# rank of matrix $A$ and result of Moore-Penrose pseudoinverse $A$ times $A$

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.

• $A^\dagger A$ 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. Jan 24, 2020 at 21:39

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 $$\operatorname{diag}(1,0,1,0)$$.