Showing that $c:=\min\{\frac{\|\alpha(v)\|}{\|v\|}:0\neq v\in Ker(\alpha)^{\perp}\}$ is a singular value of $\alpha$

Question

Let $V$ and $W$ be finite-dimensional inner product spaces, and let $\alpha:V\to W$ be a nonzero linear transformation. show that

$$c:=\min\{\frac{\|\alpha(v)\|}{\|v\|}:0\neq v\in Ker(\alpha)^{\perp}\}$$

is a nonzero singular value of $\alpha$.

I'm stuck on this problem. I appreciate any help.

If $0\in spec(\alpha^*\alpha)$, there exists a nonzero eigenvector associated with $0$ is in $Ker(\alpha)$. As, $0\neq v\in \ker(\alpha)^{\perp}$, $\alpha(v)\neq 0$. Hence, $c>0.$ Since $\alpha^*\alpha$ is positive definite, it's normal and then it's orthogonally diagonalizable. It means there exists a orthonormal basis $B=\{v_1,\dots,v_n\}$ for $V$ composed of eigenvectors of $\alpha^*\alpha$. Let $v\in ker(\alpha)^{\perp}.$ there are scalars $a_1,\dots,a_n$ satisfying $v=\sum^{n}_{i=1}a_iv_i$. Suppose $\{c_1,\dots, c_n\}$ are eigenvalues of $\alpha^*\alpha$ associated with $v_1,\dots, v_n$. If $c_i\neq 0$ for $i\in\{1,\dots,n\}$, the associated eigenvector $v_i\in \ker(\alpha)^{\perp}$. Renumbering $0\leq c_1\leq c_2\leq \dots\leq c_n$, we get $\|\alpha(v)\|= \sum^{n}_{i=1}|a_i|^2c_i$.

There are two cases:

1) If $0\notin spec(\alpha^*\alpha)$, we get $\|\alpha(v)\|= \sum^{n}_{i=1}|a_i|^2c_i\geq\sum^{n}_{i=1}|a_i|^2c_1=c_1\|v\|$.

2) If $0\in spec(\alpha^*\alpha)$, we get $\|\alpha(v)\|= \sum^{n}_{i=1}|a_i|^2c_i\geq\sum^{n}_{i=1}|a_i|^2c_j=c_j\|v\|$, where $c_j$ is the smallest nonzero eigenvalue of $\alpha^*\alpha$, for some $j$.

In either cases, we get $\min\{\frac{\|\alpha(v)\|}{\|v\|}:0\neq v\in Ker(\alpha)^{\perp}\}\geq c_j$, where $c_j$ is the smallest nonzero eigenvalue of $\alpha^*\alpha$, for some $j$.

If $v_j$ is an eigenvector associated with the smallest nonzero eigenvalue $c_j$ for some $j$, we get $v_j\in Ker(\alpha)^{\perp}$. Hence $c_j\geq \min\{\frac{\|\alpha(v)\|}{\|v\|}:0\neq v\in Ker(\alpha)^{\perp}\} $.

Answer 1

Let $(s_n,e_n,f_n)$ be the singular system of $\alpha$ with $s_n\ne0$, so we ignore zero singular values. The vectors $(e_n)$ and $(f_n)$ are bases of $\ker(\alpha)^\perp$ and $im(\alpha)$, respectively. Using the singular decomposition we get $$ \alpha(v) = \sum_n s_n \langle v,e_n\rangle f_n. $$ The vectors $(e_n)$ form on orthonormal basis of $\ker(\alpha)^\perp$. Hence it holds $v=\sum_n \langle v,e_n\rangle e_n$ for all $v\in\ker(\alpha)^\perp$. This implies $$ \frac{\|\alpha(v)\|^2}{\|v\|^2} = \frac{\sum_n s_n^2 |\langle v,e_n\rangle|^2 }{\sum_n |\langle v,e_n\rangle|^2 } \ge \min_n s_n^2. $$ Taking $v=e_j$ with $s_j = \min_n s_n$, i.e., singular vector associated to smallest singular value, shows equality.