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Let $A$ be a real $4 \times4$ matrix with rank($A$)= 2 and two columns of zeros as follows

$$ A = \begin{bmatrix}a_1&0&b_1&0\\a_2&0&b_2&0\\a_3&0&b_3&0\\a_4&0&b_4&0\\\end{bmatrix} $$

. Let the Moore-Penrose pseudoinverse of $A$ be $A^{\dagger}$. I would like to know why always

$$ A^{\dagger}A = \begin{bmatrix}1&&&\\&0&&\\&&1&\\&&&0\end{bmatrix}. $$

The https://www.quora.com/When-is-A-+-A-I-i-e-when-does-the-pseudo-inverse-yield-the-identity-matrix is very helpful but not enough.

I would be appreciated for any help.

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  • $\begingroup$ $A^\dagger A$ is the projection operator onto the column space of $A^\top$. So you are correct, $A^\dagger A = I_2$ iff rank of $A$ is 2. $\endgroup$ Jan 24 '20 at 21:39
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The linear map $x^T\mapsto x^TA^\dagger A$ is the orthogonal projection onto the row space of $A$. Since $A$ has rank $2$ and every row of $A$ lies inside $\operatorname{span}\{e_1^T,e_3^T\}$, its row space is precisely $\operatorname{span}\{e_1^T,e_3^T\}$. Therefore $A^\dagger A$ maps $e_1^T$ and $e_3^T$ to themselves and $e_2^T,e_4^T$ to zero. Hence it is equal to $\operatorname{diag}(1,0,1,0)$.

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