# Courant fischer min-max theorem

1. May 7, 2010

### math8

Assuming the corollary of the Courant Fischer Min Max theorem,

$$\sigma _{1}\geq \cdots \geq \sigma _{q}$$ are the singular values of a matrix $$M \in \textbf{C}^{m \times n}$$ in decreasing order, q = min {m,n}

$$\sigma _{k}(M)= min_{S} max \left\{\left\| Mx\right\| _{2}: x\in S, \left\| x\right\| _{2} =1 \right\}$$ , where dim(S)= n-k+1, $$S \in \textbf{C}^{n}$$.

we want to prove that

$$\sigma _{i+j-1}(A+B)\leq \sigma _{i}(A) + \sigma _{j}(B)$$ for all $$i,j=1,2,..., q$$ and $$i+j \leq q$$.

I have tried to do this using i+j-1 instead of k, and A+B instead of M, I am getting

$$\sigma _{i+j-1}(A+B) \leq min_{S} max \left\{ (\left\| Ax\right\| _{2} +\left\| Bx\right\| _{2}): x\in S, \left\| x\right\| _{2} =1 \right\}$$ , where dim(S)= n- (i+j-1) +1.
But from here, I am not sure if I can distribute the min max to get $$min max (\left\| Ax\right\| _{2} )+ min max (\left\| Bx\right\| _{2})$$, which in that case would give me the desired result since the singular values are in decreasing order.

For some reason, I am doubting that .

Last edited: May 7, 2010