# Inequality on singular values of two matrices

Let $\Sigma_1=\begin{pmatrix} \sigma_1^1 & 0\\0 &\sigma_2^1 \end{pmatrix}$ and $A=R^{-1}\Sigma_2R$, $\Sigma_2=\begin{pmatrix} \sigma_1^2 & 0\\0 &\sigma_2^2 \end{pmatrix}$, $R=\begin{pmatrix} \cos\theta & \sin\theta\\-\sin\theta &\cos\theta \end{pmatrix}$.

Suppose $\sigma_1^1\ge \sigma_2^1,\sigma_1^2,\sigma_2^2\ge 0$ and fix any $\theta$.

This is my question:

Is it true that, for every $v_1=(x_1,y_1),v_2=(x_2,y_2)\in \mathbb{R}^2$ such that all four coordinates $x_1,x_2,y_1,y_2$ are $\ge 0$, the following inequality is satisfied?

$$\frac{|\Sigma_1v_1+Av_2|}{|v_1+v_2|}\le \sigma_1^1$$

Edit: I'm quite sure that this inequality wouldn't be satisfied in case of two generic vectors $v_1,v_2\in \mathbb{R}^2$. Just consider $v_1=(1,0)$ and $v_2=(-1/2,1/2)$ and $A$ such that $A(v_2)=(1/2,1/2)$.

• You define two different $\Sigma_1$, I believe one of them should be $\Sigma_2$, could you correct it? – Daniel Cunha Jul 15 '17 at 16:37
• @DanielCunha yes of course, I've corrected it – User29983 Jul 15 '17 at 16:40
• Just to confirm, $\sigma_1^1$ is greater than the other 3, and all of them are greater than 0, is that right? – Daniel Cunha Jul 15 '17 at 16:57
• @DanielCunha yes, I thought my notation was clear, I'm sorry if it is not. please feel free to give me suggestions on how to improve it. – User29983 Jul 15 '17 at 17:01

This is not true. Let $v_1=(1,0)^T,\ v_2=(0,1)^T,\ \Sigma_1=I$ and $$A=\pmatrix{\cos\theta\\ -\sin\theta}\pmatrix{\cos\theta&-\sin\theta}=\pmatrix{\cos^2\theta&-\sin\theta\cos\theta\\ -\sin\theta\cos\theta&\sin^2\theta}.$$ Then $$\|\Sigma_1v_1+Av_2\|=\left\|\pmatrix{1-\sin\theta\cos\theta\\ \sin^2\theta}\right\|>\sqrt{2}=\sigma_1^1\|v_1+v_2\|$$ when $\theta$ is slightly greater than $\frac\pi2$.

• What do you mean with $\theta$ sufficiently small? When $\theta$ goes to zero, this norm goes to 1, which is smaller than $\sqrt{2}$. – Daniel Cunha Jul 15 '17 at 19:28
• @DanielCunha I think he means when $\theta$ is near $-1$ – User29983 Jul 15 '17 at 19:46
• @User29983 But if $\theta = -\frac{\pi}{2}$, this norm is equal to $\sqrt{2}$, still not greater. – Daniel Cunha Jul 15 '17 at 19:57
• @DanielCunha That's a typo. What I meant was that $\frac\pi2-\theta$ is a sufficiently small negative number, i.e. when $\theta$ is slightly larger than $\pi/2$. – user1551 Jul 15 '17 at 20:00
• @DanielCunha $A$ is the outer product of a unit vector with itself. So, its singular matrix is $\Sigma_2=\operatorname{diag}(1,0)$. The $\theta$ in my answer is actually the negative of the $\theta$ in the OP. – user1551 Jul 15 '17 at 20:38

[Hint]

The eigenvalues of $\Sigma_1$ are $\sigma_1^1$ and $\sigma_2^1$; the eigenvectores are $e_1^1 = \{1,0\}^T$ and $e_2^1 = \{0,1\}^T$.

The eigenvalues of $A$ are $\sigma_1^2$ and $\sigma_2^2$; the eigenvectores are $e_1^2 = \{cos\,\theta,sin\,\theta\}^T$ and $e_2^2 = \{-sin\,\theta,cos\,\theta\}^T$.

$\Sigma_1\,v_1+A\,v_2 = (v_1\cdot e_1^1)\,\Sigma_1\,e_1^1 + (v_1\cdot e_2^1)\,\Sigma_1\,e_2^1 + (v_2\cdot e_1^2)\,A\,e_1^2 + (v_2\cdot e_2^2)\,A\,e_2^2 = \\ x_1\,\sigma_1^1\,e_1^1+y_1\,\sigma_2^1\,e_2^1 + (x_2\,cos\,\theta+y_2\,sin\,\theta)\,\sigma_1^2\,e_1^2 + (-x_2\,sin\,\theta+y_2\,cos\,\theta)\,\sigma_2^2\,e_2^2$

Note that

$e_1^2 = cos\,\theta\;e_1^1 + sin\,\theta\;e_2^1$

$e_2^2 = -sin\,\theta\;e_1^1 + cos\,\theta\;e_2^1$

So, we have:

$\Sigma_1\,v_1+A\,v_2 = x_1\,\sigma_1^1\,e_1^1+y_1\,\sigma_2^1\,e_2^1 + (x_2\,cos\,\theta+y_2\,sin\,\theta)\,\sigma_1^2\,(cos\,\theta\;e_1^1 + sin\,\theta\;e_2^1) + (-x_2\,sin\,\theta+y_2\,cos\,\theta)\,\sigma_2^2\,(-sin\,\theta\;e_1^1 + cos\,\theta\;e_2^1) = \\ \boxed{(x_1\,\sigma_1^1 + x_2\,cos^2\,\theta\;\sigma_1^2+y_2\,sin\,\theta\;cos\,\theta\;\sigma_1^2 + x_2\,sin^2\,\theta\;\sigma_2^2-y_2\,sin\,\theta\;cos\,\theta\;\sigma_2^2)\,e_1^1 + \\ (y_1\,\sigma_2^1 + x_2\,sin\,\theta\;cos\,\theta\;\sigma_1^2+y_2\,sin^2\,\theta\;\sigma_1^2 - x_2\,sin\,\theta\;cos\,\theta\;\sigma_2^2+y_2\,cos^2\,\theta\;\sigma_2^2)\,e_2^1}\;\;(1)$

And also:

$\boxed{\sigma_1^1\,(v_1+v_2) = (x_1\,\sigma_1^1 + x_2\,\sigma_1^1)\,e_1^1 + (y_1\,\sigma_1^1 + y_2\,\sigma_1^1)\,e_2^1}\;\;(2)$

So, if I did not commit any mistakes, we need to show that the norm of $(2)$ is always greater (or equal) the norm of $(1)$.

• yes, I knew that. But the main difficulty of the problem is exactly showing that the norm of $(2)$ is always greater or equal to the norm of $(1)$ – User29983 Jul 15 '17 at 17:50