Matrix Analysis (Functional Analysis) Question

[Combinatorics]

Homework Statement

Let \lambda_1 ,..., \lambda_n be the eigenvalues of an nXn self-adjoint matrix A, written in increasing order.
Show that for any m \leq n one has:
\sum_{r=1}^{m} \lambda_r = min \{ tr(L) :dim(L) =m \} where L denotes any linear subspace of \mathbb {C} ^n, and tr(L):= \sum_{r=1}^{m} Q( \Phi_r) for some orthonormal basis \{ \Phi _r \} of L.

(Q is the quadratic form associated with the inner product).

Homework Equations

The Attempt at a Solution

I really have no idea on how to start this.
On the one hand, I think the trace will always be equal to m, which means I'm probably getting it wrong...

Hope you'll be able to help me

Thanks in advance

 

[algebrat]

I know perhaps next to nothing about functional analysis, but I'd like to try helping.

I think <phi,phi>=1, while Q(phi)=<A phi,phi>=lambda.

This is just a guess at the idea, I'm not sure where it would go from there.

 

[algebrat]

Then maybe, since the lambdas are written in increasing order, the minimum trace over subspaces will somehow find you the sum of lower m lambdas.

 

[micromass]

I feel the difficulty with this question is more notationalwise. Perhaps it would be a good idea to first try to prove it for m=1. The general idea is very similar.

So, we want to prove that

\sum_{r=1}^m \lambda_r = \min\{tr(L)~\vert~ \dim(L)=m\}

A good first step would be to show the existence of a subspace L with \dim(L)=m such that

\sum_{r=1}^m \lambda_r = tr(L)

We know that a hermitian matrix A always has a orthonormal base of eigenvectors. Use this base of eigenvectors to find a suitable L.

 

[Combinatorics]

I feel the difficulty with this question is more notationalwise. Perhaps it would be a good idea to first try to prove it for m=1. The general idea is very similar.

So, we want to prove that

\sum_{r=1}^m \lambda_r = \min\{tr(L)~\vert~ \dim(L)=m\}

A good first step would be to show the existence of a subspace L with \dim(L)=m such that

\sum_{r=1}^m \lambda_r = tr(L)

We know that a hermitian matrix A always has a orthonormal base of eigenvectors. Use this base of eigenvectors to find a suitable L.

Thanks a loth to you both!

Here is my attemp:
We have a Hermitian matrix A, which implies that we have an orthonormal base of eigenvectors to the entire space. Since we have n eigenvalues, we must have n eigenvectors, and by the assumption, we can choose form such that they'll form an orthonormal basis: \{ v_1 ,..., v_n \} where Av_i = \lambda_i v_i .
Now my guess was that the space L we need to choose is span\{v_1,...,v_m \}.

If the quadratic form is assumed to be the one defined by "algebrat" , then we'll indeed get the needed equality.
But how can I prove this is the minimum?Thanks a lot again ! (and hope you'll be able to help me finish this)

 

[micromass]

For the minimum case, perhaps it's better to first prove a special case to see what happens. So take m=1 and n=2.

So take a subspace L of dimension 1. This is generated by one element (of norm) 1. Call this element v.
Since there is an orthonormal basis of the entire space (call it {x,y}). We can write v=ax+by for scalar a and b.

Now, what is &lt;Av,v&gt;?? Use that v=ax+by.

 

[Combinatorics]

For the minimum case, perhaps it's better to first prove a special case to see what happens. So take m=1 and n=2.

So take a subspace L of dimension 1. This is generated by one element (of norm) 1. Call this element v.
Since there is an orthonormal basis of the entire space (call it {x,y}). We can write v=ax+by for scalar a and b.

Now, what is &lt;Av,v&gt;?? Use that v=ax+by.

OK. Let's see: We have eigenvalues \lambda_1 , \lambda_2, with corresponding eigenvectors x,y.
We'll get: &lt;Av,v&gt;= a^2 \lambda_1 + b^2 \lambda_2 . What you mean is obviously that this expression is minimal when a=1,b=0 since \lambda_1 \leq \lambda_2.
But what if we take b=0 and a smaller a ?


Thanks a lot again !

 

[micromass]

OK. Let's see: We have eigenvalues \lambda_1 , \lambda_2, with corresponding eigenvectors x,y.
We'll get: &lt;Av,v&gt;= a^2 \lambda_1 + b^2 \lambda_2 . What you mean is obviously that this expression is minimal when a=1,b=0 since \lambda_1 \leq \lambda_2.
But what if we take b=0 and a smaller a ?


Thanks a lot again !

Notice that v must have norm 1. So this places conditions on a and b.

 

[Combinatorics]

Great ! It implies that a+b=1 !

Then we're done! Thanks a lot !

I'll try generlize this idea. If I won't succeed I'll reply here aginThanks again !

 

[micromass]

Great ! It implies that a+b=1 !

That a^2+b^2=1.

Anyway, the general case is quite similar.