Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 42

We have

${\displaystyle {}\left\langle \varphi (v),\varphi (w)\right\rangle =\left\langle v,{\hat {\varphi }}(\varphi (w))\right\rangle =\left\langle v,({\hat {\varphi }}\circ \varphi )(w)\right\rangle \,}$

and, using Lemma 41.7   (3), we also have

${\displaystyle {}\left\langle {\hat {\varphi }}(v),{\hat {\varphi }}(w)\right\rangle =\left\langle v,\varphi ({\hat {\varphi }}(w))\right\rangle =\left\langle v,(\varphi \circ {\hat {\varphi }})(w)\right\rangle \,.}$

If ${\displaystyle {}\varphi }$ and ${\displaystyle {}{\hat {\varphi }}}$ commute, then we also have

${\displaystyle {}\left\langle {\hat {\varphi }}(v),{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),\varphi (w)\right\rangle \,}$

for arbitrary ${\displaystyle {}v,w\in V}$. On the other hand, if this holds, then

${\displaystyle {}\left\langle v,({\hat {\varphi }}\circ \varphi )(w))\right\rangle =\left\langle v,(\varphi \circ {\hat {\varphi }})(w))\right\rangle \,}$

for all ${\displaystyle {}v\in V}$, therefore, the endomorphisms commute. Hence, (1) and (2) are equivalent. From (2) to (3) is a restriction. But also (3) implies (2), because we can express the inner product with the norm alone, according to the polarization formula.

Suppose that ${\displaystyle {}U}$ is invariant under ${\displaystyle {}\varphi }$. Let ${\displaystyle {}v\in U^{\perp }}$ and ${\displaystyle {}u\in U}$. Then

${\displaystyle {}\left\langle u,{\hat {\varphi }}(v)\right\rangle =\left\langle \varphi (u),v\right\rangle =0\,.}$

The reverse statement follows, because the situation is symmetric, due to Corollary 32.13   (3) and Lemma 41.7   (3).

Let

${\displaystyle {}\psi$ :=\varphi -\lambda \operatorname {Id} _{V}\,;}

its kernel is the eigenspace of ${\displaystyle {}\varphi }$ for the eigenvalue ${\displaystyle {}\lambda }$. The adjoint endomorphism of ${\displaystyle {}\psi }$ is, using Lemma 41.7 ,

${\displaystyle {}{\hat {\psi }}={\hat {\varphi }}-{\left(\lambda \operatorname {Id} _{V}\right)}^{\hat {}}={\hat {\varphi }}-{\overline {\lambda }}{\left(\operatorname {Id} _{V}\right)}^{\hat {}}={\hat {\varphi }}-{\overline {\lambda }}\operatorname {Id} _{V}\,.}$

Due to Exercise 42.26 , ${\displaystyle {}\psi }$ is also normal. Therefore, due to Lemma 42.7 , we have

${\displaystyle {}\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}=\operatorname {kern} {\left({\hat {\varphi }}-{\overline {\lambda }}\operatorname {Id} _{V}\right)}=\operatorname {Eig} _{\overline {\lambda }}{\left({\hat {\varphi }}\right)}\,.}$

Suppose first that ${\displaystyle {}u_{1},\ldots ,u_{n}}$ is an orthonormal basis of ${\displaystyle {}V}$, where the ${\displaystyle {}u_{i}}$ are eigenvectors of ${\displaystyle {}\varphi }$. The describing matrix ${\displaystyle {}M}$ is a diagonal matrix, its diagonal entries are the eigenvalues. Due to Lemma 41.6 , the adjoint endomorphism is described by the conjugated-transposed matrix. Hence, this is also a diagonal matrix; therefore, it commutes with ${\displaystyle {}M}$, and ${\displaystyle {}\varphi }$ is normal.

We prove the converse statement by induction over the dimension of ${\displaystyle {}V}$. So let ${\displaystyle {}\varphi }$ be normal. The one-dimensional case is clear. Because of des Fundamental theorem of algebra, there exists an eigenvector ${\displaystyle {}v\in V}$ of ${\displaystyle {}\varphi }$, and we may assume that it has norm ${\displaystyle {}1}$. Due to Lemma 42.8   (2), ${\displaystyle {}v}$ is also an eigenvector of ${\displaystyle {}{\hat {\varphi }}}$. This implies by Lemma 42.6 that ${\displaystyle {}W=\mathbb {C} v^{\perp }}$ is invariant under ${\displaystyle {}\varphi }$. Therefore, ${\displaystyle {}\varphi }$ is the direct sum of the restrictions ${\displaystyle {}\varphi }$. Hence, the restriction of ${\displaystyle {}\varphi }$ to ${\displaystyle {}W}$ is again normal, and the induction hypothesis yields the claim.

According to Lemma 41.10 , the characteristic polynomial of ${\displaystyle {}\varphi }$ splits into real linear factors. Let ${\displaystyle {}\lambda _{1},\ldots ,\lambda _{r}}$ be the positive zeroes and let ${\displaystyle {}\lambda _{r+1},\ldots ,\lambda _{s}}$ be the negative zeroes. Due to Theorem 41.11 , we have a direct sum decomposition

${\displaystyle V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{r}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{\lambda _{r+1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{s}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{0}{\left(\varphi \right)}\,,}$

which is orthogonal with respect to the inner product (${\displaystyle {}\operatorname {Eig} _{0}{\left(\varphi \right)}}$ might be zero). For vectors ${\displaystyle {}v_{i}}$ and ${\displaystyle {}v_{j}}$ from different eigenspaces, we have

${\displaystyle {}\Psi (v_{i},v_{j})=\left\langle \varphi (v_{i}),v_{j}\right\rangle =\left\langle \lambda _{i}v_{i},v_{j}\right\rangle =\lambda _{i}\left\langle v_{i},v_{j}\right\rangle =0\,;}$

therefore, the eigenspaces are also orthogonal with respect to the form ${\displaystyle {}\Psi }$. For

${\displaystyle {}v=v_{1}+\cdots +v_{r}\neq 0\,}$

with ${\displaystyle {}v_{i}\in \operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}}$, we have

${\displaystyle {}{\begin{aligned}\Psi (v,v)&=\Psi (v_{1},v_{1})+\cdots +\Psi (v_{r},v_{r})\\&=\Psi _{\varphi }(v_{1},v_{1})+\cdots +\Psi _{\varphi }(v_{r},v_{r})\\&=\left\langle \varphi (v_{1}),v_{1}\right\rangle +\cdots +\left\langle \varphi (v_{r}),v_{r}\right\rangle \\&=\left\langle \lambda _{1}v_{1},v_{1}\right\rangle +\cdots +\left\langle \lambda _{r}v_{r},v_{r}\right\rangle \\&=\lambda _{1}\left\langle v_{1},v_{1}\right\rangle +\cdots +\lambda _{r}\left\langle v_{r},v_{r}\right\rangle \\&>0.\end{aligned}}}$

This means that on this linear subspace, the restricted form is positive definite; hence,

${\displaystyle {}p\geq \sum _{i=1}^{r}\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,.}$

If ${\displaystyle {}p}$ were strictly larger that this dimension, there existed a ${\displaystyle {}p}$-dimensional linear subspace ${\displaystyle {}W\subseteq V}$ such that the restriction of ${\displaystyle {}\Psi }$ to ${\displaystyle {}W}$ is positive definite. Because of Corollary 9.8 , we have

${\displaystyle {}W\cap {\left(\operatorname {Eig} _{\lambda _{r+1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{s}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{0}{\left(\varphi \right)}\right)}\neq 0\,.}$

This yields a contradiction, since the form ${\displaystyle {}\Psi }$ is negative semidefinite on the right-hand space. Therefore,

${\displaystyle {}p=\sum _{i=1}^{r}\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,.}$

The argument for ${\displaystyle {}q}$ is the same.

Due to Lemma 41.12   (2) and Lemma 41.12   (4), we can write

${\displaystyle {}\Psi =\Psi _{\varphi }\,}$

with some self-adjoint endomorphism

${\displaystyle \varphi \colon V\longrightarrow V.}$

Because of Theorem 41.11 , there exists an orthonormal basis ${\displaystyle {}v_{1},\ldots ,v_{n}}$ consisting of eigenvectors ot ${\displaystyle {}\varphi }$ with the eigenvalues ${\displaystyle {}\lambda _{i}}$. For this basis, we have

${\displaystyle {}\Psi (v_{i},v_{j})=\Psi _{\varphi }(v_{i},v_{j})=\left\langle \varphi (v_{i}),v_{j}\right\rangle =\left\langle \lambda _{i}v_{i},v_{j}\right\rangle =\lambda _{i}\left\langle v_{i},v_{j}\right\rangle \,.}$

Therefore, this basis is also an orthogonal basis with respect to ${\displaystyle {}\Psi }$.