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If $A$ is a $m\times n$ type matrix with $m\geq n$ then $$ \mathrm{rank}(A^*A)=\mathrm{rank} (A). $$ Is maybe also true in general that $$ \mathrm{rank}(AA^*)=\mathrm{rank}(A) ? $$ Thanks

Edit.

My question is different from the question about $\mathrm{rank}(A^TA)=\mathrm{rank}(A)$, because this concerns complex conjugate transposed matrix instead of transpose.

user26857
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Richard
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1 Answers1

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if $*$ is the transpose operation, then yes. here is why. we can show that $$Ax = 0 \, \text{iff} \, A^T Ax = 0. \tag 1$$ this is true because $$0 = |Ax|^2 = (Ax)^TAx = x^T(A^T A) x. $$

then $(1)$ implies $$N(A) = N(A^T A)$$ together will the nullity theorem $$\dim N(A) + rank(A) = n$$ should give you $$rank(A) = rank(A^T A). $$

Eugene Zhang
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abel
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