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

cross product.

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

${}a,b\in K$).

1. We have

   ${}x\times y=-(y\times x)\,.$

2. We have

   ${}(ax+by)\times z=a(x\times z)+b(y\times z)\,$

   and

   ${}z\times (ax+by)=a(z\times x)+b(z\times y)\,.$

3. We have

   ${}x\times y=0\,$

   if and only if ${}x$ and ${}y$ are linearly dependent.

4. We have

   ${}x\times (y\times z)+y\times (z\times x)+z\times (x\times y)=0\,.$

5. We have

   ${}\left\langle x\times y,z\right\rangle =\det(x\,,y\,,z)\,,$

   where ${}\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.

6. We have

   ${}\left\langle x,x\times y\right\rangle =0=\left\langle y,x\times y\right\rangle \,,$

   where ${}\left\langle -,-\right\rangle$ denotes the formal evaluation in the sense of the standard inner product.

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

${}\det {\left(u_{1},\,u_{2},\,u_{3}\right)}=1\,.$

Then the

cross product ${}x\times y$ can be computed with the coordinates of ${}x$ and ${}y$ with respect to this basis (and the formula from

Definition).

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

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is an isometry.
2. For all ${}u,v\in V$, we have ${}d(\varphi (u),\varphi (v))=d(u,v)$.
3. For all ${}v\in V$, we have ${}\Vert {\varphi (v)}\Vert =\Vert {v}\Vert$.
4. 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

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is an isometry.
2. 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 ${}W$.
3. 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

See exercise *****.

$\Box$

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

standard inner product.

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

${}{M^{\text{tr}}}M=E_{n}\,$

holds.

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

over ${}K$. It is denoted by ${}\operatorname {GL} _{n}\!{\left(K\right)}$.

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

over ${}K$. It is denoted by ${}\operatorname {SL} _{n}\!{\left(K\right)}$.

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 ${}-1$

are possible.

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

${}1$ or ${}-1$.

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

is ${}1$.

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;

it is denoted by ${}\operatorname {SO} _{n}\!{\left(K\right)}$.

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;

it is denoted by ${}\operatorname {SU} _{n}$.

Footnotes