MA22020: Advanced linear algebra

1.4 Linear maps

Definitions. A map $\phi \colon V\to W$ 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 $\psi \circ \phi :V\to U$.

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

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 1.4 (Linearity is a linear condition). $L(V,W)$ is a vector space under pointwise addition and scalar multiplication. Thus
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\begin{align*} (\phi +\psi )(v)&:=\phi (v)+\psi (v)\\ (\lambda \phi )(v)&:=\lambda \phi (v), \end{align*} for all $\phi ,\psi \in L(V,W)$, $v\in V$ and $\lambda \in \F $.

Proof. There is a lot to do here but it is all easy. First we must show that $\phi +\psi $, as defined above, really is a linear map when $\phi ,\psi \in L(V,W)$:
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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.

Now we need a zero element for our proposed vector space: 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*}

We also define $-\phi $ by
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\begin{equation*} (-\phi )(v)=-\phi (v), \end{equation*}

for $v\in V$ and check that it is also linear.

Finally, we must check all the vector space axioms which all follow from those of $W$. For example, for any $v\in V$,
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\begin{equation*} (\phi +\psi )(v)=\phi (v)+\psi (v)=\psi (v)+\phi (v)=(\psi +\phi )(v), \end{equation*}

so that $\phi +\psi =\psi +\phi $. The remaining axioms are left as a (rather boring) exercise. □

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

Proposition 1.5 (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:4} \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 $v\in V$. From Algebra 1B²,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 (1.1) 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 (1.1). 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 (1.1) for first $\phi $ and then $\phi '$. We conclude that $\phi =\phi '$ and we are done. □

² Proposition 1.3.4

Remark. In the context of Proposition 1.5, $\phi $ is an isomorphism if and only if $\lst {w}1n$ is a basis for $W$ (exercise³!).

³ This is question 6 on exercise sheet 1.

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

Theorem 1.6 (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 $\ker \phi =\set {0}$ and surjective if and only if $\im \phi =W$ we have:

Proposition 1.7. 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.