# Matrix norms and the matrix transpose

There are three parts to this question, and I'm not sure how they link together to provide answers.

$$A$$ is a linear mapping from Euclidean space $$X$$ to Euclidean space $$U$$, and the norm $$\| \cdot \|$$ is the Euclidean norm for matrices.

(i) Show that $$\|A^{T}\| = \|A\|$$.

(ii) Let $$v \in \mathbb{R}^{n}$$ be a unit vector, and $$\sigma u = Av$$, with $$\sigma = \|Av\|$$. Therefore, $$u\in \mathbb{R}^{m}$$ is also a unit vector. Does it follow that $$\sigma v = A^{T}u$$?

(iii) Now if $$v$$ is as above, but $$\sigma = \|A\|$$. Show that $$\sigma v = A^{T}u$$.

• First thing to do is to provide the definition of $\|\cdot\|$. There are many matrix norms and you did not specify which one you refer to. Commented Dec 10, 2019 at 13:31
• For (ii), it is not clear what you're asking for Commented Dec 10, 2019 at 13:39
• @GiuseppeNegro from (iii) it appears that this is the induced $2$-norm, i.e. the "spectral norm". Commented Dec 10, 2019 at 13:40
• @GiuseppeNegro A is a mapping from and into Euclidean spaces, and the norm is the Euclidean norm. I will edit the question. Commented Dec 10, 2019 at 13:42
• @Omnomnomnom I suppose that the question for (ii) is simply - is it true that $\sigma v = A^{T}u$, based on the specification of the vectors? Commented Dec 10, 2019 at 13:44

Your intuitive idea for (i) is nice, but difficult to implement since $$\|A\|$$ cannot be nicely written as a function of the entries of $$A$$. One approach to (i) is as follows. Note that $$\|A\|^2 = \max_{\|x\| = 1} \|Ax\|^2 = \max_{\|x\| = 1} (Ax)^T(Ax) = \max_{\|x\| = 1}x^T(A^TA)x.$$ By the Rayleigh-Ritz theorem, this is simply the largest eigenvalue of $$A^TA$$, and the vector at which we attain this maximum is the corresponding eigenvector. Similarly, $$\|A^T\|^2$$ must be the largest eigenvalue of $$AA^T$$. Since $$A^TA$$ and $$AA^T$$ have the same non-zero eigenvalues (this can be proven in several ways), we conclude that $$\|A\| = \|A^T\|$$.
(ii): This will not hold in general. For instance, consider $$A = \pmatrix{1&1\\0&1}, \quad u = v = \pmatrix{1\\0}, \quad \sigma = 1.$$ We indeed have $$\sigma u = Av$$, but we do not have $$A^Tu = \sigma v$$.
(iii): The key is to observe that in the case that by the Rayleigh Ritz theorem (as discussed in the first part of this answer), $$\|Av\| = \|A\|$$ implies that $$v$$ is an eigenvector of $$A^TA$$ with $$A^TAv = \|A\|^2v$$. Thus, with $$\sigma = \|A\|$$ and the definitions from (ii), we have the following.
If $$\sigma = \|A\| = 0$$, then it follows that $$A = 0$$ and the result follows trivially. In the case where $$\sigma \neq 0$$, we have $$A^Tu = A^T \left(\frac {Av}{\sigma}\right) = \frac 1{\sigma} A^TA v = \frac 1{\sigma} \sigma^2 v = \sigma v$$ as was desired.