# hermitian transformations problems

So I have a question which I do not know how to solve.

$V$ is a inner product space over $\mathbb{R}$

$S,T:V \rightarrow V$ are a linear maps

we know that $SS^{*}=S^{*}S$ and that $TT^{*}=T^{*}T$

The eigenvalues of $T$ are in $[a,b]$

The eigenvalues of $S$ are in $[c,d]$

I need to proof that the eigenvalues of $S+T$ are in $[a+c,b+d]$

So how can I do that ?

If the field was $\mathbb{C}$ it was more easy to solve that.

but in $\mathbb{R}$ $T,S$ are not necessary diagonalizable.

So how ?

Suppose $\lambda$ is an eigenvalue, and $v$ a corresponding unit eigenvector. $\lambda$ is real since $S+T$ is Hermitian.

Now $$\lambda = v^*(S+T)v = v^*Sv + v^*Tv.$$ Since $S$ and $T$ are Hermitian, they have a full eigenbasis over $\mathbb{C}$ and we can decompose $v$ in that basis. Let's look at $S$ first. We can write

$$v = \alpha_1 w_1 + \alpha_2 w_2 + \ldots + \alpha_n w_n$$ where $\|v\|^2 = \sum \|\alpha_i\|^2 = 1$. Then \begin{align*} v^*Sv &= (\alpha_1 w_1 + \alpha_2 w_2 + \ldots + \alpha_n w_n)^*(\alpha_1 \lambda_1 w_1 + \alpha_2\lambda_2 w_2 + \ldots + \alpha_n\lambda_n w_n)\\ &= \lambda_1 \|\alpha_1\|^2 + \lambda_2 \|\alpha_2\|^2 + \ldots + \lambda_n\|\alpha_n\|^2. \end{align*}

In other words, $v^*Sv$ is a (real) convex combination of $S$'s (real) eigenvalues, whose minimum value is $\min(\lambda_i) = c$ and whose maximum value is $\max(\lambda_i) = d$.

The exact same result holds for $T$.

Actually $T,S$ are diagonalizable over $\mathbb{R}$:

Let $T=PDP^{-1}$ where $D$ is diagonal and $P\in GL(\mathbb{C})$ (Notice that we are given that $D$ has real entries, and this is all that we are going to use now). Let $X$ and $Y$ be the real matrices such that $P=X+iY$, so we have $TX=XD$ and $TY=YD$ which would give us what we need if only $X$ or $Y$ were invertible. Therefore we will look at $X+aY$ (for $a$ scalar), and try to find a value of $a$ for which it is invertible. It is enough to find such $a$ so that this matrix wont vanish on any of the basis vectors.

So let $v\in V$. Considering $v$ as a complex vector with real coordinates we see that $(X+iY)v\ne 0$ and so $v\not\in \ker{X}\cap\ker{Y}$. Therefore, if there is some $a$ for which $(X+aY)v=0$, then it is unique. so there are only a finite number of "bad" $a$ we wont choose from, and for any other $a$, we get a diagonalization $T=(X+aY)D(X+aY)^{-1}$.