# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 33

    The cross product
A special feature of $\mathbb {R} ^{3}$ is the so-called cross product. This assigns, to two given vectors, a vector that is orthogonal to them.

## Definition

Let $K$ be a field. The operation on $K^{3}$, defined by
    $x\times y={\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}:={\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}\,,$
is called the
The cross product is also called the vector product. To remember this formula, one might think
    $x\times y=\det {\begin{pmatrix}e_{1}&x_{1}&y_{1}\\e_{2}&x_{2}&y_{2}\\e_{3}&x_{3}&y_{3}\end{pmatrix}}\,,$
where $e_{1},e_{2},e_{3}$ are the standard vectors, and where we expand formally with respect to the first column. In this way, the cross product is defined with respect to the standard basis.

## Example

The cross product of the vectors ${\begin{pmatrix}5\\8\\-2\end{pmatrix}},{\begin{pmatrix}3\\4\\1\end{pmatrix}}\in \mathbb {R} ^{3}$ is
    ${\begin{pmatrix}5\\8\\-2\end{pmatrix}}\times {\begin{pmatrix}3\\4\\1\end{pmatrix}}={\begin{pmatrix}8\cdot 1-4\cdot (-2)\\-5\cdot 1-2\cdot 3\\5\cdot 4-8\cdot 3\end{pmatrix}}={\begin{pmatrix}16\\-11\\-4\end{pmatrix}}\,.$

## Lemma

The cross product on $K^{3}$ fulfills the following properties (where $x,y,z\in K^{3}$ and
$a,b\in K$
* We have
    $x\times y=-(y\times x)\,.$
* We have and
    $(ax+by)\times z=a(x\times z)+b(y\times z)\,$
    $z\times (ax+by)=a(z\times x)+b(z\times y)\,.$
* We have if and only if $x$ and $y$ are linearly dependent.
    $x\times y=0\,$
* We have
    $x\times (y\times z)+y\times (z\times x)+z\times (x\times y)=0\,.$
* We have where $\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.
    $\left\langle x\times y,z\right\rangle =\det(x\,,y\,,z)\,,$
* We have where $\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.
    $\left\langle x,x\times y\right\rangle =0=\left\langle y,x\times y\right\rangle \,,$

### Proof

(1) is clear from the definition.
(2). We have
    ${\begin{aligned}{\left(a{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}}+b{\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}\right)}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}&={\begin{pmatrix}ax_{1}+by_{1}\\ax_{2}+by_{2}\\ax_{3}+by_{3}\end{pmatrix}}\times {\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\\&={\begin{pmatrix}(ax_{2}+by_{2})z_{3}-(ax_{3}+by_{3})z_{2}\\-(ax_{1}+by_{1})z_{3}+(ax_{3}+by_{3})z_{1}\\(ax_{1}+by_{1})z_{2}-(ax_{2}+by_{2})z_{1}\end{pmatrix}}\\&={\begin{pmatrix}ax_{2}z_{3}-ax_{3}z_{2}\\-ax_{1}z_{3}+ax_{3}z_{1}\\ax_{1}z_{2}-ax_{2}z_{1}\end{pmatrix}}+{\begin{pmatrix}by_{2}z_{3}-by_{3}z_{2}\\-by_{1}z_{3}+by_{1}z_{3}\\by_{1}z_{2}-by_{2}z_{1}\end{pmatrix}}\\&=a{\begin{pmatrix}x_{2}z_{3}-x_{3}z_{2}\\-x_{1}z_{3}+x_{3}z_{1}\\x_{1}z_{2}-x_{2}z_{1}\end{pmatrix}}+b{\begin{pmatrix}y_{2}z_{3}-y_{3}z_{2}\\-y_{1}z_{3}+y_{1}z_{3}\\y_{1}z_{2}-y_{2}z_{1}\end{pmatrix}}\\&=a(x\times z)+b(y\times z).\end{aligned}}$
The second equation follows from this and from (1).
(3). If $x$ and $y$ are linearly dependent, then we can write $x=cy$ (or the other way round). In this case,
    ${\begin{pmatrix}cy_{1}\\cy_{2}\\cy_{3}\end{pmatrix}}\times {\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}}={\begin{pmatrix}cy_{2}y_{3}-cy_{2}y_{3}\\-cy_{1}y_{3}+cy_{3}y_{1}\\cy_{1}y_{2}-cy_{2}y_{1}\end{pmatrix}}=0\,.$
If the cross product is $0$, then all entries of the vectors ${\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}}$ equal $0$. For example, let $y_{1}\neq 0$. From $x_{1}=0$, we can deduce directly
    $x_{2}=x_{3}=0\,,$
and $x$ is the zero vector. So suppose that $x_{1}\neq 0$. Then $y_{2}={\frac {y_{1}}{x_{1}}}x_{2}$ and $y_{3}={\frac {y_{1}}{x_{1}}}x_{3}$; therefore, we get
    $y={\frac {y_{1}}{x_{1}}}x\,.$
(4). See exercise *****.
(5). We have
    ${\begin{aligned}\left\langle x\times y,z\right\rangle &=\left\langle {\begin{pmatrix}x_{2}y_{3}-x_{3}y_{2}\\-x_{1}y_{3}+x_{3}y_{1}\\x_{1}y_{2}-x_{2}y_{1}\end{pmatrix}},{\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}}\right\rangle \\&=z_{1}x_{2}y_{3}-z_{1}x_{3}y_{2}-z_{2}x_{1}y_{3}+z_{2}x_{3}y_{1}+z_{3}x_{1}y_{2}-z_{3}x_{2}y_{1}.\end{aligned}}$
This coincides with the determinant, due to Sarrus.
(6) follows from (5).
$\Box$
The expression $\left\langle x\times y,z\right\rangle$ from (5), that is, the determinant of the three vectors, considered as a column vector, is also called triple product.

## Lemma

Let $u_{1},u_{2},u_{3}$ be an orthonormal basis of $\mathbb {R} ^{3}$ with
    $\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,.$
cross product $x\times y$ can be computed with the coordinates of $x$ and $y$ with respect to this basis (and the formula from

### Proof

Let
    $x=c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\,$
and
    $y=d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\,.$
Due to Fact (2) *****, we have
    $x\times y={\left(c_{1}u_{1}+c_{2}u_{2}+c_{3}u_{3}\right)}\times {\left(d_{1}u_{1}+d_{2}u_{2}+d_{3}u_{3}\right)}=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\,.$
Due to Fact (3) *****, we have
    $u_{i}\times u_{i}=0\,,$
and, because of Fact (1) *****, we have
    $u_{i}\times u_{j}=-u_{j}\times u_{i}\,.$
According to Fact (6) *****, $u_{1}\times u_{2}$ is perpendicular to $u_{1}$ and to $u_{2}$; therefore,
    $u_{1}\times u_{2}=\lambda u_{3}\,$
with some $\lambda \in \mathbb {R}$, as this orthogonality condition defines a line. Because of Fact (5) ***** and the condition, we get
    $\lambda =\left\langle \lambda u_{3},u_{3}\right\rangle =\left\langle u_{1}\times u_{2},u_{3}\right\rangle =\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,;$
hence,
    $u_{1}\times u_{2}=u_{3}\,.$
Using Lemma 17 2.  (3), we obtain $u_{1}\times u_{3}=-u_{2}$ and $u_{2}\times u_{3}=u_{1}$. Altogether we get
    ${\begin{aligned}x\times y&=\sum _{1\leq i,j\leq 3}c_{i}d_{j}{\left(u_{i}\times u_{j}\right)}\\&=\sum _{i<j}(c_{i}d_{j}-c_{j}d_{i}){\left(u_{i}\times u_{j}\right)}\\&=(c_{1}d_{2}-c_{2}d_{1})u_{3}-(c_{1}d_{3}-c_{3}d_{1})u_{2}+(c_{2}d_{3}-c_{3}d_{2})u_{1},\end{aligned}}$
and this is the claim.
$\Box$
    Isometries

## Definition

Let $V,W$ be vector spaces over ${\mathbb {K} }$, endowed with inner products, and let
    $\varphi \colon V\longrightarrow W$
be s linear mapping. Then $\varphi$ is called an isometry if
    $\left\langle \varphi (v),\varphi (w)\right\rangle =\left\langle v,w\right\rangle \,$
holds for all
$v,w\in V$
An isometry is always injective. For ${\mathbb {K} }=\mathbb {C}$, we also talk about an unitary mapping. As there are also affine isometries, we talk about a linear isometry.

## Lemma

Let $V$ and $W$ be vector spaces over ${\mathbb {K} }$, both endowed with an inner product, and let $\varphi \colon V\rightarrow W$ be a
* $\varphi$ is an isometry.
* For all $u,v\in V$, we have $d(\varphi (u),\varphi (v))=d(u,v)$.
* For all $v\in V$, we have $\Vert {\varphi (v)}\Vert =\Vert {v}\Vert$.
* For all $v\in V$ fulfilling $\Vert {v}\Vert =1$, we have $\Vert {\varphi (v)}\Vert =1$.

### Proof

The implications $(1)\Rightarrow (2)$, $(2)\Rightarrow (3)$ and $(3)\Rightarrow (4)$ are restrictions. $(4)\Rightarrow (3)$. For the zero vecto r,the statement $(3)$ is clear; so suppose that $v\neq 0$. Then, ${\frac {v}{\Vert {v}\Vert }}$ has norm $1$, and, because of
    $\varphi (v)=\varphi {\left(\Vert {v}\Vert {\frac {v}{\Vert {v}\Vert }}\right)}=\Vert {v}\Vert \varphi {\left({\frac {v}{\Vert {v}\Vert }}\right)}\,,$
we have
    $\Vert {\varphi (v)}\Vert =\Vert {v}\Vert \,.$
$(3)\Rightarrow (1)$ follows from Fact *****.
$\Box$
Therefore, an isometry is just a (linear) mapping that preserves distances. The set of all the vectors with norm $1$ in a Euclidean vector space is also called the sphere. Hence, an isometry is characterized by the property that it maps the sphere to the sphere.

## Lemma

Let $V$ and $W$ be euclidean vector spaces, and let
    $\varphi \colon V\longrightarrow W$
denote a
* $\varphi$ is an isometry.
* For every orthonormal basis $u_{i}$, $i=1,\ldots ,n$, of $V$, $\varphi (u_{i})$, $i=1,\ldots ,n$, is part of an orthonormal basis of $W$.
* There exists an orthonormal basis $u_{i}$, $i=1,\ldots ,n$, of $V$ such that $\varphi (u_{i})$, $i=1,\ldots ,n$, is part of an orthonormal basis of $W$.

### Proof

$\Box$

## Theorem

For every euclidean vector space $V$, there exists a bijective isometry
    $\varphi \colon \mathbb {R} ^{n}\longrightarrow V,$
where $\mathbb {R} ^{n}$ carries the

### Proof

Let $u_{1},\ldots ,u_{n}$ be an orthonormal basis of $V$, and let
    $\varphi \colon \mathbb {R} ^{n}\longrightarrow V$
be the linear mapping given by
    $\varphi (e_{i})=u_{i}\,.$
Because of Fact (3) *****, this is an isometry.
$\Box$
    Isometries on a Euclidean vector space
We now discuss isometries of a Euclidean vector space in itself. These are always stets bijective. With respect to any orthonormal basis of $V$, they are described in the following way.

## Lemma

Let $V$ be a Euclidean vector space, and let $u_{1},\ldots ,u_{n}$ denote an orthonormal basis of $V$. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping, and let $M$ be the describing matrix of $\varphi$ with respect to the given basis. Then $\varphi$ is an isometry if and only if
    ${M^{\text{tr}}}M=E_{n}\,$

### Proof

Suppose first that $\varphi$ is an isometry. Then, $v_{i}=\varphi (u_{i})$ is an orthonormal basis due to Fact *****. The coordinates of $v_{i}$ with respect to $u_{i}$ constitute the columns of the describing matrix $M$. Therefore, using exercise *****, we have
    ${v_{i}^{\text{tr}}}v_{j}=\left\langle v_{i},v_{j}\right\rangle =\delta _{ij}\,.$
Read as a matrix equation, this means
    ${M^{\text{tr}}}M={\begin{pmatrix}1&0&\cdots &\cdots &0\\0&1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&0\\0&\cdots &\cdots &0&1\end{pmatrix}}\,.$
This argument can be read backwards to get the reverse implication.
$\Box$
The set of isometries on a Euclidean vector space form a group; in fact, it is a subgroup of the group of all bijective linear mappings. We recall briefly the corresponding definitions.
For a field $K$ and $n\in \mathbb {N} _{+}$, the set of all invertible $n\times n$-matrices with entries in $K$ is called the general linear group
$K$
$\operatorname {GL} _{n}\!{\left(K\right)}$
For a field $K$, and $n\in \mathbb {N} _{+}$, the set of all invertible $n\times n$-matrices over $K$ with
    $\det M=1\,$
is called the special linear group
$K$
$\operatorname {SL} _{n}\!{\left(K\right)}$

## Definition

Let $K$ be a field, and $E_{n}$ the identity matrix of length $n$. A matrix $M\in \operatorname {GL} _{n}\!{\left(K\right)}$ fulfilling
    ${M^{\text{tr}}}M=E_{n}\,$
is called an orthogonal matrix. The set of all orthogonal matrices is called orthogonal group; it is denoted by
    $\operatorname {O} _{n}\!{\left(K\right)}={\left\{M\in \operatorname {GL} _{n}\!{\left(K\right)}\mid {M^{\text{tr}}}M=E_{n}\right\}}\,.$

## Definition

A matrix $M\in \operatorname {GL} _{n}\!{\left(\mathbb {C} \right)}$ fulfilling
    ${{\overline {M}}^{\text{tr}}}M=E_{n}\,$
is called a unitary matrix. The set of all unitary matrices is called unitary group; it is denoted by
    $\operatorname {U} _{n}\!={\left\{M\in \operatorname {GL} _{n}\!{\left(\mathbb {C} \right)}\mid {{\overline {M}}^{\text{tr}}}M=E_{n}\right\}}\,.$
    Eigenvalues of an isometry

## Theorem

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, and let
    $\varphi \colon V\longrightarrow V$
denote a linear isometry. Then the modulus of every eigenvalue of $\varphi$ is $1$. In case ${\mathbb {K} }=\mathbb {R}$, only the eigenvalues $1$ and $-1$

### Proof

Let $\varphi (v)=\lambda v$ with $v\neq 0$, that is, $v$ is an eigenvector for the eigenvalue $\lambda \in {\mathbb {K} }$. Due to the isometry property, we have
    $\Vert {v}\Vert =\Vert {\varphi (v)}\Vert =\Vert {\lambda v}\Vert =\vert {\lambda }\vert \cdot \Vert {v}\Vert \,.$
Because of $\Vert {v}\Vert \neq 0$, this implies $\vert {\lambda }\vert =1$. In the real case this means $\lambda =\pm 1$.
$\Box$
In general, an isometry does not have necessarily an eigenvalue; however, if the dimension is odd, then there is an eigenvalue, see the following lecture.

## Lemma

The determinant of a linear isometry
    $\varphi \colon V\longrightarrow V$
on a euclidean vector space $V$ is
$1$
$-1$

### Proof

Due to Fact *****, we have
    ${M^{\text{tr}}}M={\begin{pmatrix}1&0&\cdots &\cdots &0\\0&1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&0\\0&\cdots &\cdots &0&1\end{pmatrix}}\,.$
Therefore, the statement follows from the multiplication theorem for the determinant and from Fact *****.
$\Box$
    Proper isometries

## Definition

An isometry on a euclidean vector space is called proper if its determinant
$1$
An isometry that is not proper, that is, its determinant is $-1$, is also called an improper isometry.

## Definition

Let $K$ be a field, and $n\in \mathbb {N} _{+}$. An orthogonal $n\times n$-matrix $M$ fulfilling
    $\det M=1\,$
is called a special orthogonal matrix. The set of all special orthogonal matrices is called special orthogonal group;
$\operatorname {SO} _{n}\!{\left(K\right)}$

## Definition

A unitary $n\times n$-matrix $M\in \operatorname {GL} _{n}\!{\left(\mathbb {C} \right)}$ fulfilling
    $\det M=1\,$
is called a special unitary matrix. The set of all special unitary matrices is called Special unitary group;
$\operatorname {SU} _{n}$
    Footnotes
* ↑ This formulation is due the fact that an inner product is defined only over $\mathbb {R}$ and $\mathbb {C}$. However, the formula that describes the standard inner product in the real case, works over every field.
* ↑ We say that such an orthonormal basis represents the orientation.