Suppose $$A=\begin{bmatrix}A_1\\A_2\end{bmatrix}$$ where $A_1\in\mathbb{C}^{n\times n}$ is non-singular and $A_2\in \mathbb{C}^{(m-n)\times n}$ is arbitrary.
Prove $\left\Vert A^+\right\Vert_2 \leq \left\Vert A_1^{-1}\right\Vert_2$, where $A^+ = (A^*A)^{-1}A^*$ is the pseudo-inverse of $A$.
(out of Trefethen and Bau's "Numerical Linear Algebra")
A bit of help with this?
Consider a QR factorization of $A$ in the form $$ A=\begin{bmatrix} A_1\\A_2 \end{bmatrix} = \begin{bmatrix} Q_1\\Q_2 \end{bmatrix} R=:QR $$ where $Q$ has orthogonal columns and is partitioned in the same way as $A$. The matrix $A$ has full column rank so $R$ is nonsingular.
We have $A^+=R^{-1}Q^*$ and hence $\|A^+\|_2=\|R^{-1}\|_2$. We have to show that $\|R^{-1}\|_2\leq\|A_1^{-1}\|_2$. From $A_1=Q_1R$, we get $R^{-1}=A_1^{-1}Q_1$ and $$ \|R^{-1}\|_2\leq\|Q_1\|_2\|A_1^{-1}\|_2. $$ But $Q_1$ is a submatrix of $Q$ so $\|Q_1\|_2\leq\|Q\|_2=1$.