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I am facing difficulties in calculating the Moore-Pensore pseudoinverse of a positive semidefinite matrix $A$, where $A$ is self-adjoint and $\langle A u, u \rangle \geq 0$ for all $u \in \mathcal{H}$, where $\mathcal{H}$ is a complex Hilbert space.

For example,

$$A = \begin{bmatrix} 1&-1\\ -1&1\end{bmatrix}$$

is a positive semidefinite matrix. How to calculate the Moore-Penrose pseudoinverse of $A$?

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  • $\begingroup$ If the inverse of $(A^{\top}A)$ exists, the Moore-Penrose pseudo-inverse of $A$ is given by: $$ A^{+} = (A^{\top}A)^{-1}A^{\top}. $$ $\endgroup$ – pitchounet Jul 27 '17 at 9:49
  • $\begingroup$ One way: orthogonally diagonalize, take the reciprocal of the positive eigenvalues. $\endgroup$ – Ian Jul 27 '17 at 9:52
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Computing the singular value decomposition (SVD) of symmetric, rank-$1$ matrix $\rm A$,

$$\mathrm A = \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix} = \begin{bmatrix} 1\\ -1\end{bmatrix} \begin{bmatrix} 1\\ -1\end{bmatrix}^\top = \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix} \right) \begin{bmatrix} 2 & 0\\ 0 & 0\end{bmatrix} \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix} \right)^\top$$

Hence, the pseudoinverse of $\rm A$ is

$$\mathrm A^+ = \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix} \right) \begin{bmatrix} \frac12 & 0\\ 0 & 0\end{bmatrix} \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ -1 & 1\end{bmatrix} \right)^\top = \color{blue}{\frac14 \mathrm A}$$

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  • $\begingroup$ Thank you. But unfortunately I don't understand you method. What mens by the SVD of A?Thank you $\endgroup$ – Student Jul 27 '17 at 10:03
  • $\begingroup$ I edited my answer. Note that the pseudoinverse of a diagonal matrix is the inverse of the nonzero diagonal entries. $\endgroup$ – Rodrigo de Azevedo Jul 27 '17 at 10:07
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Since $A$ is selfadjoint we have

$A^{+}= \lim_{t \to 0}(A^2+tE)^{-1} A$.

In the case of $A=\left(\begin{array}{cc}1&-1\\-1&1\end{array}\right)$ an easy coputation gives $A^{+}= \frac{1}{4} A$.

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  • $\begingroup$ Thank you but I don't understand what is $E$? For $3\times3$- matrix this method remains true?Thank you $\endgroup$ – Student Jul 27 '17 at 9:56
  • $\begingroup$ The formula $A^{+}= \lim_{t \to 0}(A^2+tE)^{-1} A$ is valid for each selfadjoint $ A \in \mathbb C^{n \times n}$. $E$ denotes the identity - matrix. $\endgroup$ – Fred Jul 27 '17 at 10:00
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Given a rank-one matrix of any dimension $$\eqalign{ &A = xy^H \quad\;{\rm where}\; &x\in{\mathbb C}^{m\times 1},\;y\in{\mathbb C}^{n\times 1} }$$ there is a closed-form expression for its Moore-Penrose inverse $$ A^+ = \frac{yx^H}{(x^Hx)(y^Hy)}$$ $A$ is obviously a rank-one matrix, therefore $$\eqalign{ x &= y= z= \left(\begin{array}{rr}1\\-1\end{array}\right), \quad z^Hz = 2 \\ A^+ &= \frac{zz^H}{(2)(2)} = \frac{A}{4} \\ }$$

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