# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 38

    Bilinear forms
Real inner products are positive definite symmetric bilinear forms. In the following, we discuss bilinear forms in general. Beside inner products, there are the Hesse-forms, which are important in higher dimensional analysis in order to determine extrema, and the Minkowski-forms, which are used to describe special relativity theory (see Lecture 40).

## Definition

Let $K$ be a field, and let $V$ denote a $K$-vector space. A mapping
    $V\times V\longrightarrow K,(v,w)\longmapsto \left\langle v,w\right\rangle ,$
is called a bilinear form, if, for all $v\in V$, the induced mappings
    $V\longrightarrow K,w\longmapsto \left\langle v,w\right\rangle ,$
and for all $w\in V$, the induced mappings
    $V\longrightarrow K,v\longmapsto \left\langle v,w\right\rangle ,$
are
$K$
Bilinear simply means being multilinear in two components. An extreme example is the zero form, which assigns to every pair the value $0$. It is easy to describe many different bilinear forms on $K^{n}$.

## Example

Let $V=K^{n}$, and let $a_{ij}\in K$ denote fixed numbers, for $1\leq i,j\leq n$. Then, the assignment
    $({\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}},{\begin{pmatrix}y_{1}\\\vdots \\y_{n}\end{pmatrix}})\longmapsto \Psi (x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{n})=\sum _{i,j}a_{ij}x_{i}y_{j}$
is a bilinear form. In case
    $a_{ij}=0\,$
for all $i,j$, this is the zero form; in case
    $a_{ij}=\delta _{ij}={\begin{cases}1,{\text{ if }}i=j\,,\\0{\text{ else}}\,,\end{cases}}\,$
we have the standard inner product (where the expression makes sense for every field; the property of being positive definite does only make sense for an ordered field). In case $n=4$ and
    $\Psi (x_{1},\ldots ,x_{4},y_{1},\ldots ,y_{4})=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}-x_{4}y_{4}\,,$
we talk about a Minkowski-form. For $n=2$ and
    $\Psi (x_{1},x_{2},y_{1},y_{2})=x_{1}y_{2}-x_{2}y_{1}\,,$
this is the determinant in the two-dimensional case.
An important property of a bilinear form (which inner products fulfill) is formulated in the next definition.

## Definition

Let $K$ be a field, and let $V$ denote a $K$-vector space. A bilinear form
    $V\times V\longrightarrow K,(v,w)\longmapsto \left\langle v,w\right\rangle ,$
is called nondegenerate, if for every $v\in V$, $v\neq 0$, the induced mapping
    $V\longrightarrow K,w\longmapsto \left\langle v,w\right\rangle ,$
and, for every $w\in V$, $w\neq 0$, the induced mapping
    $V\longrightarrow K,v\longmapsto \left\langle v,w\right\rangle ,$
is not the
We will prove in Lemma 38.5 , for a vector space endowed with a nondegenerate bilinear form, the existence of a natural bijective relation between vectors and linear forms. This holds in particular for inner products. In general, there is a strong relation between bilinear forms and linear mappings to the dual space.

## Lemma

Let $K$ be a field, let $V$ denote a $K$-vector space with its dual space ${V}^{*}$. Let
    $\Theta \colon V\longrightarrow {V}^{*}$
be a linear mapping. Then
    $\Psi (u,v)=(\Theta (u))(v)\,$
defines a bilinear form
$V$

### Proof

Since $\Theta (u)\in {V}^{*}$, the evaluation at a vector $v\in V$ yields an element of the base field. The linearity in the second component rests on the fact that $\Theta (u)$ belongs to the dual space. The linearity in the first component rest on the linearity of $\Theta$.
$\Box$
    The gradient

## Lemma

Let $K$ be a field, and let $V$ denote a $K$-vector space, endowed with a bilinear form
$\left\langle -,-\right\rangle$
* For every vector $u\in V$, the assignments and are $K$-linear.
    $V\longrightarrow K,v\longmapsto \left\langle u,v\right\rangle ,$
    $V\longrightarrow K,v\longmapsto \left\langle v,u\right\rangle ,$
* The assignment is $K$-linear.
    $V\longrightarrow {V}^{*},u\longmapsto \left\langle u,-\right\rangle ,$
* If $\left\langle -,-\right\rangle$ is nongenerate, then the assignment in (2) is injective. If, moreover, $V$ is finite-dimensional, then this assignment is bijective.

### Proof

(1) follows immediately from bilinearity. (2). Let $u_{1},u_{2}\in V$ and $a_{1},a_{2}\in K$. Then, for every vector $v\in V$, we have
    $\left\langle a_{1}u_{1}+a_{2}u_{2},v\right\rangle =a_{1}\left\langle u_{1},v\right\rangle +a_{2}\left\langle u_{2},v\right\rangle \,,$
and this means the linearity of the assignment. (3). Since the assignment is linear by part (2), we have to show that its kernel is not trivial. So let $u\in V$ be such that $\left\langle u,-\right\rangle$ is the zero mapping. This means that $\left\langle u,v\right\rangle =0$ for all $v\in V$. This implies, by the definition of nondegenerated, that $u=0$. If $V$ has finite dimension, then we have an injective linear mapping between vector spaces of the same dimension. Such a mapping is also bijective, by Corollary 11.9 .
$\Box$
If a finite-dimensional vector space is endowed with a fixed nondegenerate bilinear form, then there exists, for every linear form, a uniquely determined vector that describes this linear form. More precisely: there exists a vector $y\in V$ such that
    $L(v)=\left\langle y,v\right\rangle \,$
holds for all $v\in V$, and a vector $z\in V$ such that
    $L(v)=\left\langle v,z\right\rangle \,$
holds. In this situation, $y$ is called the left gradient for $L$ with respect to the bilinear form, and $z$ is called the right gradient. For an inner product and, more generally, for any nondegenerate symmetric bilinear form (see below), the two concepts coincide, and we just talk about the gradient. For a Euclidean vector space, we formulate this relation explicitly.

## Corollary

Let $(V,\left\langle -,-\right\rangle )$ be a euclidean space, and let
    $f\colon V\longrightarrow \mathbb {R}$
denote a linear form. Then there exists a uniquely determined vector $w\in V$ such that
    $f(v)=\left\langle w,v\right\rangle \,.$
$u_{1},\ldots ,u_{n}$
orthonormal basis of $V$, and $f(u_{i})=a_{i}$, then this vector equals
$w=\sum _{i=1}^{n}a_{i}u_{i}$

### Proof

This follows immediately from Lemma 38.5  (3). The extra statement is clear because of
    $\left\langle w,u_{i}\right\rangle =\left\langle \sum _{j=1}^{n}a_{j}u_{j},u_{i}\right\rangle =a_{i}=f(u_{i})\,.$
$\Box$
    The Gram matrix

## Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a bilinear form on $V$. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$. The the $n\times n$-matrix
    $\left\langle v_{i},v_{j}\right\rangle _{1\leq i,j\leq n}$
$\left\langle -,-\right\rangle$
In Example 38. , the matrix ${\left(a_{ij}\right)}_{ij}$ is the Gram matrix with respect to the standard basis of $K^{n}$. If the Gram matrix of a bilinear form $\left\langle -,-\right\rangle$ with respect to a basis $v_{1},\ldots ,v_{n}$ is given, then one can compute $\left\langle v,w\right\rangle$ for arbitrary vectors. For this, just write $v=\sum _{i=1}^{n}b_{i}v_{i}$ and $w=\sum _{i=1}^{n}c_{i}v_{i}$; then we get, using the general distributive law,
    ${\begin{aligned}\left\langle v,w\right\rangle &=\left\langle \sum _{i=1}^{n}b_{i}v_{i},\sum _{j=1}^{n}c_{j}v_{j}\right\rangle \\&=\sum _{1\leq i,j\leq n}b_{i}c_{j}\left\langle v_{i},w_{j}\right\rangle \\&=\sum _{i=1}^{n}b_{i}{\left(\sum _{j=1}^{n}c_{j}\left\langle v_{i},w_{j}\right\rangle \right)}\\&=(b_{1},\ldots ,b_{n})G{\begin{pmatrix}c_{1}\\\vdots \\c_{n}\end{pmatrix}}.\end{aligned}}$
Thus we obtain the value of the bilinear form at two vectors in applying the Gram matrix to the coordinate tuple of the second vector, and multiplying the result (which is a column vector) with the coordinate tuple of the first vector, considered as a row tuple. Put briefly,
    $\left\langle v,w\right\rangle ={v^{\text{tr}}}Gw\,.$

## Lemma

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a bilinear form on $V$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ be two bases of $V$, and let $G$ and $H$ be the Gram matrices of $\left\langle -,-\right\rangle$ with respect to these bases. Suppose that we have the relations
    $w_{j}=\sum _{i=1}^{n}a_{ij}v_{i}\,$
between the basis elements, which we encode in the transformation matrix $A={\left(a_{ij}\right)}_{i,j}$. Then, we have the relation
    $H={A^{\text{tr}}}GA\,$

### Proof

We have
    ${\begin{aligned}\left\langle w_{r},w_{s}\right\rangle &=\left\langle \sum _{i=1}^{n}a_{ir}v_{i},\sum _{k=1}^{n}a_{ks}v_{k}\right\rangle \\&=\sum _{1\leq i,k\leq n}a_{ir}a_{ks}\left\langle v_{i},v_{k}\right\rangle \\&=\sum _{1\leq i\leq n}a_{ir}{\left(\sum _{1\leq k\leq n}a_{ks}\left\langle v_{i},v_{k}\right\rangle \right)}\\&=\sum _{1\leq i\leq n}a_{ir}{\left(G\circ A\right)}_{is}\\&={\left({A^{\text{tr}}}\circ {\left(G\circ A\right)}\right)}_{rs}.\end{aligned}}$
$\Box$
    Symmetric bilinear forms

## Definition

Let $K$ be a field, let $V$ be a $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a bilinear form on $V$. The bilinear form is called symmetric, if
    $\left\langle v,w\right\rangle =\left\langle w,v\right\rangle \,$
holds for all
$v,w\in V$
As in the case of an inner product, we have again a polarization formula.

## Lemma

Let $K$ be a field, and suppose that its characteristic is not $2$. Let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on the $K$-vector space $V$. Then the relation
    $\left\langle v,w\right\rangle ={\frac {1}{2}}{\left(\left\langle v+w,v+w\right\rangle -\left\langle v,v\right\rangle -\left\langle w,w\right\rangle \right)}\,$

### Proof

$\Box$

## Definition

Let $V$ be a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a nondegenerate symmetric bilinear form on $V$. For a given linear form
    $L\colon V\longrightarrow K,$
the uniquely determined vector $z\in V$ fulfilling
    $L(v)=\left\langle z,v\right\rangle =\left\langle v,z\right\rangle \,$
for all $v\in V$, is called the gradient
$L$

## Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on $V$. Two vectors $v,w\in V$ are called orthogonal, if
    $\left\langle v,w\right\rangle =0\,$

## Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on $V$. A basis $v_{i}$, $i\in I$, of $V$ is called orthogonal basis, if
    $\left\langle v_{i},v_{j}\right\rangle =0\,$
holds for all
$i\neq j$
Far a symmetric bilinear form, it is possible, different to the case of an inner product, that a vector $\neq 0$ is orthogonal to itself. It is possible, at least in the degenerate case, that a vector $\neq 0$ is orthogonal to all vectors. Like in the case of an inner product, there exist orthogonal bases.

## Definition

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space, and let $\left\langle -,-\right\rangle$ denote a symmetric bilinear form on $V$. The linear subspace
    ${\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in V\right\}}$
is called the degeneracy space
The degeneracy space is indeed a linear subspace of $V$, see exercise *****.

## Theorem

Let $K$ be a field, let $V$ be a finite-dimensional $K$-vector space endowed with a symmetric bilinear form $\left\langle -,-\right\rangle$. Then there exists an orthogonal basis
$V$

### Proof

$\Box$
    The vector space of bilinear forms
Let $V$ be a vector space over a field $K$, and let $\Psi _{1}$ and $\Psi _{2}$ denote bilinear forms on $V$. The sum of these two bilinear forms is defined pointwisely, that is,
    $(\Psi _{1}+\Psi _{2})(u,v):=\Psi _{1}(u,v)+\Psi _{2}(u,v)\,.$
In the same way, for a scalar $c\in K$, the form $c\Psi$ is defined via
    $(c\Psi )(u,v)=c\Psi (u,v)\,.$
These functions are again bilinear, see exercise *****. With these definition, we obtain a vector space structure on the set of all bilinear forms on $V$.

## Definition

Let $V$ be a vector space over the field $K$. The set of all bilinear forms on $V$, endowed with pointwise addition and scalar multiplication, is called the vector space of bilinear forms.
$\operatorname {Bilin} _{\left(V\right)}$

## Lemma

Let $V$ be a finite-dimensional $K$-vector space. For every basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$, the mapping
    $\operatorname {Bilin} _{\left(V\right)}\longrightarrow \operatorname {Mat} _{n}(K),\Psi \longmapsto G_{\mathfrak {v}}(\Psi ),$
which assigns to a bilinear form $\Psi$ its Gram matrix with respect to the given basis, is an isomorphism

### Proof

The injectivity of the mapping follows from Lemma 16.6 . The surjectivity follows from the fact that an arbitrary matrix might b interpret as a bilinear form in the sense of Example 38. . The linearity follows immediately from the pointwise definition of the vector space structure on $\operatorname {Bilin} _{\left(V\right)}$.
$\Box$
    Sesquilinear forms

## Definition

Let $V$ and $W$ be vector spaces over the complex numbers $\mathbb {C}$. A mapping
    $\varphi \colon V\longrightarrow W$
is called antilinear, if
    $\varphi (u+v)=\varphi (u)+\varphi (v)\,$
holds for all $u,v\in V$, and if
    $\varphi (\lambda v)={\overline {\lambda }}\varphi (v)\,$
holds for all $v\in V$ and
$\lambda \in \mathbb {C}$
If we consider the complex vector spaces as a real vector space, then this is, in particular, a real-linear mapping. We have encountered this property already in the context of a complex inner product.

## Definition

Let $V$ be a $\mathbb {C}$-vector space. A mapping
    $V\times V\longrightarrow \mathbb {C} ,(v,w)\longmapsto \left\langle v,w\right\rangle ,$
is called a sesquilinear form, if, for all $v\in V$, the induced mappinga
    $V\longrightarrow \mathbb {C} ,w\longmapsto \left\langle v,w\right\rangle ,$
are $\mathbb {C}$-antilinear, and, for all $w\in V$, the induced mappings
    $V\longrightarrow \mathbb {C} ,v\longmapsto \left\langle v,w\right\rangle ,$
are
$\mathbb {C}$
We impose linearity in the first and antilinearity in the second component. There exists also the other convention.
Many concepts and statements carry over, with some minor changes, from the real to the complex situation.

## Definition

Let $V$ be a finite-dimensional $\mathbb {C}$-vector space, endowed with a sesquilinear form $\left\langle -,-\right\rangle$. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$. The $n\times n$-matrix
    $\left\langle v_{i},v_{j}\right\rangle _{1\leq i,j\leq n}$
$\left\langle -,-\right\rangle$
If the Gram matrix of a Sesquilinear form $\left\langle -,-\right\rangle$ with respect to a basis $v_{1},\ldots ,v_{n}$ is given, then we can compute $\left\langle v,w\right\rangle$ for arbitrary vectors. We write $v=\sum _{i=1}^{n}b_{i}v_{i}$ and $w=\sum _{i=1}^{n}c_{i}v_{i}$, and we obtain, using the general distributive law,
    ${\begin{aligned}\left\langle v,w\right\rangle &=\left\langle \sum _{i=1}^{n}b_{i}v_{i},\sum _{j=1}^{n}c_{j}v_{j}\right\rangle \\&=\sum _{1\leq i,j\leq n}b_{i}{\overline {c_{j}}}\left\langle v_{i},w_{j}\right\rangle \\&=\sum _{i=1}^{n}b_{i}{\left(\sum _{j=1}^{n}{\overline {c_{j}}}\left\langle v_{i},w_{j}\right\rangle \right)}\\&=(b_{1},\ldots ,b_{n})G{\begin{pmatrix}{\overline {c_{1}}}\\\vdots \\{\overline {c_{n}}}\end{pmatrix}}.\end{aligned}}$
Thus, we get the value of the sesquilinear form at two vectors by applying the Gram matrix to the coordinate tuple of the complex-conjugated second vector, and by multiplying the result (which is a column vector) with the coordinate tuple of the first vector, considered as a row tuple. Therefore,
    $\left\langle v,w\right\rangle ={v^{\text{tr}}}G{\overline {w}}\,.$

## Lemma

Let $V$ be a finite-dimensional $\mathbb {C}$-vector space, endowed with a sesquilinear form $\left\langle -,-\right\rangle$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote two bases of $V$, and let $G$ and $H$ denote the Gram matrices of $\left\langle -,-\right\rangle$ with respect to these bases. Suppose that the basis elements are related by
    $w_{j}=\sum _{i=1}^{n}a_{ij}v_{i}\,,$
which we encode in the transformation matrix $A={\left(a_{ij}\right)}_{i,j}$. Then the Gram matrices are related by
    $H={A^{\text{tr}}}G{\overline {A}}\,.$

### Proof

We have
    ${\begin{aligned}\left\langle w_{r},w_{s}\right\rangle &=\left\langle \sum _{j=1}^{n}a_{rj}v_{j},\sum _{k=1}^{n}a_{sk}v_{k}\right\rangle \\&=\sum _{1\leq j,k\leq n}a_{rj}{\overline {a_{sk}}}\left\langle v_{j},v_{k}\right\rangle \\&=\sum _{1\leq j\leq n}a_{rj}{\left(\sum _{1\leq k\leq n}{\overline {a_{sk}}}\left\langle v_{j},v_{k}\right\rangle \right)}\\&={\left({A^{\text{tr}}}\circ {\left(G\circ {\overline {A}}\right)}\right)}_{rs}.\end{aligned}}$
$\Box$

## Remark

The set of all sesquilinear forms on a $\mathbb {C}$-vector space $V$ form a $\mathbb {C}$-vector space. It is denoted by $\operatorname {Sesq} _{\left(V\right)}$.
    Hermitian forms

## Definition

A sesquilinear form $\left\langle -,-\right\rangle$ on a complex vector space $V$ is called Hermitian, if
    $\left\langle v,u\right\rangle ={\overline {\left\langle u,v\right\rangle }}\,$
holds for all
$u,v\in V$

## Definition

A complex square matrix
    $M={\left(a_{ij}\right)}_{ij}\,$
is called Hermitian, if
    $a_{ij}={\overline {a_{ji}}}\,$
$i,j$