Let $\mbox{cond} (M) := \frac{\sigma_1 (M)}{\sigma_n (M)}$ be the condition number of matrix $M$. Is
$$\mbox{cond} ([A,B]) \leq \mbox{cond}(A) + \mbox{cond}(B)$$
true? And is this true for $n \times m$ rectangular matrices? Let's consider $3$ different cases:
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$n<m$
-
$n=m$
-
$n>m$
Best Answer
This could be true if $n\leq m$ at least in the case when all matrices involved have full rank equal to $n$. Because the singular values of $[A,B]$ are the square roots of the eigenvalues of $AA^T+BB^T$, then we have $$ \begin{split} \sigma_n^2([A,B]) &= \lambda_n(AA^T+BB^T) =\min_{\|x\|_2=1}x^T(AA^T+BB^T)x\\&\geq\min_{\|x\|_2=1}x^TAA^Tx=\lambda_n(AA^T)=\sigma_n^2(A) \end{split} $$ Similarly, $$ \begin{split} \sigma_n^2([A,B]) &= \lambda_n(AA^T+BB^T) =\min_{\|x\|_2=1}x^T(AA^T+BB^T)x\\&\geq\min_{\|x\|_2=1}x^TBB^Tx=\lambda_n(BB^T)=\sigma_n^2(B) \end{split} $$ and hence $$ \sigma_n([A,B])\geq\max\{\sigma_n(A),\sigma_n(B)\}. $$ Similarly, variational characterisation of singular values also implies that $$ \sigma_1^2([A,B])\leq\sigma_1^2(A)+\sigma_1^2(B). $$ Indeed, $$ \begin{split} \sigma_1^2([A,B]) &= \lambda_1(AA^T+BB^T) =\max_{\|x\|_2=1}x^T(AA^T+BB^T)x\\&\leq\max_{\|x\|_2=1}x^TAA^Tx+\max_{\|x\|_2=1}x^TBB^Tx=\lambda_1(AA^T)+\lambda_1(BB^T)\\&=\sigma_1^2(A)+\sigma_1^2(B) \end{split} $$ Hence $$ \mathrm{cond}^2([A,B])\leq\frac{\sigma_1^2(A)+\sigma_1^2(B)}{\max\{\sigma_n^2(A),\sigma_n^2(B)\}}\leq\frac{\sigma_1^2(A)}{\sigma_n^2(A)}+\frac{\sigma_1^2(B)}{\sigma_n^2(B)}\leq\mathrm{cond}^2(A)+\mathrm{cond}^2(B) $$ and $$ \mathrm{cond}([A,B])\leq\sqrt{\mathrm{cond}^2(A)+\mathrm{cond}^2(B)} \leq\mathrm{cond}(A)+\mathrm{cond}(B). $$
For $n>m$, it's generally not true (still assuming full rank of both $A$, $B$, and $[A,B]$). As an extreme example, take $m=1$. Then $\mathrm{cond}(A)=\mathrm{cond}(B)=1$ but $\mathrm{cond}([A,B])$ can be arbitrarily large. Consider, e.g., $A=[1,0]^T$ and $B=[1,0.001]^T$.
The rank-deficient case is a bit more difficult if you considered $\mathrm{cond}(A)=\sigma_1(A)/\sigma_r(A)$, where $r$ is the rank of $A$ (this is how the condition number is usually defined when $A$ does not have full rank). The trouble here is not with the upper bound on $\sigma_1([A,B])$ but with the lower bound which does not generally hold. Again, as an extreme example, consider $A$ and $B$ of the case 1) and 2) and augment them with a sufficient number of zero rows to convert it to the case 3) (which we know where the statement is false).