# Self-adjoint and positive operator minimal polynomial on complex inner product spaces

Suppose that T is a self-adjoint operator on the 2-dimensional complex inner product space. Suppose that the minimal polynomial of T is $$T^2-(a+c)T+(ac-|b|^2)I$$ a)Given that a, c are real numbers and b is complex. deduce from this that T is positive if and only if $$a+c\le0$$ and $$ac\le b^2$$ B) deduce from the polynomial that the eigenvalues a of T are real.

I have finished all the questions from my maths tutorial except these. We don't get answers and I am really stuck. Can somebody help me?

For part B I was thinking that eigenvalues are the roots to the minimal polynomial and using http://m.wolframalpha.com/input/?i=x%5E2-%28a%2Bc%29x%2Bac-%7Cb%7C%5E2&x=0&y=0 we can see that the roots are real so there for the eigenvalues are real. Is this a correct approach?

I have absolutely no idea how to do part a.

Thanks for any help in advance

## 1 Answer

Consider the quadratic $$t^2-(a+c)t+(ac-|b|^2)=\left(t-\frac{a+c}2\right)^2+ac-|b|^2-\frac{(a+c)^2}4\\ =\left(t-\frac{a+c}2\right)^2-\left(|b|^2+\frac{(a-c)^2}4\right).$$ The eigenvalues of $T$ are the roots of this equation (this is where we use that the dimension of the space is 2), namely $$\tag{1} \frac{a+c}2\pm\left(|b|^2+\frac{(a-c)^2}4\right)^{1/2}.$$ For both to be positive we need $$\left(|b|^2+\frac{(a-c)^2}4\right)^{1/2}\leq\frac{a+c}2.$$ This forces $a+c\geq0$; and, comparing squares, $$|b|^2+\frac{(a-c)^2}4\leq\frac{(a+c)^2}4,$$ which immediately simplifies to $|b|^2\leq ac$.

As the two eigenvalues of $T$ in $(1)$ are real, this also answers part b).