MA20216: Algebra 2A

1.4 Linear maps

Definitions. A map $\phi :V\to W$ of vector spaces over $\F $ is a linear map (or, in older books, linear transformation) if
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\begin{align*} \phi (v+w)&=\phi (v)+\phi (w)\\ \phi (\lambda v)&=\lambda \phi (v), \end{align*} for all $v,w\in V$, $\lambda \in \F $.

The kernel of $\phi $ is $\ker \phi :=\set {v\in V\st \phi (v)=0}\leq V$.

The image of $\phi $ is $\im \phi :=\set {\phi (v)\st v\in V}\leq W$.

Remark. $\phi $ is linear if and only if
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\begin{equation*} \phi (v+\lambda w)=\phi (v)+\lambda \phi (w), \end{equation*}

for all $v,w\in V$, $\lambda \in \F $, which has the virtue of being only one thing to prove.

Examples.
- (1) $A\in M_{m\times n}(\F )$ determines a linear map $\phi _A:\F ^n\to \F ^m$ by $\phi _A(x)=y$ where, for $1\leq i\leq m$,
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  \begin{equation*} y_i=\sum _{j=1}^nA_{ij}x_j. \end{equation*}
  
  Otherwise said, $y$ is given by matrix multiplication: $\mathbf {y}=A\bx $.
- (2) For any vector space $V$, the identity map $\id _V:V\to V$ is linear.
- (3) If $\phi :V\to W$ and $\psi :W\to U$ are linear then so is $\psi \circ \phi :V\to U$.
- (4) Recall that $c$ is the vector space of convergent sequences.
  
  The map $\lim _{n\to \infty }:(a_n)_{n\in \N }\mapsto \lim _{n\to \infty }a_n:c\to \R $ is linear thanks to the Algebra of Limits Theorem in Analysis 1.
- (5) $\int _a^b:f\mapsto \int _a^bf:C^0[a,b]\to \R $ is also linear.

Definition. A linear map $\phi :V\to W$ is a (linear) isomorphism if there is a linear map $\psi :W\to V$ such that
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\begin{equation*} \psi \circ \phi =\id _V,\qquad \phi \circ \psi =\id _W. \end{equation*}

If there is an isomorphism $V\to W$, say that $V$ and $W$ are isomorphic and write $V\cong W$.

In Algebra 1B, we saw:

Lemma 1.5 (Algebra 1B, lemma 1.2.3). $\phi :V\to W$ is an isomorphism if and only if $\phi $ is a linear bijection (and then $\psi =\phi ^{-1}$).

1.4.1 Vector spaces of linear maps

Notation. For vector spaces $V,W$ over $\F $, denote by $L_{\F }(V,W)$ (or simply $L(V,W)$) the set $\set {\phi :V\to W\st \text {$\phi $ is linear}}$ of linear maps from $V$ to $W$.

Theorem 1.6 (Linearity is a linear condition). $L(V,W)$ is a vector space under pointwise addition and scalar multiplication. Otherwise said, $L(V,W)\leq W^V$.

Proof. It is enough to show that $L(V,W)$ is a vector subspace of $W^V$, that is, is non-empty and closed under addition and scalar multiplication.

First observe that the zero map $0:v\mapsto 0\in W$ is linear:
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\begin{equation*} 0(v+\lambda w)=0=0+\lambda 0=0(v)+\lambda 0(w). \end{equation*}

In particular, $L(V,W)$ is non-empty.

Now let $\phi ,\psi \in L(V,W)$ and show that $\phi +\psi $ is linear:
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\begin{align*} (\phi +\psi )(v+\lambda w) &=\phi (v+\lambda w)+\psi (v+\lambda w)\\ &=\phi (v)+\lambda \phi (w)+\psi (v)+\lambda \psi (w)\\ &=(\phi (v)+\psi (v))+\lambda (\phi (w)+\psi (w))\\ &=(\phi +\psi )(v)+\lambda (\phi +\psi )(w), \end{align*} for all $v,w\in V$, $\lambda \in \F $. Here the first and last equalities are just the definition of pointwise addition while the middle equalities come from the linearity of $\phi ,\psi $ and the vector space axioms of $W$.

Similarly, it is a simple exercise to see that if $\mu \in \F $ and $\phi \in L(V,W)$ then $\mu \phi $ is also linear. □

1.4.2 Linear maps and matrices

Recall from Algebra 1B §1.5:

Definition. Let $V,W$ be finite-dimensional vector spaces over $\F $ with bases $\cB :\lst {v}1n$ and $\cB ':\lst {w}1m$ respectively. Let $\phi \in L(V,W)$. The matrix of $\phi $ with respect to $\cB ,\cB '$ is the matrix $A=(A_{ij})\in M_{m\times n}(\F )$ defined by:
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\begin{equation} \label {eq:27} \phi (v_j)=\sum _{i=1}^mA_{ij}w_{i}, \end{equation}

for all $\bw 1jn$.

In the special case where $V=W$ and $\cB =\cB '$, we call $A$ the matrix of $\phi $ with respect to $\cB $.

Thus the recipe for computing $A$ is: expand $\phi (v_j)$ in terms of $\lst {w}1m$ to get the $j$-th column of $A$.

Equivalently, $\phi (\lc {x}v1n)=\lc {y}w1m$ where

\begin{equation*} \by =A\bx . \end{equation*}

There is a fancy way to say all this: recall that a basis $\cB :\lst {v}1n$ of $V$ gives rise to a linear isomorphism $\phi _{\cB }:\F ^n\to V$ via

\begin{equation} \label {eq:3} \phi _{\cB }\vec \lambda 1n=\sum _{i=1}^n\lambda _iv_i. \end{equation}

Now the relation between $\phi $ and $A$ is that

\begin{equation*} \phi =\phi _{\cB '}\circ \phi _A\circ \phi _{\cB }^{-1} \end{equation*}

or, equivalently, $\phi _{\cB '}\circ \phi _A=\phi \circ \phi _{\cB }$ so that the following diagram commutes:

(math image)

(The assertion that such a diagram commutes is simply that the two maps one builds by following the arrows in two different ways coincide. However, the diagram also helps us keep track of where the various maps go!)

The map $\phi \mapsto A$ is a linear isomorphism $L(V,W)\cong M_{m\times n}(\F )$ which also plays well with composition and matrix multiplication: if $U$ is a third vector space with basis $\cB ''$ and $\psi \in L(W,U)$ has matrix $B$ with respect to $\cB ',\cB ''$ then $\psi \circ \phi $ has matrix $BA$ with respect to $\cB ,\cB ''$. This gives us a compelling dictionary between linear maps and matrices.

1.4.3 Extension by linearity

A linear map of a finite-dimensional vector space is completely determined by its action on a basis. More precisely:

Proposition 1.7 (Extension by linearity). Let $V,W$ be vector spaces over $\F $. Let $\lst {v}1n$ be a basis of $V$ and $\lst {w}1n$ any vectors in $W$.

Then there is a unique $\phi \in L(V,W)$ such that
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\begin{equation} \label {eq:2} \phi (v_i)=w_i,\qquad 1\leq i\leq n. \end{equation}

Proof. We need to prove that such a $\phi $ exists and that there is only one. We prove existence first.

Let $v\in V$. By Proposition 1.1, we know there are unique $\lst \lambda 1n\in \F $ for which
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\begin{equation*} v=\lc {\lambda }{v}1n \end{equation*}

and so we define $\phi (v)$ to be the only thing it could be:
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\begin{equation*} \phi (v):=\lc \lambda {w}1n. \end{equation*}

Let us show that this $\phi $ does the job. First, with $\lambda _i=1$ and $\lambda _j=0$, for $i\neq j$, we see that
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\begin{equation*} \phi (v_i)=\sum _{j\neq i}0w_j+1w_i=w_i \end{equation*}

so that (1.4) holds. Now let us see that $\phi $ is linear: let $v,w\in V$ with
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\begin{align*} v&=\lc \lambda {v}1n\\ w&=\lc \mu {v}1n. \end{align*} Then, for $\lambda \in \F $,
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\begin{equation*} v+\lambda w=(\lambda _1+\lambda \mu _1)v_1+\dots +(\lambda _n+\lambda \mu _n)v_n \end{equation*}

whence
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\begin{align*} \phi (v+\lambda w) &=(\lambda _1+\lambda \mu _1)w_1+\dots +(\lambda _n+\lambda \mu _n)w_n\\ &=(\lc \lambda {w}1n)+\lambda (\lc \mu {w}1n)\\ &=\phi (v)+\lambda \phi (w). \end{align*}

For uniqueness, suppose that $\phi ,\phi '\in L(V,W)$ both satisfy (1.4). Let $v\in V$ and write $v=\lc \lambda {v}1n$. Then
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\begin{align*} \phi (v)&=\lambda _1\phi (v_1)+\dots +\lambda _n\phi (v_n)\\ &= \lc \lambda {w}1n\\ &=\lambda _1\phi '(v_1)+\dots +\lambda _n\phi '(v_n)\\ &=\phi '(v), \end{align*} where the first and last equalities come from the linearity of $\phi ,\phi '$ and the middle two from (1.4) for first $\phi $ and then $\phi '$. We conclude that $\phi =\phi '$ and we are done. □

Remark. In the context of Proposition 1.7, $\phi $ is an isomorphism if and only if $\lst {w}1n$ is a basis for $W$ (exercise³!).

³ This is question 2 on exercise sheet 2.

1.4.4 The rank-nullity theorem

Among the most important results in Algebra 1B is the famous rank-nullity theorem:

Theorem 1.8 (Rank-nullity). Let $\phi :V\to W$ be linear with $V$ finite-dimensional. Then
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\begin{equation*} \dim \im \phi +\dim \ker \phi =\dim V. \end{equation*}

Using this, together with the observation that $\phi $ is injective if and only if $\ker \phi =\set {0}$, we have:

Proposition 1.9. Let $\phi :V\to W$ be linear with $V,W$ finite-dimensional vector spaces of the same dimension: $\dim V=\dim W$.

Then the following are equivalent:
- (1) $\phi $ is injective.
- (2) $\phi $ is surjective.
- (3) $\phi $ is an isomorphism.

Remark. Proposition 1.9 is flat-out false for infinite-dimensional $V,W$. For example: let $S:\R ^{\N }\to \R ^{\N }$ be the shift operator:
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\begin{equation*} S((a_0,a_1,\dots )):=(a_1,\dots ). \end{equation*}

We readily check that:
- • $S$ is linear;
- • $S$ surjects;
- • $S$ is not injective. For example: $S((1,0,0,\dots ))=0$.