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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 $m \geq n$.

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2 Answers 2

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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}

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  • $\begingroup$ Unfortunately, it is not obvious for me why the largest singular value of $A$ or $A^T$ equals one? $\endgroup$
    – user355419
    Commented Jul 7, 2019 at 8:21
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    $\begingroup$ @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 $\|A\|_2=1$. $\endgroup$
    – user1551
    Commented Jul 7, 2019 at 12:31
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    $\begingroup$ There is a small error in my previous comment. The singular values of $A$ should be the nonnegative square roots of the eigenvalues of $A^TA$. $\endgroup$
    – user1551
    Commented Jul 9, 2019 at 9:55
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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 $\|A\|\leq\|B\|=1$.

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  • $\begingroup$ Thanks for your answer. Is there any general result such as $\| A \|_2 = 1$? If so, how do you prove that? $\endgroup$
    – user355419
    Commented Jul 6, 2019 at 13:05

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