1.4 Linear maps
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Definitions. A map \(\phi \colon V\to W\) of vector spaces over \(\F \) is a linear map (or, in older books, linear transformation) if
\(\seteqnumber{0}{1.}{0}\)\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 \(\im \phi :=\set {\phi (v)\st v\in V}\leq W\).
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(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 \(1\leq i\leq m\),
\(\seteqnumber{0}{1.}{0}\)\begin{equation*} y_i=\sum _{j=1}^nA_{ij}x_j. \end{equation*}
Otherwise said, \(y\) is given by matrix multiplication: \(\mathbf {y}=A\bx \).
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(2) For any vector space \(V\), the identity map \(\id _V:V\to V\) is linear.
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(3) If \(\phi :V\to W\) and \(\psi :W\to U\) are linear then so is \(\psi \circ \phi :V\to U\).
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Definition. A linear map \(\phi :V\to W\) is a (linear) isomorphism if there is a linear map \(\psi :W\to V\) such that
\(\seteqnumber{0}{1.}{0}\)\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:
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Theorem 1.4 (Linearity is a linear condition). \(L(V,W)\) is a vector space under pointwise addition and scalar multiplication. Thus
\(\seteqnumber{0}{1.}{0}\)\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 \(\lambda \in \F \).
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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 \(\phi ,\psi \in L(V,W)\):
\(\seteqnumber{0}{1.}{0}\)\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:
\(\seteqnumber{0}{1.}{0}\)\begin{equation*} 0(v+\lambda w)=0=0+\lambda 0=0(v)+\lambda 0(w). \end{equation*}
We also define \(-\phi \) by
\(\seteqnumber{0}{1.}{0}\)\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 \(v\in V\),
\(\seteqnumber{0}{1.}{0}\)\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:
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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
\(\seteqnumber{0}{1.}{0}\)\begin{equation} \label {eq:4} \phi (v_i)=w_i,\qquad 1\leq i\leq n. \end{equation}
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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
\(\seteqnumber{0}{1.}{1}\)\begin{equation*} v=\lc {\lambda }{v}1n \end{equation*}
and so we define \(\phi (v)\) to be the only thing it could be:
\(\seteqnumber{0}{1.}{1}\)\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
\(\seteqnumber{0}{1.}{1}\)\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
\(\seteqnumber{0}{1.}{1}\)\begin{align*} v&=\lc \lambda {v}1n\\ w&=\lc \mu {v}1n. \end{align*} Then, for \(\lambda \in \F \),
\(\seteqnumber{0}{1.}{1}\)\begin{equation*} v+\lambda w=(\lambda _1+\lambda \mu _1)v_1+\dots +(\lambda _n+\lambda \mu _n)v_n \end{equation*}
whence
\(\seteqnumber{0}{1.}{1}\)\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
\(\seteqnumber{0}{1.}{1}\)\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. □
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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!).
Among the most important results in Algebra 1B is the famous rank-nullity theorem:
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: