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 $\sigma v = A^{T}u$.

  • $\begingroup$ 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. $\endgroup$ Commented Dec 10, 2019 at 13:31
  • $\begingroup$ For (ii), it is not clear what you're asking for $\endgroup$ Commented Dec 10, 2019 at 13:39
  • $\begingroup$ @GiuseppeNegro from (iii) it appears that this is the induced $2$-norm, i.e. the "spectral norm". $\endgroup$ Commented Dec 10, 2019 at 13:40
  • $\begingroup$ @GiuseppeNegro A is a mapping from and into Euclidean spaces, and the norm is the Euclidean norm. I will edit the question. $\endgroup$
    – F McA
    Commented Dec 10, 2019 at 13:42
  • $\begingroup$ @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? $\endgroup$
    – F McA
    Commented Dec 10, 2019 at 13:44

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .