# Minimum singular value of composite matrix

Let $$A$$ be a $$m\times n_1$$ matrix, $$B$$ a $$m\times n_2$$ matrix and $$C=[A B]$$ the $$m\times(n_1+n_2)$$ matrix obtained by concatenating the columns of $$A$$ and $$B$$.

I have observed numerically for thousands of examples (taking $$A$$ and $$B$$ to be Gaussian random matrices) that, if $$m> (n_1+n_2)$$, then $$\sigma_1(C)<{\rm min}(\sigma_1(A),\sigma_1(B))$$ where $$\sigma_1$$ is the smallest singular value.

Can this be proved?

EDIT: I have found the paper "Principal Submatrices IX: Interlacing Inequalities for Singular Values of Submatrices", by R.C.Thompson, which states the following theorem:

Let $$C$$ be an $$m \times n$$ matrix with singular values $$\alpha_1\ge \alpha_2\ge...$$. Let $$B$$ be a $$p \times q$$ submatrix of $$C$$, with singular values $$\beta_1\ge \beta_2\ge...$$. Then $$\beta_i\ge \alpha_{i+(m-p)+(n-q)}$$ for $$i\leq {\rm min}(p+q-m,p+q-n)$$.

If I take $$p=m$$ and $$i=q$$, I have $$\beta_q\geq \alpha_n$$ which means the smallest singular value of $$C$$ is smaller than the smallest singular value of $$B$$, which is what I want.

However, I have only verified this in numerical simulations when $$m\gg n$$. Why?

• But the theorem should hold both for $A$ and for $B$... – thedude Nov 19 '19 at 18:31

You need $$i\le\min(p+q-m,p+q-n)$$. When $$p=m$$, this means $$i\le\min(q,\,m+q-n)$$. You cannot put $$i=q$$ unless $$q\le m+q-n$$ (i.e. unless $$n\le m$$).
• You are right, with the observation that, in the notation of the original question, it should be $n_1+n_2<m$. – thedude Nov 19 '19 at 23:32