How can I show that $(A^TA)^{-1} A^T$ minimizes $\mathbf{Tr}(X^TX)$ over all matrices $X$ such that $XA = I$?

[Note that $A$ is an $m \times n$ matrix.]

I've tried rearranging the trace, and expanding using eigenvalues, and haven't had any luck yet!


We have the following least-norm problem

$$\begin{array}{ll} \text{minimize} & \mathrm \| \mathrm X \|_{\text F}^2\\ \text{subject to} & \mathrm X \mathrm A = \mathrm I\end{array}$$

where $\mathrm A \in \mathbb R^{m \times n}$ is given. Let the Lagrangian be

$$\mathcal L (\mathrm X, \Lambda) := \frac 12 \mathrm \| \mathrm X \|_{\text F}^2 + \langle \Lambda, \mathrm X \mathrm A - \mathrm I \rangle$$

Taking the derivatives of the Lagrangian with respect to $\rm X$ and $\Lambda$ and finding where they vanish, we obtain a system of coupled matrix equations

$$\begin{aligned} \mathrm X + \Lambda \mathrm A^\top &= \mathrm O\\ \mathrm X \mathrm A &= \,\mathrm I\end{aligned}$$

Right-multiplying the first matrix equation by $\rm A$, we obtain $\mathrm I + \Lambda \mathrm A^\top \mathrm A = \mathrm O$. Assuming that $\rm A$ has full column rank, then $\mathrm A^\top \mathrm A$ is invertible and, thus, $\Lambda = - ( \mathrm A^\top \mathrm A )^{-1}$ and $\mathrm X_{\text{LN}} := \color{blue}{( \mathrm A^\top \mathrm A )^{-1} \mathrm A^\top}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.