# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 41

A bilinear form or a sesquilinear form on an $n$-dimensional ${\mathbb {K} }$-vector space $V$ is described, with respect to a given basis, by its Gram matrix. Also, a linear mapping from $V$ to $V$ is described by a matrix. Altogether, we have a correspondence (for ${\mathbb {K} }=\mathbb {R}$)
    $\operatorname {Bilin} _{\left(V\right)}\longleftrightarrow \operatorname {Mat} _{n}(\mathbb {R} )\longleftrightarrow \operatorname {End} _{\mathbb {R} }{\left(V\right)}.$
On the left-hand side, properties like symmetric, hermitian, positive definite are relevant; on the right-hand side, we have eigenvalues, eigenspaces, characteristic polynomial. How are these two worlds related? In the next lectures we will deal with these questions. For this, we will work with a direct correspondence between the left-hand and the right-hand side which comes from an inner product, not from a fixed basis.
    Adjoint endomorphism

## Definition

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$, and let
    $\varphi \colon V\longrightarrow V$
denate an
endomorphism. An endomorphism
    $\psi \colon V\longrightarrow V$
is called adjoint to $\varphi$, if
    $\left\langle \varphi (v),w\right\rangle =\left\langle v,\psi (w)\right\rangle \,$
holds for all
$v,w\in V$

## Example

For an isometry
    $\varphi \colon V\longrightarrow V$
on a Euclidean vector space, the inverse mapping $\varphi ^{-1}$ is the adjoint endomorphism. In this case, we have indeed
    $\left\langle \varphi (v),w\right\rangle =\left\langle \varphi ^{-1}(\varphi (v)),\varphi ^{-1}(w)\right\rangle =\left\langle v,\varphi ^{-1}(w)\right\rangle \,.$

## Example

For a homothety $V\rightarrow V$ with scaling factor $s\in {\mathbb {K} }$ on a ${\mathbb {K} }$-vector space $V$, endowed with an inner product, the homothety with scaling factor ${\overline {s}}$ is the adjoint mapping. Indeed, we have
    $\left\langle sv,w\right\rangle =s\left\langle v,w\right\rangle ={\overline {\overline {s}}}\left\langle v,w\right\rangle =\left\langle v,{\overline {s}}w\right\rangle \,.$

## Example

Suppose that for the linear mapping
    $\varphi \colon {\mathbb {K} }^{n}\longrightarrow {\mathbb {K} }^{n},$
there exists an orthonormal basis (with respect to the standard inner product) $u_{1},\ldots ,u_{n}$ consisting of eigenvectors; that is, the describing matrix with respect to this basis 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 by the complex-conjugated matrix
    $\psi ={\begin{pmatrix}{\overline {\lambda }}_{1}&0&\cdots &\cdots &0\\0&{\overline {\lambda }}_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&{\overline {\lambda }}_{n-1}&0\\0&\cdots &\cdots &0&{\overline {\lambda }}_{n}\end{pmatrix}}\,.$
Indeed, on one hand we have
    $\left\langle \varphi (u_{i}),u_{j}\right\rangle =\left\langle \lambda _{i}u_{i},u_{j}\right\rangle =\lambda _{i}\left\langle u_{i},u_{j}\right\rangle \,,$
and on the other hand we have
    $\left\langle u_{i},\psi (u_{j})\right\rangle =\left\langle u_{i},{\overline {\lambda _{j}}}u_{j}\right\rangle =\lambda _{j}\left\langle u_{i},u_{j}\right\rangle \,.$
For $i\neq j$, we have $0$ on both sides, and for $i=j$, we have $\lambda _{i}$ on both sides.

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$, and let
    $\varphi \colon V\longrightarrow V$
denote an endomorphism. Then there exists a uniquely determined adjoint endomorphism
$\varphi$

### Proof

Let
    $\varphi \colon V\longrightarrow V$
be given, and let $w\in V$ be fixed. Then, the mapping
    $V\longrightarrow {\mathbb {K} },v\longmapsto \left\langle \varphi (v),w\right\rangle ,$
is a linear form on $V$. Therefore, there exists (due to Fact ***** in the real case; for the complex case see exercise *****) a right gradient $r=:{\hat {\varphi }}(w)$ in $V$ (uniquely determined by $\varphi$ and $w$) fulfilling
    $\left\langle v,{\hat {\varphi }}(w)\right\rangle =\left\langle \varphi (v),w\right\rangle \,.$
We have to show that the assignment
    $w\longmapsto {\hat {\varphi }}(w)$
is linear. We have
    ${\begin{aligned}\left\langle v,{\hat {\varphi }}(w_{1}+w_{2})\right\rangle &=\left\langle \varphi (v),w_{1}+w_{2}\right\rangle \\&=\left\langle \varphi (v),w_{1}\right\rangle +\left\langle \varphi (v),w_{2}\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})\right\rangle +\left\langle v,{\hat {\varphi }}(w_{2})\right\rangle \\&=\left\langle v,{\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\right\rangle .\end{aligned}}$
As this holds for all $v\in V$, we have
    ${\hat {\varphi }}(w_{1}+w_{2})={\hat {\varphi }}(w_{1})+{\hat {\varphi }}(w_{2})\,.$
Moreover,
    ${\begin{aligned}\left\langle v,{\hat {\varphi }}(sw)\right\rangle &=\left\langle \varphi (v),sw\right\rangle \\&={\overline {s}}\left\langle \varphi (v),w\right\rangle \\&={\overline {s}}\left\langle v,{\hat {\varphi }}(w)\right\rangle \\&=\left\langle v,s{\hat {\varphi }}(w)\right\rangle .\end{aligned}}$
As this holds for all $v\in V$, we get
    ${\hat {\varphi }}(sw)=s{\hat {\varphi }}(w)\,.$
$\Box$
As in the proof of this theorem, we denote the adjoint endomorphism by ${\hat {\varphi }}$. If we denote the assignment that sends a vector $w\in V$ to the linear form $v\mapsto \left\langle v,w\right\rangle$, by $\Theta$, then we have
    ${\hat {\varphi }}=\Theta ^{-1}\circ {\varphi }^{*}\circ \Theta \,,$
where
    ${\varphi }^{*}\colon {V}^{*}\longrightarrow {V}^{*}$
is the dual mapping.

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Let
    $\varphi \colon V\longrightarrow V$
be an endomorphism, and suppose that it is described by the matrix $M$ with respect to the orthonormal basis $u_{1},\ldots ,u_{n}$. Then the adjoint endomorphism
${\hat {\varphi }}$
${{\overline {M}}^{\text{tr}}}$

### Proof

Let $u_{1},\ldots ,u_{n}$ be the orthonormal basis, and let
    $M={\left(a_{ij}\right)}_{ij}\,$
and
    $N={\left(b_{ij}\right)}_{ij}\,$
be the matrices of $\varphi$ and of ${\hat {\varphi }}$ with respect to this basis. This means that we have in particular
    $\varphi (u_{i})=\sum _{k=1}^{n}a_{ki}u_{k}\,$
and
    ${\hat {\varphi }}(u_{i})=\sum _{k=1}^{n}b_{ki}u_{k}\,.$
Due to adjointness, the relation
    ${\begin{aligned}a_{ji}&=\sum _{k=1}^{n}a_{ki}\left\langle u_{k},u_{j}\right\rangle \\&=\left\langle \sum _{k=1}^{n}a_{ki}u_{k},u_{j}\right\rangle \\&=\left\langle \varphi (u_{i}),u_{j}\right\rangle \\&=\left\langle u_{i},{\hat {\varphi }}(u_{j})\right\rangle \\&=\left\langle u_{i},\sum _{k=1}^{n}b_{kj}u_{k}\right\rangle \\&=\left\langle u_{i},b_{ij}u_{i}\right\rangle \\&={\overline {b_{ij}}}\end{aligned}}$
holds. That is
    ${{\overline {N}}^{\text{tr}}}=M\,,$
and vice versa.
$\Box$

## Lemma

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$. Then the adjoint endomorphism fulfills the following properties
$\varphi ,\psi$
    ${\left(\varphi +\psi \right)}^{\hat}={\hat {\varphi }}+{\hat {\psi }}\,.$
    ${\left(s\varphi \right)}^{\hat}={\overline {s}}{\hat {\varphi }}\,.$
    ${\left({\hat {\varphi }}\right)}^{\hat}=\varphi \,.$
    ${\left(\varphi \circ \psi \right)}^{\hat}={\hat {\psi }}\circ {\hat {\varphi }}\,.$

### Proof

$\Box$
    Self-adjoint endomorphisms

## Definition

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product, and let
    $\varphi \colon V\longrightarrow V$
be an endomorphism. Then $\varphi$ is called self-adjoint if
    $\left\langle \varphi (u),v\right\rangle =\left\langle u,\varphi (v)\right\rangle \,$
holds for all
$u,v\in V$
This property simply means
    $\varphi ={\hat {\varphi }}\,.$
A homothety is self-djoint if and only if the scaling factor is real.

## Theorem

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product $\left\langle -,-\right\rangle$, and let
    $\varphi \colon V\longrightarrow V$
denote an endomorphism. Then $\varphi$ is self-adjoint if and only if it is described by an Hermitian matrix with respect to some (every) orthonormal basis
$V$

### Proof

If $\varphi$ is self-adjoint, then the statement follows from Lemma 41.6 . If $\varphi$ is described by an Hermitian matrix $M$ with respect to a orthonormal basis, then, again by Lemma 41.6 , the adjoint endomorphism ${\hat {\varphi }}$ is described by
    ${{\overline {M}}^{\text{tr}}}=M\,$
with respect to this basis. Therefore, it coincides with $\varphi$.
$\Box$

## Lemma

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product, and let
    $\varphi \colon V\longrightarrow V$
be a
* For a $\varphi$-invariant linear subspace $U\subseteq V$, also the orthogonal complement $U^{\perp }$ is $\varphi$-invariant.
* All eigenvalues are real.
* The eigenspaces for different eigenvalues are orthogonal to each other.
* Let $V$ be finite-dimensional. Then the characteristic polynomial of $\varphi$ splits into linear factors.

### Proof

* Let $v\in U^{\perp }$, and $u\in U$. Because of the invariance of $U$, we have $\varphi (u)\in U$. Therefore, This means that $\varphi (v)$ is orthogonal to $U$; thus, it belongs to $U^{\perp }$, and this means the invariance.
    $\left\langle \varphi (v),u\right\rangle =\left\langle v,\varphi (u)\right\rangle =0\,.$
* This is only relevant in case ${\mathbb {K} }=\mathbb {C}$. Let $\lambda \in \mathbb {C}$ be an eigenvalue, and let $v\in V$ be an eigenvector, that is, We may assume that this eigenvector is normed. We have therefore, $\lambda$ is real.
    $\varphi (v)=\lambda v\,.$
    $\lambda =\left\langle \lambda v,v\right\rangle =\left\langle \varphi (v),v\right\rangle =\left\langle v,\varphi (v)\right\rangle =\left\langle v,\lambda v\right\rangle ={\overline {\lambda }}\,;$
* Let $v_{1}$ be an eigenvector for the eigenvalue $\lambda _{1}$, and let $v_{2}$ be an eigenvector for the eigenvalue $\lambda _{2}\neq \lambda _{1}$. Then This is only possible for
    $\lambda _{1}\left\langle v_{1},v_{2}\right\rangle =\left\langle \lambda _{1}v_{1},v_{2}\right\rangle =\left\langle \varphi (v_{1}),v_{2}\right\rangle =\left\langle v_{1},\varphi (v_{2})\right\rangle =\left\langle v_{1},\lambda _{2}(v_{2})\right\rangle ={\overline {\lambda _{2}}}\left\langle v_{1},v_{2}\right\rangle =\lambda _{2}\left\langle v_{1},v_{2}\right\rangle \,.$
    $\left\langle v_{1},v_{2}\right\rangle =0\,.$
* We may assume that $V={\mathbb {K} }^{n}$, endowed with the standard inner product. In case ${\mathbb {K} }=\mathbb {C}$, the statement is known; so suppose that ${\mathbb {K} }=\mathbb {R}$. We may consider the mapping also as a mapping from $\mathbb {C} ^{n}$ to $\mathbb {C} ^{n}$. This map is again self-adjoint, and the characteristic polynomial does not change. Therefore, it splits into linear factors, and the zeroes are real by (2).
$\Box$
The following statement is called Spectral theorem for self-adjoint endomorphisms.

## Theorem

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product, and let
    $\varphi \colon V\longrightarrow V$
be a self-adjoint endomorphism. Then there exists an orthonormal basis of $V$ consisting of eigenvectors
$\varphi$

### Proof

We do induction over the dimension of $V$. Because of Lemma 41.10  (4), $\varphi$ has an eigenvector $v$; we may assume that this vector is normed. Due to Lemma 41.10  (1), the orthogonal complement
    $W={\mathbb {K} }v^{\perp }\,$
is also invariant. Hence, we have a direct sum decomposition
    $V={\mathbb {K} }v\oplus W\,.$
The restriction of $\varphi$ to $W$ is also self-adjoint. Therefore, the induction hypothesis yields the claim.
$\Box$
In particular, a self-adjoint endomorphism is diagonalizable.
    Self-adjoint endomorphisms and hermitian forms
Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product. An endomorphism
    $\varphi \colon V\longrightarrow V$
induces, with the help of the inner product, a form $\Psi _{\varphi }$ defined by
    $\Psi _{\varphi }(v,w)=\left\langle \varphi (v),w\right\rangle \,.$
The following properties hold for this form.

## Lemma

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an
* The assignment assigns to an endomorphism a sesquilinear form. Hence,
    $\varphi \longmapsto \Psi _{\varphi }$
    $\operatorname {End} _{\left(V\right)}\longrightarrow \operatorname {Sesq} _{\left(V\right)},\varphi \longmapsto \Psi _{\varphi }.$
* This assignment is linear; it is bijective if $V$ has finite dimension.
* Let $V$ be finite-dimensional. The endomorphism $\varphi$ is bijective if and only if $\Psi _{\varphi }$ is not degenerate.
* Let $V$ be finite-dimensional. The endomorphism $\varphi$ is self-adjoint if and only if $\Psi _{\varphi }$ is Hermitian.

### Proof

* We have and that is, the assignment is linear in the first component, and antilinear in the second component. Therefore, $\Psi _{\varphi }$ is a sesquilinear form.
    ${\begin{aligned}\psi _{\varphi }(av_{1}+bv_{2},w)&=\left\langle \varphi (av_{1}+bv_{2}),w\right\rangle \\&=\left\langle a\varphi (v_{1})+b\varphi (v_{2}),w\right\rangle \\&=a\left\langle \varphi (v_{1}),w\right\rangle +b\left\langle \varphi (v_{2}),w\right\rangle \\&=a\psi _{\varphi }(v_{1},w)+b\psi _{\varphi }(v_{2},w)\end{aligned}}$
    ${\begin{aligned}\psi _{\varphi }(v,aw_{1}+bw_{2})&=\left\langle \varphi (v),aw_{1}+bw_{2}\right\rangle \\&={\overline {a}}\left\langle \varphi (v),w_{1}\right\rangle +{\overline {b}}\left\langle \varphi (v),w_{2}\right\rangle \\&={\overline {a}}\psi _{\varphi }(v,w_{1})+{\overline {b}}\psi _{\varphi }(v,w_{2}),\end{aligned}}$
* The linearity follows from the linearity of the inner product in the first component. In the finite-dimensional case, we have on the left-hand side and on the right-hand side vector spaces of the dimension ${\left(\dim _{K}{\left(V\right)}\right)}^{2}$; therefore, it is enough to show injectivity. If $\Psi _{\varphi }=0$ is the zero form, then $\left\langle \varphi (v),w\right\rangle =0$ for all $v,w$. In particular, $\left\langle \varphi (v),\varphi (v)\right\rangle =0$, which implies $\varphi (v)=0$.
* If $\varphi$ is not bijective, then let $v\in \operatorname {kern} \varphi$, $v\neq 0$. Then, $\Psi _{\varphi }(v,-)$ is the zero mapping in the second component, and the form is degenerate. To prove the converse, suppose that $\Psi _{\varphi }(-,-)$ is degenerate. Then there exists a vector $v\in V$, $v\neq 0$, such that $\left\langle \varphi (v),-\right\rangle$ is the zero-mapping. Since an inner product is nondegenerate, this implies $\varphi (v)=0$, and $\varphi$ is not bijective.
* In the self-adjoint case, we have The converse follows from
    $\Psi _{\varphi }(v,w)=\left\langle \varphi (v),w\right\rangle =\left\langle v,\varphi (w)\right\rangle ={\overline {\left\langle \varphi (w),v\right\rangle }}={\overline {\Psi _{\varphi }(w,v)}}\,.$
    $\left\langle \varphi (v),w\right\rangle =\Psi _{\varphi }(v,w)={\overline {\Psi _{\varphi }(w,v)}}={\overline {\left\langle \varphi (w),v\right\rangle }}=\left\langle v,\varphi (w)\right\rangle \,.$
$\Box$