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

    Normal endomorphisms
Due to Theorem 34.2 , an isometry over $\mathbb {C}$ posses an orthonormal basis consisting of eigenvectors. Due to Theorem 41.11 , a self-adjoint mapping (over $\mathbb {R}$ or $\mathbb {C}$) has also an orthonormal basis of eigenvectors. We are looking for a generalization of these two statements over $\mathbb {C}$. The result is the Spectral theorem for normal endomorphisms; see Theorem 42.9 . As this theorem provides an equivalent characterization for the existence of an orthonormal basis of eigenvectors, a further generalization is not possible.

## Definition

We say that linear mappings
    $\varphi ,\psi \colon V\longrightarrow V$
on a $K$-vector space commute if
    $\psi \circ \varphi =\varphi \circ \psi \,$

## Definition

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. An endomorphism
    $\varphi \colon V\longrightarrow V$
is called normal if $\varphi$ and its adjoint endomorphism ${\hat {\varphi }}$
Thus, the condition is
    $\varphi \circ {\hat {\varphi }}={\hat {\varphi }}\circ \varphi \,.$
A self-adjoint endomorphism is normal. In the case of an isometry $\varphi$, the adjoint endomorphism is $\varphi ^{-1}$, due to Example 41.2 . Therefore, an isometry is normal. If the endomorphism $\varphi$ is described by the matrix $M$, with respect to an orthonormal basis, then the condition for normality means
    $M{{\overline {M}}^{\text{tr}}}={{\overline {M}}^{\text{tr}}}M\,.$

## Example

In the two-dimensional situation, the condition for normality is
    ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}{\begin{pmatrix}{\overline {a}}&{\overline {c}}\\{\overline {b}}&{\overline {d}}\end{pmatrix}}={\begin{pmatrix}{\overline {a}}&{\overline {c}}\\{\overline {b}}&{\overline {d}}\end{pmatrix}}{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,;$
looking at the entries, this translates to the conditions
    $b{\overline {b}}=c{\overline {c}}\,$
and
    $a{\overline {c}}+b{\overline {d}}={\overline {a}}b+{\overline {c}}d\,.$
Besides diagonal matrices and rotation matrices, also real matrices of the form
    ${\begin{pmatrix}a&b\\-b&a\end{pmatrix}}$
have this property.

## Example

Suppose that the linear mapping
    $\varphi \colon {\mathbb {K} }^{n}\longrightarrow {\mathbb {K} }^{n}$
has an orthonormal basis (with respect to the standard inner product) $u_{1},\ldots ,u_{n}$ consisting of eigenvectors; that means that the describing matrix is in diagonal form
    ${\begin{pmatrix}\lambda _{1}&0&\cdots &\cdots &0\\0&\lambda _{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda _{n-1}&0\\0&\cdots &\cdots &0&\lambda _{n}\end{pmatrix}}.$
Then the adjoint endomorphism is described, due to Example 41.4 , by the complex-conjugated matrix
    ${\begin{pmatrix}{\overline {\lambda }}_{1}&0&\cdots &\cdots &0\\0&{\overline {\lambda }}_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&{\overline {\lambda }}_{n-1}&0\\0&\cdots &\cdots &0&{\overline {\lambda }}_{n}\end{pmatrix}}.$
These two matrices commute, that is, we have a normal endomorphism.
The third property of the following lemma explains the word "normal“.

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
denote an
* $\varphi$ is normal.
* For all $v,w\in V$, the identity holds.
    $\left\langle {\hat {\varphi }}(v),{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),\varphi (w)\right\rangle \,$
* For all $v\in V$, the identity holds.
    $\Vert {{\hat {\varphi }}(v)}\Vert =\Vert {\varphi (v)}\Vert \,$

### Proof

We have
    $\left\langle \varphi (v),\varphi (w)\right\rangle =\left\langle v,{\hat {\varphi }}(\varphi (w))\right\rangle =\left\langle v,({\hat {\varphi }}\circ \varphi )(w))\right\rangle \,$
and, using Lemma 41.7  (3), we also have
    $\left\langle {\hat {\varphi }}(v),{\hat {\varphi }}(w)\right\rangle =\left\langle v,\varphi ({\hat {\varphi }}(w))\right\rangle =\left\langle v,(\varphi \circ {\hat {\varphi }})(w))\right\rangle \,.$
If $\varphi$ and ${\hat {\varphi }}$ commute, then we also have
    $\left\langle {\hat {\varphi }}(v),{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),\varphi (w)\right\rangle \,$
for arbitrary $v,w\in V$. On the other hand, if this holds, then
    $\left\langle v,({\hat {\varphi }}\circ \varphi )(w))\right\rangle =\left\langle v,(\varphi \circ {\hat {\varphi }})(w))\right\rangle \,$
for all $v\in V$, 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.
$\Box$

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
denote an endomorphism. Then a linear subspace $U\subseteq V$ is $\varphi$-invariant if and only if the orthogonal complement
$U^{\perp }$
${\hat {\varphi }}$

### Proof

Suppose that $U$ is invariant under $\varphi$. Let $u\in U$ and $v\in U^{\perp }$. Then
    $\left\langle u,{\hat {\varphi }}(v)\right\rangle =\left\langle \varphi (u),v\right\rangle =0\,.$
The reverse statement follows, because the situation is symmetric, due to Corollary 32.13  (3) and Lemma 41.7  (3).
$\Box$

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
denote a normal endomorphism. Then
    $\operatorname {kern} \varphi =\operatorname {kern} {\hat {\varphi }}\,.$

### Proof

$\Box$

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
denote a
* $\lambda \in {\mathbb {K} }$ is an eigenvalue of $\varphi$ if and only if ${\overline {\lambda }}$ is an eigenvalue of ${\hat {\varphi }}$.
* A vector $v\in V$ is an eigenvector of the eigenvalue $\lambda$ if and only if $v$ is an eigenvector of ${\hat {\varphi }}$ of the eigenvalue${\overline {\lambda }}$.

### Proof

Let
    $\psi :=\varphi -\lambda \operatorname {Id} _{V}\,;$
its kernel is the eigenspace of $\varphi$ for the eigenvalue $\lambda$. The adjoint endomorphism of $\psi$ is, using Lemma 41.7 ,
    ${\hat {\psi }}={\hat {\varphi }}-{\left(\lambda \operatorname {Id} _{V}\right)}^{\hat}={\hat {\varphi }}-{\overline {\lambda }}{\left(\operatorname {Id} _{V}\right)}^{\hat}={\hat {\varphi }}-{\overline {\lambda }}\operatorname {Id} _{V}\,.$
Due to exercise *****, $\psi$ is also normal. Therefore, due to Lemma 42.7 , we have
    $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}=\operatorname {kern} {\left({\hat {\varphi }}-{\overline {\lambda }}\operatorname {Id} _{V}\right)}=\operatorname {Eig} _{\overline {\lambda }}{\left({\hat {\varphi }}\right)}\,.$
$\Box$

## Theorem

Let $V$ be a finite-dimensional complex vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
denote an endomorphism. Then $\varphi$ is normal if and only if there exists an orthonormal basis consisting of eigenvectors
$\varphi$

### Proof

Suppose first that $u_{1},\ldots ,u_{n}$ is an orthonormal basis of $V$, where the $u_{i}$ are eigenvectors of $\varphi$. The describing matrix $M$ 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 $M$, and $\varphi$ is normal.
We prove the converse statement by induction over the dimension of $V$. So let $\varphi$ be normal. The one-dimensional case is clear. Because of des Fundamental theorem of algebra, there exists an eigenvector $v\in V$ of $\varphi$, and we may assume that it has norm $1$. Due to Lemma 42.8  (2), $v$ is also an eigenvector of ${\hat {\varphi }}$. This implies by Lemma 42.6
that $W=\mathbb {C} v^{\perp }$ is invariant under $\varphi$. The restriction of $\varphi$ to $W$ is again normal, and the induction hypothesis yields the claim.
$\Box$
The preceding statement does not hold over the real numbers, as every plane rotation (with the exception of the identity and the point reflection) shows.
    Principal axis transformation
We want to apply the results of the previous lecture to Hermitian forms. This technique is called principal axis transformation, these terms will become clear in the next lecture. As we are working over the complex numbers, we have to mention briefly that the concepts positive definite, negative definite, and type, which we have defined for a real-symmetric bilinear form, carry over directly to complex-hermitian sesquilinear forms. In the same spirit, Sylvester's law of inertia holds again, with the same proof; also the minor criterion holds.

## Theorem

Let $V$ be a finite-dimensional complex vector space, endowed with an Hermitian sesquilinear form $\left\langle -,-\right\rangle$ of type $(p,q)$. Then the Gram matrix of $\left\langle -,-\right\rangle$ with respect to every orthogonal basis is a diagonal matrix
$p$
$q$
This implies immediately that, if we start with a real-symmetric bilinear form on $\mathbb {R} ^{n}$, and if we consider it as an Hermitian sesquilinear form on $\mathbb {C} ^{n}$, then the real type coincides with the complex type. The big advantage of the complex situation is that the fundamental theorem of algebra is available, and this guarantees the existence of eigenvalues. Even if we know, like in Lemma 41.10 , that all eigenvalues are real, their existence is only clear when working over the complex numbers.

## Theorem

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product. Let $\Psi$ denote an Hermitian form on $V$, corresponding to the self-adjoint endomorphism
    $\varphi \colon V\longrightarrow V$
in the sense of Lemma 41.12 . Let $(p,q)$ be the type of $\Psi$. Then $p$ is the number of positive eigenvalues, and $q$ is the number of negative eigenvalues of $\varphi$, where we have to take this numbers with their (algebraic or geometric)

### Proof

According to Lemma 41.10 , the characteristic polynomial of $\varphi$ splits into real linear factors. Let $\lambda _{1},\ldots ,\lambda _{r}$ be the positive zeroes and let $\lambda _{r+1},\ldots ,\lambda _{s}$ be the negative zeroes. Due to Theorem 41.11 , we have a direct sum decomposition
    $V=\operatorname {Eig} _{\lambda _{1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{r}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{\lambda _{r+1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{s}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{0}{\left(\varphi \right)}\,,$
which is orthogonal with respect to the inner product ($\operatorname {Eig} _{0}{\left(\varphi \right)}$ might be zero). For vectors $v_{i}$ and $v_{j}$ from different eigenspaces, we have
    $\Psi (v_{i},v_{j})=\left\langle \varphi (v_{i}),v_{j}\right\rangle =\left\langle \lambda _{i}v_{i},v_{j}\right\rangle =\lambda _{i}\left\langle v_{i},v_{j}\right\rangle =0\,;$
therefore, the eigenspaces are also orthogonal with respect to the form $\Psi$. For
    $v=v_{1}+\cdots +v_{r}\neq 0\,$
with $v_{i}\in \operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}$, we have
    ${\begin{aligned}\Psi (v,v)&=\Psi (v_{1},v_{1})+\cdots +\Psi (v_{r},v_{r})\\&=\Psi _{\varphi }(v_{1},v_{1})+\cdots +\Psi _{\varphi }(v_{r},v_{r})\\&=\left\langle \varphi (v_{1}),v_{1}\right\rangle +\cdots +\left\langle \varphi (v_{r}),v_{r}\right\rangle \\&=\left\langle \lambda _{1}v_{1},v_{1}\right\rangle +\cdots +\left\langle \lambda _{r}v_{r},v_{r}\right\rangle \\&=\lambda _{1}\left\langle v_{1},v_{1}\right\rangle +\cdots +\lambda _{r}\left\langle v_{r},v_{r}\right\rangle \\&>0.\end{aligned}}$
This means that on this linear subspace, the restricted form is positive definite; hence,
    $p\geq \sum _{i=1}^{r}\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,.$
If $p$ were strictly larger that this dimension, there existed a $p$-dimensional linear subspace $W\subseteq V$ such that the restriction of $\Psi$ to $W$ is positive definite. Because of Corollary 9.8 , we have
    $W\cap {\left(\operatorname {Eig} _{\lambda _{r+1}}{\left(\varphi \right)}\oplus \cdots \oplus \operatorname {Eig} _{\lambda _{s}}{\left(\varphi \right)}\oplus \operatorname {Eig} _{0}{\left(\varphi \right)}\right)}\neq 0\,.$
This yields a contradiction, since the form $\Psi$ is negative semidefinite on the right-hand space. Therefore,
    $p=\sum _{i=1}^{r}\dim _{K}{\left(\operatorname {Eig} _{\lambda _{i}}{\left(\varphi \right)}\right)}\,.$
The argument for $q$ is the same.
$\Box$
We are now able to present the proof of zum eigenvalue criterion about the type of a real-symmetric bilinear form. This follows immediately from Theorem 42.11 .
The following Theorem is called the principle axis theorem.

## Theorem

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product. Let $\Psi$ denote an Hermitian form on $V$. Then there exists an orthonormal basis of $V$ (with respect to the inner product), which is an orthogonal basis
$\Psi$

### Proof

Due to Lemma 41.12  (2) and Lemma 41.12  (4), we can write
    $\Psi =\Psi _{\varphi }\,$
with some self-adjoint endomorphism
    $\varphi \colon V\longrightarrow V.$
Because of Theorem 41.11 , there exists an orthonormal basis $v_{1},\ldots ,v_{n}$ consisting of eigenvectors ot $\varphi$ with the eigenvalues $\lambda _{i}$. For this basis, we have
    $\Psi (v_{i},v_{j})=\Psi _{\varphi }(v_{i},v_{j})=\left\langle \varphi (v_{i}),v_{j}\right\rangle =\left\langle \lambda _{i}v_{i},v_{j}\right\rangle =\lambda _{i}\left\langle v_{i},v_{j}\right\rangle \,.$
Therefore, this basis is also an orthogonal basis with respect to $\Psi$.
$\Box$
In this context, the Eigenlines of $\varphi$ are also called principal axes of $\Psi _{\varphi }$, and the eigenvalues are also called principal values.