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I'm trying to prove that rank is invariant under similarity. My approach is as follows: suppose I have some matrix $B=P^{-1} AP$. I can simplify this to $PB = AP$. Using the identity

$$rank(AB) \le \min(rank(A), rank(B))$$

we conclude the two following properties:

$$rank(AP) \le \min(rank(A), q)$$

and

$$rank(PB) \le \min(rank(B), q)$$

$AP$ and $PB$ are equivalent, thus we have

$$rank(AP) = rank(PB) \le \min(rank(A),rank(B), q)$$

and we conclude that $rank(A) = rank(B)$.

Is this correct? If not, why? How do we prove that rank is invariant under similarity?

Siong Thye Goh
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  • I would suggest trying to use the fact that similar matrices share eigenvalues (including the geometric and algebraic multiplicities), and looking at the multiplicities of the 0 eigenvalue, which relates to the dimension of the kernel of the matrix (then you can use the rank-nullity theorem). Also, if 0 is not an eigenvalue of either, then both are invertible and they both will have full rank. – Dave Apr 26 '16 at 21:26

1 Answers1

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I agree that $rank(AP)=rank(PB) \leq \min(rank(A),rank(B),q)$ but I don't see how you can conclude that $rank(A)=rank(B)$.

One possible approach is to prove that if a matrix $A$ is of rank $k$, and $Q$ is nonsingular, prove that $QA$ is still of rank $k$.

Similarly, prove that $AQ$ is still of rank $k$. (or you can see this by taking tranpose)

With these results, given a non-singular matrices $Q_1$ and $Q_2$, you can show that $$rank(Q_1AQ_2)=rank(A)$$ which covers the special case of similar matrices.

Siong Thye Goh
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