Verify that if $A\in\mathbb R^{m\times n}_n$, then $A^+ = R^{-1}Q_1^T$.

least squareslinear algebraorthogonal matricespseudoinversesvd

Given $A\in\mathbb R ^{m\times n}_n,Q^T\in\mathbb R^{ m\times m} , Q=\left[\begin{matrix} Q_1 & Q_2 \end{matrix} \right]$ orthogonal, $ R\in\mathbb R^{n\times n}_n$ upper triangular. Householder transformations can be done to reduce $A$ into
$$\begin{align}
Q^TA&=\left[\begin{matrix}R\\0\end{matrix}\right] \\
A &= Q\left[\begin{matrix}R\\0\end{matrix}\right] \\
&=\left[\begin{matrix}Q_1 & Q_2\end{matrix}\right]\left[\begin{matrix}R \\ 0\end{matrix}\right]\\
(1)\quad &= Q_1R
\end{align}$$
Question: Use the for Penrose conditions and the fact that $Q_1$ has orthonormal columns to verify that if $A\in\mathbb R ^{m\times n}_n$ can be factored into the form (1), then $A^+ = R^{-1}Q_1^T$.

I am assuming that the questions is asking me to verify that if $A^+$ satisfies the four Penrose conditions of a pseudoinverse then, indeed $A^+ = R^{-1}Q_1^T$ is indeed the pseudoinverse of $A$.

Since $A$ is full column rank, let the compact SVD of $A=U_1 SV_1^T$. Then we have the four Penrose conditions:

  1. $(U_1 SV_1^T)(R^{-1}Q_1^T)(U_1 SV_1^T)$
  2. $(R^{-1}Q_1^T)(U_1 SV_1^T)(R^{-1}Q_1^T)$
  3. $\big((R^{-1}Q_1^T)(U_1 SV_1^T)\big)^T =(U_1 SV_1^T)^T(R^{-1}Q_1^T)^T = V_1SU_1^TQ_1(R^{-1})^T$
  4. $\big((U_1 SV_1^T)(R^{-1}Q_1^T)\big)^T=(R^{-1}Q_1^T)^T(U_1 SV_1^T)^T = Q_1(R^{-1})^TV_1SU_1^T$

Is this what the questions is asking me to confirm? If so, I am not sure how to evaluate these expressions to satisfy the Penrose conditions. Help is appreciated!

Best Answer

The first of the four Penrose conditions is actually $AA^+A=A$.
See Moore–Penrose inverse.

If we find a matrix that satisfies all 4 conditions, then that matrix is the pseudo inverse, which is guaranteed to exist, and is guaranteed to be unique.

So let's check. The hypothesis is that the pseudo inverse of $Q_1R$ is $R^{-1}Q_1^T$: $$AA^+A = (Q_1R)(R^{-1}Q_1^T)(Q_1R)=Q_1 (RR^{-1}) (Q_1^T Q_1) R = Q_1 R = A$$ So the first condition checks out.

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