)AP for some invertible n x n mat. P. I looked at similar question on here but I do not understand what they are saying.

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Product of invertible matrices is invertible.

So if $PBP^{-1}$ is invertible then so is $P^{-1}PBP^{-1}$ and so is $P^{-1}PBP^{-1}P$. The latter is equal to $B$.

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if you are going to show that similarity preserves the rank, then use the facts (a) the eigenvalues stays the same under similarity, (b) rank of a matrix is the number(including algebraic multiplicity) of nonzero eigenvalues.

the two together should give you what u are after.

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To show that two matrices have the same rank, it suffices to show that their image (column space) is of the same dimension.

Take and $y$ in the image of $A$. We may write $y = Ax$ for some $x$. Then $$ P^{-1}y = P^{-1}Ax = P^{-1}APP^{-1}x = (P^{-1}AP)(P^{-1}x) = B(P^{-1}x) $$ which means that $P^{-1}y$ is in the image of $B$. Thus, $P^{-1}:im(A) \to im(B)$ is a linear transformation with inverse $P:im(B) \to im(A)$. It follows that $im(A)$ and $im(B)$ have the same dimension, which is to say that $A$ and $B$ have the same rank.


Alternatively, one can more generally prove that if $P$ is invertible, then $rank(AP) = rank(A)$ and $rank(PA) = rank(A)$ (whenever those products make sense). With that, we have $$ rank(B) = rank([P^{-1}A]P) = rank(P^{-1}A) = rank(A) $$ as desired.


Yet another approach: note that $im(A) = im(AP)$, and $\ker(AP) = \ker(P^{-1}AP)$. Apply the rank nullity theorem to confirm that $\dim im(P^{-1}AP) = \dim im(A)$.

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