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Linear Algebra: Companion notes

Abhay Soman

Section 6.3 Fundamental Homomorphism Theorem

We prove the Fundamental Homomorphism Theorem, a similar result as in the Group Theory course, in context of vector spaces. We begin with the following result.
Define \(\overline{T}\colon V/W\to X\) by \(\overline{T}(v+W)= T(v)\text{.}\) Since, \(W\leq\ker(T)\text{,}\) the map \(\overline{T}\) is well-defined. Checking of \(F\)-linearity, uniqueness of \(\overline{T}\text{,}\) and \(\overline{T}\circ\pi_W=T\) is left to the reader.
By Theorem 6.3.1, there is an \(F\)-linear map \(\overline{T}\colon V/\ker(T)\to\Im(T)\) given by \(v+\ker(T)\mapsto T(v)\text{.}\) This map is surjective (verify!). We show that \(\overline{T}\) is injective. If \(\overline{T}(v+\ker(T))=T(v)=0\) then, \(v\in\ker(T)\text{,}\) i.e., \(v+\ker(T)=0+\ker(T)\text{.}\) Hence, \(\overline{T}\) is injective. Therefore, \(\overline{T}\) is an \(F\)-isomorphism between \(V/\ker(T)\) and \(\Im(T)\text{.}\)
Since, \(V\) is finite-dimensional, its subspace \(\ker(T)\) is also finite-dimensional. Let \(\{v_1,\ldots,v_s\}\) be a basis for \(\ker(T)\text{.}\) Extend it to a basis for \(V\text{,}\) say \(\mathfrak{B}=\{v_1,\ldots,v_s,u_1,\ldots,u_r\}\text{.}\) Then, \(\{u_1+\ker(T),\ldots,u_r+\ker(T)\}\) is a basis for \(V/\ker(T)\text{.}\) Indeed, if \(\sum\alpha_i(u_i+\ker(T))=0+\ker(T)\) then \(\sum\alpha_iu_i\in\ker(T)\text{.}\) Therefore, there exists \(\beta_j\in F\) such that \(\sum\alpha_iu_i=\sum\beta_jv_j\text{.}\) By linear independence of \(\mathfrak{B}\text{,}\) \(\alpha_i=0\) for every \(i\text{.}\) Therefore, \(\{u_1+\ker(T),\ldots,u_r+\ker(T)\}\) is linearly independent over \(F\text{.}\) Given a vector \(v\in V\) there are uniquely determined \(\beta_i,\gamma_j\) such that \(v=\sum\beta_iv_i+\gamma_ju_j\text{,}\) i.e., \(v+\ker(T)=\big(\sum\gamma_ju_j\big)+\ker(T)\text{.}\) This shows that \(\{u_1+\ker(T),\ldots,u_r+\ker(T)\}\) spans \(V/\ker(T)\text{.}\) By Theorem 6.3.2 and Lemma 5.1.6,
\begin{equation*} \{\overline{T}(u_1+\ker(T)),\ldots,\overline{T}(u_r+\ker(T))\}=\{T(u_1),\ldots,T(u_r)\} \end{equation*}
is a basis for \(\Im(T)\text{.}\)
We define an \(F\)-linear map
\begin{equation*} \varphi\colon\ker(T)\bigoplus\Im(T)\to V \end{equation*}
by
\begin{equation*} (v_i,0)\mapsto v_i\quad\text{and}\quad \left(0,T(u_j)\right)\mapsto u_j. \end{equation*}
Therefore, \(\varphi\) maps a basis of \(\ker(T)\bigoplus\Im(T)\) onto a basis of \(V\text{,}\) and hence, \(\varphi\) is an \(F\)-isomorphism.
Both the assertions of the corollary now follows.

Remark 6.3.4.

Let \(T\colon\R^2\to\R^2\) be a linear map given by
\begin{equation*} e_1\mapsto 0\quad\text{and}\quad e_2\mapsto e_1. \end{equation*}
As \(T(e_1)=0\) we get that \(e_1\in\ker(T)\text{.}\) Since \(T(e_2)=e_1\text{,}\) we have \(e_1\in\Im(T)\text{.}\) By rank-nullity theorem (Theorem 4.5.3) we obtain the following.
\begin{equation*} \ker(T)=\langle e_1\rangle=\Im(T). \end{equation*}
So, \(V\neq\ker(T)\bigoplus\Im(T)\) (see Theorem 3.7.5). However, as Corollary 6.3.3 shows, \(V\) is isomorphic as a vector space to \(\ker(T)\bigoplus\Im(T)\text{.}\)
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