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I have been wondering if the following statement is corrrect?

Assume that $A$ is $n \times m$ matrix where $m \leq n$ and rank of $A$ is $m$, $X$ and $B$ are $m \times m$ square matrix and it is known that $B$ is full rank, i.e, rank of $B$ is $m$. If, $$A X A^T = A B A^T$$ then we can say that $X = B$. If $m > n$, we can not reach to the same conclusion and $X \neq B$ anymore.

My proof is based on multipliying the expression from left by $A^T$ and observing $A^TA$ invertible and then multiplying from the right by $A$ concluding that $X$ should be equal to $B$.

My reasoning is correct? Am I missing something? Thanks in advance for your help.

Rodrigo de Azevedo
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Osman Coskun
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1 Answers1

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Yes, your reasoning looks correct, assuming your matrices are real. Following the argument in this or this explanation, we have $\text{rank}(A^TA) = \text{rank}(A)$.

Therefore if $\text{rank}(A) = m$, we have $\text{rank}(A^TA) = \text{rank}(A) = m$, which means that $A^TA$ is invertible, because it is an $m \times m$ matrix with rank $m$.

As you noted, this argument does not hold if $m > n$, because in this case $A$ cannot have rank $m$, because $\text{rank}(A) \leq \min(n, m)$ for any $n \times m$ matrix $A$, and neither can $A^TA$. Therefore $A^TA$ is an $m \times m$ matrix with rank less than $m$, so it is not invertible, and this argument does not hold.

Stratis Markou
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