# Moore Condition for m x n matrix

I'm not sure how to prove any of this. The only thing that seems to be related to this is: How does the SVD solve the least squares problem? but it does not answer any of the questions below

3.32. Prove that the pseudoinverse $$\boldsymbol{A}^+$$ of an $$m\times n$$ matrix $$\boldsymbol A$$, as defined using the SVD in Section 3.6.1, satisfies the following four properties, known as the Moore-Penrose conditions.

(a) $$\boldsymbol{A}\boldsymbol{A}^+\boldsymbol{A}=\boldsymbol{A}$$.

(b) $$\boldsymbol{A}^+\boldsymbol{A}\boldsymbol{A}^+=\boldsymbol{A}^+$$.

(c) $$(\boldsymbol{A}\boldsymbol{A}^+)^T=\boldsymbol{A}\boldsymbol{A}^+$$.

(d) $$(\boldsymbol{A}^+\boldsymbol{A})^T=\boldsymbol{A}^+\boldsymbol{A}$$.

• A little bit difficult to answer without having section 3.6.1. I suppose $A^+$ and $A$ are defined in terms of the matrices commonly called $U,\Sigma,V$ in the SVD and that you are supposed to substitute in left and right hand sides and see if you can get them to be equal. – mathreadler Mar 31 '17 at 2:18
• Please try and use less images of text in questions and more typeset content even if there probably was a good thought behind this time. – mathreadler Mar 31 '17 at 2:20
• Please consider updating the title to use 'Moore-Penrose conditions'. This helps others to find your question. – dantopa Mar 31 '17 at 3:30

$$\mathbf{A} = \mathbf{U} \, \Sigma \, \mathbf{V}^{*}, \qquad \mathbf{A}^{+} = \mathbf{U} \, \Sigma^{+} \, \mathbf{V}^{*}$$

$\Sigma$ gymnastics

The problem distills down to an exercise in $\Sigma$ gymnastics. The diagonal matrix $\mathbf{S}$ contains the ordered singular values has size $\rho \times \rho.$ Note $\mathbf{S}^{T} = \mathbf{S}.$ The inverse matrix has reciprocal values on the diagonal.

First step: \begin{align} \Sigma = \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{m\times n}, \qquad % \Sigma^{T} = \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{n\times m}, \qquad % \Sigma^{+} = \left[ \begin{array}{cc} \mathbf{S}^{-1} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{n\times m} % \end{align} Second step: \begin{align} \Sigma \, \Sigma^{T} = \left[ \begin{array}{cc} \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{m\times m}, \qquad % \Sigma^{T} \, \Sigma &= \left[ \begin{array}{cc} \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{n\times n} % \end{align} Third step: \begin{align} \Sigma \, \Sigma^{+} = \left[ \begin{array}{cc} \mathbf{I}_{\rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{m\times m}, \qquad % \Sigma^{+} \, \Sigma &= \left[ \begin{array}{cc} \mathbf{I}_{\rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{n\times n} % \end{align} Fourth step: \begin{align} \Sigma \, \Sigma^{+} \, \Sigma &= \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{m\times n} = \Sigma, \qquad % \Sigma^{+} \, \Sigma \, \Sigma^{+} = \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]_{n\times m} = \Sigma^{+} % \end{align}

a

\begin{align} \mathbf{A} \, \mathbf{A}^{+} \, \mathbf{A} &= % A \left( \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \right) % A \left( \mathbf{V} \, \Sigma^{+} \, \mathbf{U}^{*} \right) % A+ \left( \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \right) \\ % & = \mathbf{U} \, \Sigma \, \Sigma^{+} \, \Sigma \, \mathbf{V}^{*} \\ % & = \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % & = \mathbf{A} % \end{align}

b

\begin{align} \mathbf{A}^{+} \mathbf{A} \mathbf{A}^{+} & = \mathbf{V} \, \Sigma^{+} \, \Sigma \, \Sigma^{+} \, \mathbf{U}^{*} \\ & = \mathbf{V} \, \Sigma^{+} \, \mathbf{U}^{*} \\ % & = \mathbf{A}^{+} % \end{align}

c

\begin{align} \left( \mathbf{A} \, \mathbf{A}^{+} \right)^{*} &= % A \left( \mathbf{U} \, \Sigma \, \Sigma^{+} \, \mathbf{U}^{*} \right)^{*} \\ % A &= \left( \mathbf{U} \, \left[ \begin{array}{cc} \mathbf{I}_{\rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] \, \mathbf{U}^{*} \right)^{*} \\ &= \mathbf{U} \, \left[ \begin{array}{cc} \mathbf{I}_{\rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] \, \mathbf{U}^{*} \\ &= \mathbf{A} \, \mathbf{A}^{\dagger} % \end{align}

d

\begin{align} \left( \mathbf{A}^{+} \, \mathbf{A} \right)^{*} &= % A \left( \mathbf{V} \, \Sigma^{+} \, \Sigma \, \mathbf{V}^{*} \right)^{*} \\ % A &= \mathbf{V} \, \left[ \begin{array}{cc} \mathbf{I}_{\rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] \, \mathbf{V}^{*} \\ &= \mathbf{A}^{+} \, \mathbf{A} % \end{align}

I'm going to assume that the SVD definition of the pseudo-inverse is $$A^+ = V\Sigma^+U^*$$ where the SVD of $A$ is $A=U\Sigma V^*$ and $\Sigma^+$ is a diagonal matrix with diagonal entries $$[\Sigma^+]_{ii}=\left\{ \begin{array} 00 & \text{if} [\Sigma]_{ii}=0\\ ([\Sigma]_{ii})^{-1}&\text{otherwise}\end{array}\right.$$

It's then just a matter of plugging in the SVD of $A$ and $A^{+}$ into the equations (a)-(d). For example for (a) you have $$AA^{+}A=(U\Sigma V^*)(V\Sigma^{+}U^*)(U\Sigma V^*)=U\Sigma\Sigma^+\Sigma V^*=U\Sigma V^*=A$$ 

• How would the transpose be used in this? – steve Mar 31 '17 at 2:39
• According to wikipedia the transposes should be conjugate transposes (Hermitian transpose), but when the matrices are real these can be changed to transposes. – Peder Mar 31 '17 at 2:53