# Rank of product of a matrix and its transpose

How do we prove that

$rank(A) = rank(AA^T) = rank(A^TA)$ ?

Is it always true?

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Is there a reason you have not accepted any answer recently ? this brought down your accept rate and it seems that you are unsatisfied with the answers given to you. –  Belgi Oct 16 '12 at 21:29
You may also be interested in this answer. –  EuYu Oct 16 '12 at 21:48
@Belgi, Thanks for curiosity. Is it a rule to accept one of the answers to all your questions? Does it not close the question? For some of the answer, the correctness of responses is not verifiable for me. But I always increase the vote of responses that increase something to my understanding, after I see them. –  user25004 Oct 16 '12 at 23:11

It is always true. One of the important theorems one learns in linear algebra is that $$\mathrm{Nul}(A^T)^{\perp} = \mathrm{Col}(A), \quad \mathrm{Nul}(A)^{\perp} = \mathrm{Col}(A^T).$$

Therefore $\mathrm{Nul}(A^T) \cap \mathrm{Col}(A) = \{0\}$, and so forth. Now consider the matrix $A^TA$. Then $\mathrm{Col}(A^TA) = \{A^TAx\} = \{A^Ty: y \in \mathrm{Col}(A)\}$. But since the null space of $A^T$ only intersects trivially with $\mathrm{Col}(A)$, then $\mathrm{Col}(A^TA)$ must have the same dimension as $\mathrm{Col}(A)$, which gives us the equality of ranks.

We can replace $A$ with $A^T$ to prove the other equality.

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This is only true for real matrices. For instance $\begin{bmatrix} 1 & i \\ 0 &0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ i &0 \end{bmatrix}$ has rank zero. For complex matrices, you'll need to take the conjugate transpose.