# Inequality on self adjoint operator eigenvalues

I'm studying for a upcoming exam and have found the following problem:

Let $$\phi$$ be a self adjoint operator in an $$n$$-dimensional Hermitian space $$(V, \left \langle\space , \space\right \rangle)$$. If the eigenvalues of $$\phi$$ are $$\lambda_1 \leq ... \leq \lambda_n$$, show that $$\lambda_1 \leq \frac{\left \langle\phi(a),a \right \rangle}{\left \langle a,a \right \rangle} \leq \lambda_n$$ for every non zero $$a \in V$$.

I have zero ideas on how to proceed. I'm especially interested in hints for this question, but any help would be deeply appreciated.

Since you requested a hint, I'll provide a modest one, then put a spoiler of the solution at the end (hover your cursor over the block).

Hint: Write $$a$$ in terms of the orthonormal eigenbasis guaranteed by $$\phi$$, then just see what pops out when you compute $$\frac{\langle \phi(a),a\rangle}{\langle a,a\rangle}$$.

Spoiler:

By the spectral theorem, there exists an orthonormal basis of eigenvectors of $$\phi$$, and all of the eigenvalues are real. If $$a=\sum_{j=1}^n a_je_j,$$ where $$e_j$$ are an orthonormal basis of eigenvectors, then $$\langle a,a\rangle=\sum_{j=1}^n |a_j|^2,$$ and $$\langle \phi(a),a\rangle=\left\langle \sum_{j=1}^n a_j\lambda_je_j,\sum_{j=1}^na_je_j\right\rangle=\sum_{j=1}^n |a_j|^2\lambda_j,$$ where we have used that $$\phi$$ is linear and that $$e_j$$ are eigenvectors of $$\phi,$$ so $$\phi(a_je_j)=a_j\phi(e_j)=a_j\lambda_je_j.$$ Putting this all together, $$\frac{\langle \phi(a),a\rangle}{\langle a,a\rangle}=\frac{\sum_{j=1}^n |a_j|^2\lambda_j}{\sum_{j=1}^n |a_j|^2}.$$ To make this as larger, bound each $$\lambda_j$$ from above by $$\lambda_n$$, and to make it smaller, bound each $$\lambda_j$$ from below by $$\lambda_1$$. This will provide you with the desired result.

For self-adjoint $$\phi$$,

$$\phi^\dagger = \phi, \tag 1$$

the eigenvaues are all real:

$$\lambda_i \in \Bbb R, \; 1 \le i \le n; \tag 2$$

furthermore, there exists a basis $$\mathbf e_i$$ such that

$$\phi \mathbf e_i = \lambda_i \mathbf e_i, \tag 3$$

with

$$\langle \mathbf e_i, \mathbf e_j \rangle = \delta_{ij},\; 1 \le i, j \le n; \tag 4$$

then we may write

$$a = \displaystyle \sum_1^n a_i \mathbf e_i, \tag 5$$

from which

$$\langle a, a \rangle = \langle \displaystyle \sum_1^n a_i \mathbf e_i, \sum_1^n a_j \mathbf e_j \rangle = \sum_{i, j = 1}^n \bar a_i a_j \langle \mathbf e_i, \mathbf e_j \rangle = \sum_1^n \bar a_i a_j \delta_{ij} = \sum_1^n \bar a_i a_i \ne 0 \tag 6$$

since $$a \ne 0$$; applying $$\phi$$ to $$a$$ using (5) amd (3):

$$\phi a = \displaystyle \sum_1^n a_i \phi\mathbf e_i = \sum_1^n a_i \lambda_i \mathbf e_i; \tag 7$$

now a calculation similar to (6) yields. by virtue of (2),

$$\langle \phi a, a \rangle = \displaystyle \sum_1^n \lambda_i \bar a_i a_i, \tag 8$$

from which

$$\dfrac{\langle \phi a, a \rangle}{\langle a, a \rangle} = \displaystyle \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} \lambda_i; \tag 9$$

then

$$\lambda_1 \displaystyle \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} = \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} \lambda_1 \le \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} \lambda_i = \dfrac{\langle \phi a, a \rangle}{\langle a, a \rangle}$$ $$\le \displaystyle \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} \lambda_n = \lambda_n \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} \tag{10}$$

since

$$\lambda_1 \le \lambda_i \le \lambda_n, \forall i, \; 1 \le i \le n; \tag{11}$$

now,

$$\displaystyle \sum_1^n \dfrac{\bar a_i a_i}{\sum_1^n \bar a_i a_i} = \dfrac{1}{\sum_1^n \bar a_i a_i}\sum_1^n \bar a_i a_i = 1, \tag{12}$$

and in light of this, (10) yields

$$\lambda_1 \le \dfrac{\langle \phi a, a \rangle}{\langle a, a \rangle} \le \lambda_n, \tag{13}$$

the required inequality.

• Obviously, this is correct, but it seems unnecessary to me to add a third answer that has all of the same points as the other two answers, just with more simple calculations filled in, especially when the OP only asked for a hint. Again, that's just my opinion. – cmk Jul 24 at 17:37

Let $$\left|n\right>$$ be the eigenfunctions of $$\phi$$. Since the operator is self-adjoint, we can choose them to be orthonormal, and write $$\left|a \right> = \sum_{i = 1}^n \left\left|n\right>$$. Then we have $$\left = \sum_{i=1}^n\left\left = \sum_{i=1}^n|\left|^2$$ and So $$\frac{\left}{\left} = \frac{\sum_{i=1}^n\lambda_i\left|\left\right|^2}{\sum_{i=1}^n|\left|^2}$$ is just a weighted average of the eigenvalues, and a weighted average always falls between the highest and lowest values.