How is the spectral norm of a row matrix calculated?

Question

If $x \in \mathbb{R}^{1\times n}$ is a row matrix, is it true that $\| x \|_2 = \| x^T \|_2 = \sqrt{x x^T}$?

Note that $\| x \|_2$ is the induced matrix norm and $\| x^T \|_2$ is the vector $2$-norm.

Answer 1

Hint: See that when you multiply a row vector with a column vector it is a number in $\Bbb R$.

Answer 2

Yes, it's true. For clarity, let us denote the Euclidean norm of a row or column vector by $\|\cdot\|_E$. Pick any orthogonal matrix $V$ whose first column is $\frac{1}{\|x^T\|_E}x^T$. Then $$ x=1\pmatrix{\|x^T\|_E&0&\ldots&0}V^T $$ is a singular value decomposition of $x$ and it follows that the induced $2$-norm of $x$ is $\|x^T\|_E$.

Alternatively, since $$ \|x\|_2 =\max_{\substack{\|v^T\|_E=1\\ v^T\in\mathbb R^{n\times1}}}\|xv^T\|_E =\max_{\substack{\|v^T\|_E=1\\ v^T\in\mathbb R^{n\times1}}}\langle x^T,v^T\rangle \le\max_{\substack{\|v^T\|_E=1\\ v^T\in\mathbb R^{n\times1}}}\|x^T\|_E\|v^T\|_E =\|x^T\|_E $$ and equality holds when $v$ is a unit vector pointing in the same direction of $x$, we conclude that $\|x\|_2=\|x^T\|_E$.