# Can we simplify this proof?

I'm reading Hoffman and Kunze's linear algebra and on page 313 they prove the following theorem:

Theorem 16 On a finite-dimensional inner product space of positive dimension, every self-adjoint operator has a (non-zero) characteristic vector.

They proved this in the following way: I didn't understand:

1. where do they use the fact $A=A^*$
2. Why does $c$ be a real scalar matters?

To sum up: Why don't they simply say the characteristic polynomial, $\det(xI-A)$, is a polynomial of degree $n$ over the complex numbers then there is $c$ such that $\det (cI-A)=0$ and then there is a non-zero $X$ such that $AX=cX$ which follows there is a non-zero vector $\alpha\in V$ such that $T\alpha=c\alpha$?

Q: Where do they use the fact that $A=A^{*}$

Author says "$U(X)=AX$ defines a self adjoint linear operator $U$ on $W$". To show that $U$ is self adjoint, you will need the fact that $A$ is self adjoint.

Hint: $$<U(X)| Y> \;= \;<AX|Y>\;=Y^{*}AX\;=(A^{*}Y)^{*}X^{*}\;=\;<X|A^{*}Y>$$

And for the second question. We are using the fact that All eigenvalues of a self-adjoint operator are real . That's why real scalar matters. The self ad-jointness of $U$ forces $c$ to be real.

Read the comments on page 313 just below the proof. It says "The argument shows that characteristic polynomial of a self adjoint matrix has real coefficients, in spite of the fact that the matrix may not have real entries."

• Do you mean: $Y^{*}AX\;=(A^{*}Y)^{*}X$? – user42912 Oct 23 '16 at 14:15
• I understood the answer for the first question, but I'm still struggling to follow the answer to the second one. I know why all eigenvalues of $U$ are real, what I don't understand is where we are using this fact to prove the statement of the theorem. Thank you for your answer! – user42912 Oct 23 '16 at 14:20
• I think I know, is it because the case where $V$ is real? – user42912 Oct 23 '16 at 14:35