I have proved $A^{\dagger}A = I$ for a $m$ by $n$ matrix with $m\geq n$ and $\text{rank}(A) = n$ I am trying to find a counter example which shows that $AA^{\dagger} \not= I$ but to no avail. The only one I have is the 0 matrix, but I guess that's trivial
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Any non orthogonal matrix will do. – krirkrirk Mar 12 '15 at 22:11
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Any such matrix with $m>n$ works. – Algebraic Pavel Mar 13 '15 at 11:12
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First of all, what you said is not true, since if $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$$ then $A^\dagger = A^T$, and $$A^\dagger A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ Let $$ A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$$ then $A^\dagger = A^T$ and $A^\dagger A = I$, but $$AA^\dagger = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$
Victor Liu
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In that case, then the second part of my answer with the counterexample still works. – Victor Liu Mar 12 '15 at 22:23
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