0.4 Linear algebra

A field is a commutative ring in which every nonzero element is invertible (colloquially, a place where we can do normal arithmetic). Linear algebra takes place over fields. We will use the following examples:

• •

  the field $\mathbb{Q}$ of rational numbers

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  the field $ℝ$ of real numbers

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  the field $ℂ$ of complex numbers

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  the fields ${𝔽}_p$ of integers modulo $p$, for $p$ prime.

If $V,W$ are vector spaces over a field $k$, then a linear map $T:V\to W$ is a function such that

$$T(\lambda v+\mu w)=\lambda T(v)+\mu T(w)$$

for all $\lambda ,\mu \in k$ and all $v,w\in V$. If bases have been chosen for $V$ and $W$ (and they are finite dimensional), then every linear map can be written as a matrix. The linear map $T$ is invertible if it is a bijection, in which case its inverse $T^{-1}$ is also a linear map. We write $\mathrm{Hom}(V,W)$ for the set of linear maps from $V$ to $W$, which is also vector space over $k$. If $V=W$, we write $\mathrm{End}(V)=\mathrm{Hom}(V,V)$. We write $\mathrm{GL}(V)$ for the group of invertible linear maps from $V$ to itself ($\mathrm{GL}$ stands for ’general linear’). If a basis of $V$ is chosen, and $dimV=n$, then $\mathrm{End}(V)$ is given by the $n\times n$ matrices over $k$ while $\mathrm{GL}(V)={\mathrm{GL}}_n(k)$ is the group of $n\times n$ invertible matrices.

If $T:V\to W$ is a linear map, then its kernel and image are

$$\ker (T)=\{v\in V:Tv=0\}$$

and

$$\mathrm{im}(T)=\{w\in W:w=Tv\text{ for some }v\in V\}.$$

A subspace of $V$ is a subset $W$ closed under addition and scalar multiplication.

If $W$ is a subspace of $V$, then the quotient space $V/W$ is the set of cosets (for addition) of $W$ in $V$. We denote its elements by

$$\overline{v}=v+W=\{v+w:w\in W\}.$$

In this situation, the map $V\to V/W$ sending $v$ to $\overline{v}$ is a surjective linear map whose kernel is $W$. If $T:V\to U$ is a linear map, then the map $T:V/\ker (T)\to U$ taking $\overline{v}$ to $T(v)$ gives a well-defined isomorphism from $V/\ker (T)$ to $\mathrm{im}(T)$. Compare the first isomorphism theorem in group theory, and also the rank-nullity theorem

$$dim\ker (T)+dim\mathrm{im}(T)=dimV.$$

If $V$ and $W$ are two vector spaces, then their (external) direct sum is

$$V\oplus W=\{(v,w):v\in V,w\in W\}$$

with componentwise addition and scalar multiplication. If $V$ and $W$ are both subspaces of some common space $U$, we say that $U$ is the internal direct sum of $V$ and $W$ if every element of $U$ can be written uniquely as $u=v+w$ for $v\in V$, $w\in W$. This is equivalent to requiring $U=V+W$ and $V\cap W=\{0\}$, or to requiring that the map

$$V\oplus W\to U$$

sending

$$(v,w)\mapsto v+w$$

is an isomorphism. Often in this situation we will simply say that $U$ is the direct sum of $V$ and $W$.

We can generalise this to more than one subspace. If $V_1,\mathrm{\dots },V_n$ are subspaces of $V$, then we say that $V$ is their internal direct sum if every element of $V$ can be written uniquely as $v_1+\mathrm{\dots }+v_n$ with $v_i\in V_i$ for all $i$. Equivalently, if the map

	$V_1\oplus \mathrm{\dots }\oplus V_n$	$\to V$
	$(v_1,\mathrm{\dots },v_n)$	$\mapsto v_1+\mathrm{\dots }+v_n$

is an isomorphism.

If $T:V\to V$ is a linear map from a vector space to itself, then an eigenvector of $T$ with eigenvalue $\lambda$ is a non-zero vector $v\in V$ such that $Tv=\lambda v$.

The linear map $T$ is diagonalizable if there is a basis of $V$ consisting of eigenvectors of $T$. This is equivalent to there being a basis for which the matrix of $T$ is diagonal.

For later use, we record the following theorem from linear algebra: if $T_1,\mathrm{\dots },T_n$ are linear maps $V\to V$ that commute with each other and that are diagonalizable, then there is a basis of $V$ consisting of simultaneous eigenvectors of the $T_i$. Equivalently, a basis for which the matrices of the $T_i$ are all diagonal.