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Suppose $A$ and $B$ are similar matrices over $\mathbb{C}^n$. Why do we have $\text{rank}(A) = \text{rank}(B)$?

Unit
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usser-L
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3 Answers3

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There is an invertible matrix $P$ such that $PA = BP$, so there is a well-defined map $\operatorname{Col}(A) \to \operatorname{Col}(B)$ sending $x\mapsto Px$. It is a linear transformation that is bijective (since $P$ is invertible). Therefore the columnspaces $\operatorname{Col}(A)$ and $\operatorname{Col}(B)$ have the same dimension, i.e., $\operatorname{rank}(A) = \operatorname{rank}(B)$.

kobe
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  • https://math.stackexchange.com/questions/77650/is-the-rank-of-ab-always-equal-to-the-rank-of-ba –  Oct 01 '17 at 08:06
  • Rank(XY)=Rank(YX), How? –  Oct 01 '17 at 08:07
  • That was an error @N.Maneesh. Thanks for the comment. It's fixed now. I think at the time I thought about trace. Before this question I gave a proof of the result https://math.stackexchange.com/questions/1154998/dim-cab-dim-cb-dim-operatornamenulla-cap-cb which gives a formula for $\operatorname{rank}(AB)$. – kobe Oct 01 '17 at 15:10
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I will show that by premultiplication and postmultiplication by invertible matrices, rank is unchanged.

Let $A=PBQ$ where $P,Q$ are invertible.

$Rank(A)=Rank(PBQ)\leq Rank(PB)\leq Rank(B)= Rank(P^{-1}AQ^{-1})\leq Rank(A)$ so all inequalities must be equalities.

So $Rank(A)=Rank(B)$.

In particular, if $A$ and $B$ are similar, then $Q=P^{-1}$ which is still invertible.

Landon Carter
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If $A$ and $B$ are similar then $A = PBP^{-1}$ for some invertible matrix $P$. Because $P$ is invertible its null space is trivial, i.e. $nullity(P) = \{0\}$.

Now suppose that $Ax = PBP^{-1}x = 0$. Using the result above $PBP^{-1}x = 0$ if and only if $PBx = 0$; Again $PBx = 0$ if $Bx = 0$. The latter result means that $nullity(A) = nullity(B)$. Using rank-nullity-theorem we can conclude that $rank(A) = rank(B)$.

The proof is taken from here.

therealak12
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