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:
- where do they use the fact $A=A^*$
- 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: $$\langle U(X)\mid Y\rangle \;= \langle AX\mid Y\rangle\;=Y^{*}AX\;=(A^{*}Y)^{*}X^{*}\;=\;\langle X\mid A^{*}Y\rangle$$
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."