Fundamental Homomorphism Theorem

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.

Theorem 6.3.1. (Universal Property of Quotient Space).

Let \(V, X\) be vector spaces over a field \(F\) and \(W\) be a subspace of \(V\text{.}\) Let \(T\colon V\to X\) be an \(F\)-linear map such that \(W\leq\ker(T)\text{.}\) Then there is a unique linear map \(\overline{T}\colon V/W\to X\) such that \(\overline{T}\circ\pi_W=T\text{.}\)

Diagrammatically we require an \(F\)-linear map \(\overline{T}\colon V/W\to X\) such that \(\overline{T}\circ\pi_W=T\text{.}\)

\begin{equation*} \begin{CD}V@>\pi_W>>V/W\\@V T VV@.\\X@.\end{CD} \end{equation*}

Proof.

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.

Theorem 6.3.2. (Fundamental Homomorphism Theorem).

Let \(V, W\) be vector spaces over a field \(F\) and let \(T\colon V\to W\) be an \(F\)-linear map. Then,

\begin{equation*} V/\ker(T)\simeq\Im(T). \end{equation*}

Proof.

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 \(\Im(T)\text{.}\)

Corollary 6.3.3.

We keep notations of Theorem 6.3.2. We further assume that \(V\) is a finite-dimensional vector space. We have

\begin{equation*} \dim_F\big(V/\ker(T)\big)=\dim_FV-\dim_F\ker(T). \end{equation*}

Moreover,

\begin{equation*} V\simeq\ker(T)\bigoplus\Im(T). \end{equation*}

Proof.

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*}

\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 \(\ker(T)\bigoplus\Im(T)\text{.}\)