# Operator norm of rectangular $(m \times n)$-matrix with $m < n$ orthonormal rows?

Let $$m \leq n$$ and let $$A \in \mathbb{R}^{m \times n}$$ be a matrix with $$m$$ orthonormal rows. Apparently, it can happen that $$A^T A \neq I$$. Is there any general result such as $$\|A\|_2 = 1$$? If so, how do you prove it?

PS: Note that in this question the role of $$m$$ and $$n$$ is interchanged, that is $$m \geq n$$.

This is obvious if you know that $$A$$ and $$A^T$$ have the same set of nonzero singular values and $$\|A\|_2$$ is just the largest singular value of $$A$$. Alternatively, if $$Q$$ is any orthogonal matrix whose first $$m$$ rows are identical to $$A$$'s, we have $$\pmatrix{A\\ 0_{(n-m)\times n}}=\pmatrix{I_m\\ &0_{(n-m)\times(n-m)}}Q.$$ It follows that \begin{aligned} \|Ax\|_2 =\left\|\pmatrix{Ax\\ 0}\right\|_2 =\left\|\pmatrix{A\\ 0}x\right\|_2 =\left\|\pmatrix{I\\ &0}Qx\right\|_2 =\left\|\pmatrix{I\\ &0}u\right\|_2 \end{aligned} where $$u=Qx\in\mathbb R^n$$. Consequently, \begin{aligned} \|A\|_2 =\max_{\|x\|_2=1}\|Ax\|_2 =\max_{\|u\|_2=1}\left\|\pmatrix{I\\ &0}u\right\|_2 =1. \end{aligned}

• Unfortunately, it is not obvious for me why the largest singular value of $A$ or $A^T$ equals one? Commented Jul 7, 2019 at 8:21
• @user355419 The singular values of $A$ are the eigenvalues of $A^TA$. But $A^TA$ and $AA^T=I$ share the same set of nonzero eigenvalues. Therefore the singular values of $A$ are $1$ and $0$. Hence $\|A\|_2=1$. Alternatively, since $A=I_m\pmatrix{I_m&0}Q$ is a singular value decomposition of $A$, we have $\|A\|_2=1$. Commented Jul 7, 2019 at 12:31
• There is a small error in my previous comment. The singular values of $A$ should be the nonnegative square roots of the eigenvalues of $A^TA$. Commented Jul 9, 2019 at 9:55

In terms of computation: simply add extra zero-filled rows to A to make it square and compute the usual induced norm for a $$n\times n$$ matrix.

In terms of definition: it is still the maximal value of $$\|Ax\|_2$$ for an unit norm $$x$$, that is $$\|x\|_2$$.

Going further, let $$x$$ be the first row of $$A$$. Then $$Ax^T=(\|x\|_2,0,\ldots,0)$$, thus by definition $$\|A\|\geq \|Ax^T\|_2 = 1$$. On the other hand, let $$B$$ be the $$n\times n$$ matrix obtained by completing $$A$$ with $$n-m$$ orthonormal rows. Then $$\|A\|\leq\|B\|=1$$.

• Thanks for your answer. Is there any general result such as $\| A \|_2 = 1$? If so, how do you prove that? Commented Jul 6, 2019 at 13:05