# $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$

In this post norm denotes a matrix norm, i.e. it is sub-multiplicative. All matrices are real. $A$ is of size $n \times k$ with independent columns ($k \leq n$). $B$ is of size $k \times n$.

Let $\| \cdot \|_\square$ be some arbitrary matrix norm. Assume $\| AB\|_\square \leq 1$. Then we know $\rho(AB) \leq 1$. Then we know from this post that $\rho(BA) \leq 1$.

Edit: thanks to @loupblanc I realized this was incorrect. From this we know that there must be a matrix norm $\| \cdot \|_\triangle$ such that $\| BA\|_\triangle \leq 1$. . In fact we only know that there is a norm which is arbitrarily close to the spectral radius.

I am interested in whether there is an easy way of defining $\| \cdot \|_\triangle$ from $\| \cdot \|_\square$. For example, is it true that $\| \cdot \|_\triangle = \|F (\cdot) Q\|_\square$ for some choice of $F$ and $Q$?

Edit: when I say the norm $\| \cdot \|_\square$ is sub-multiplicative I mean it is sub-multiplicative in the space of $n \times n$ matrices. I do not mean $\| A B \|_\square \leq \|A\|_\square \| B \|_\square$, which indeed doesn't make sense. What I mean that if, say, one decomposed $AB$ into a product of square matrices, i.e. $AB = C_1C_2$, where $C_1,C_2$ are square, then we would have $\| A B\|_\square = \| C_1 C_2 \|_\square \leq \| C_1 \|_\square \| C_2 \|_\square$.

Submultiplicative norms for non-square matrices obviously do not make sense, but with square matrices, I don't think what you originally wrote was wrong. More specifically, we have the following

Proposition. Suppose $A,B\in M_n(\mathbb C)$. If $\rho(AB)=\|AB\|_\square\le1$ for some submultiplicative norm $\|\cdot\|_\square$, then there exists some submultiplicative matrix norm $\|\cdot\|_\triangle$ such that $\|BA\|_\triangle\le1$.

If $\rho(AB)<1$, the proposition is obvious because $\rho(AB)=\rho(BA)$ and the spectral radius is the infinum of all submultiplicative matrix norms. So, it suffices to consider only the case $\rho(AB)=\rho(BA)=\|AB\|_\square=1$, and we need to find some norm such that $\|BA\|_\triangle=1$.

As $\rho(AB)=1$, we have $1 = \left(\rho(AB)\right)^k = \rho((AB)^k) \le \|(AB)^k\|_\square \le \|AB\|_\square^k = 1$ and in turn $\|(AB)^k\|_\square=1$ for all natural number $k$. Hence $AB$ is power bounded. As $(BA)^{k+1}=B(AB)^kA$, the product $BA$ and in turn its Jordan form are power bounded too. Since all matrix norms (submultiplicative or not) on $M_n(\mathbb C)$ are equivalent, the Frobenius norms of the powers of the Jordan form of $BA$ are uniformly bounded. It follows that for each unit eigenvalue of $BA$, its algebraic and geometric multiplicities coincide. That is, in the Jordan decomposition $BA=PJP^{-1}$, we may assume that $$J=\pmatrix{T&0\\ 0&D},$$ where $T$ is a direct sum of (upper triangular) Jordan blocks for eigenvalues whose magnitudes are strictly smaller than $1$, and $D$ is a diagonal matrix whose diagonal entries have unit moduli.

Now suppose the size of $T$ is $m\times m$. Let $\epsilon>0$ and $\Lambda=\operatorname{diag}(1,\epsilon,\epsilon^2,\ldots,\epsilon^{m-1})\oplus I_{n-m}$. When $\epsilon$ is sufficiently small, the superdiagonal of $\Lambda^{-1}J\Lambda$ can be made arbitrarily small and the diagonal entries remain invariant. Therefore the spectral norm $\|\Lambda^{-1}J\Lambda\|_2=\sigma(\Lambda^{-1}J\Lambda)$ is equal to $1$. Consequently, if we define $$\|X\|_\triangle = \|\Lambda^{-1}P^{-1}XP\Lambda\|_2,$$ we get $\|BA\|_\triangle=1$.

• @ user1551 , pretty proof. In fact you show that: there is a sub-mult. norm $||.||$ s.t. $\rho(U)=||U||$ iff the eigenvalues of $U$ of maximal modulus are semi-simple. – user91684 Dec 7 '14 at 23:26
• @loupblanc Thanks. Initially I thought this is a known result, but I couldn't find it anywhere in the books offhand. Google showed nothing either. The closest thing I can find is the characterisation of radial matrices. Do you know if this is a known result? – user1551 Dec 7 '14 at 23:46
• @ user1551 , I did not know this equivalence. Householder speaks about this result in one of his paper; you can read for free the first 2 pages as follows: with Google, do "S. Householder, Minimal matrix norms" and click on the Springer link. Note that it is a reference cited by Goldberg-Zwaz. – user91684 Dec 8 '14 at 11:04
• @loupblanc Thanks a lot. Minimal norms (for fixed matrices!) ... of course ... that makes much sense. I have looked for useful results about minimal norms before answering the question, but quickly abandoned the search because such norms (the usual ones, not Householder's) are minimal for all matrices. – user1551 Dec 8 '14 at 11:37
• @user1551: The book Numerical Analysis by Ortega has a proof of the same. In it such matrices are said to be " of class M". – me10240 Mar 3 '16 at 4:39

@ ziutek , a matrix norm $||.||$ is defined on a space $M_{n,m}$. If $||.||$ is sub-multiplicative, then necessarily $m=n$ and $A$ cannot be rectangular. Moreover there is a big mistake in your line 5. Indeed if $\rho(BA)\leq 1$, then, we have only the following: for every $\epsilon>0$, there is an induced norm $N()$ s.t. $N(BA)\leq \rho(BA)+\epsilon$.

Assume for example that $B$ is invertible and let $N(U)=||B^{-1}UB||$. Then $N()$ is sub-multiplicative norm (easy). Moreover $N(BA)=||AB||=1$. It remains the case when $A,B$ are non-invertible.

• Thanks for the useful answer. I clarified the question a little bit and stated explicitly what I mean by sub-multiplicative. – ziutek Dec 7 '14 at 10:44