7B Spectral Theorem

Suppose $T$ is a self-adjoint operator on a finite-dimenaional real inner product space $V$ and $b, c \in \mathbb{R}$ are scalars such that $b^2 < 4c$. Then $p(T) = T^2 + bT + cI$ is an invertible operator.

We will show that for every non-zero $v \in V$, the inner product $\langle p(T)(v), v \rangle$ is a positive scalar multiple of the norm of $v$. Let $v \in V$ be non-zero. Then

with the first inequality come from Cauchy-Schwatz and the last coming from both $c - \frac{|b|^2}{4}$ and $\Vert v \Vert^2$ being positive. Because $(T^2 + bT + cI)(v) = 0$ would make $\langle (T^2 + bT + cI)(v), v \rangle = 0$, there can be no non-zero vectors in the kernel of $T$. Therefore $T$ is invertible.

Suppose $T$ is a self-adjoint operator on a real inner product space $V$. Then the minimal polynomial of $T$ splits into linear factors.

We know that monic polynomials over $\mathbb{R}$ factor completely into linear and quadratic factors. However, no such quadratic factors can exist in the minimal polynomial because of their positive discriminants, which would make the corresponding operator $T^2 + bT + cI$ with $b^2 - 4c > 0$ invertible, which would contradict the minimality of the degree of the minimal polynomial.

Suppose $T$ is an operator on a real inner product space $V$. Then the following are equivalent. a) $T$ is self-adjoint. b) $T$ has a diagonal matrix with respect to an orthonormal basis of $V$. c) $V$ has an orthonormal basis consisting of eigenvectors of $T$.

(b) and (c) are clearly equivalent, so we will prove the equivalence between (a) and (b). (a)$\Rightarrow$(b): Suppose $T$ is self-adjoint. Then from the last result, $T$ has an upper triangular matrix with respect to an orthonormal basis because it has a minimal polynomial that splits into linear factors. Now because $T$ is self-adjoint, this matrix must be symmetric and therefore diagonal. (a)$\Leftarrow$(b): If $T$ has a diagonal matrix with respect to an orthonormal basis then the matrix of $T^*$ is just the transpose of this diagonal matrix, which is of course symmetric. So $T = T^*$.

Suppose $T$ is an operator on a complex inner product space $V$. Then the following are equivalent. a) $T$ is normal. b) $T$ has a diagonal matrix with respect to an orthonormal basis of $V$. c) $V$ has an orthonormal basis consisting of eigenvectors of $T$.

(b) and (c) are clearly equivalent, so we will prove the equivalence between (a) and (b). (a)$\Rightarrow$(b): Suppose $T$ is normal. Because $V$ is a complex vector space, there is an orthonormal basis with respect to which the matrix of $T$ is upper-triangular. Then $\Vert T(e_1) \Vert^2$ is the square of the absolute value of the top right matrix entry, while $\Vert T^*(e_1) \Vert^2$ is the sum of the squares of the absolute values of the top row's entries. Now, $\Vert T(e_1) \Vert = \Vert T^*(e_1) \Vert$ means that all the other top row entries must be 0. This allows us to say that $Vert T(e_2) \Vert^2$ is the square of the absolute value of the second diagonal entry of the matrix, which means we can argue the other entries on the second row are also all zero. Continuing in this fashion, all entries in all rows other than each diagonal must be 0. (a)$\Leftarrow$(b): Because diagonal matrices commute, $T$ and $T^*$ must also commute.