Kernel, Image and the Rank-Nullity Theorem

Let \(\,V\,\) and \(\,W\,\) be vector spaces over the field \(\,K.\)

Definition. \(\\\)

The kernel of a linear transformation \(\,F\in\text{Hom}(V,W)\ \) is the set of all elements of the space \(\,V,\ \) which are mapped into the zero vector of the space \(\,W:\)

\[\text{Ker}\,F\ :\,=\ \{\,v\in V:\ F(v)=\theta_W\}\,.\]

In other words, the kernel of a homomorphism \(\,F\in\text{Hom}(V,W)\ \) is the preimage (inverse image) of the one-element set consisting of the zero vector of the space \(\,W: \,\text{Ker}\,F\,=\,F^{-1}\{\,\theta_W\}.\)

Definition. \(\\\)

The image of a linear transformation \(\,F\in\text{Hom}(V,W)\ \) is the image of the space \(\,V\ \) under the map \(\,F,\ \ \) i.e. \(\,\) it is the set of values of the map \(\,F:\)

\[\text{Im}\,F\ :\,=\ F(V)\ \equiv\ \{\,F(v):\ v\in V\,\}\,.\]

Obviously, \(\ \,\text{Ker}\,F\subset V,\ \,\text{Im}\,F\subset W.\ \) \(\,\) Moreover

Theorem 6. \(\\\)

The kernel of a linear transformation \(\,F\in\text{Hom}(V,W)\ \) is a subspace of the vector space \(\,V,\ \) whereas the image of \(\,F\,\) is a subspace of the vector space \(\,W:\)

\[\text{Ker}\,F < V,\qquad\text{Im}\,F < W.\]

The Proof is based on the following criterion for a subspace (see Chapter 1.). A subset \(\,X\,\) of the vector space \(\,V\,\) is a subspace of \(\,V\,\) if, and only if, \(\,X\,\) is closed under vector addition and scalar multiplication, that is iff for any two vectors from \(\,X,\,\) every linear combination of these vectors also belongs to \(\,X.\)

So let’s assume that \(\ v_1,\,v_2\in\text{Ker}\,F:\ \) \(\ F(v_1)=F(v_2)=\theta_W\,.\)

Then \(\ \ F(a_1\,v_1+a_2\,v_2)\ =\ a_1\,F(v_1)+a_2\,F(v_2)\ =\ \theta_W\,,\) \(\\\) meaning that \(\ \,a_1\,v_1+a_2\,v_2\in\text{Ker}\,F\ \,\) \(\ \forall\ \ a_1,a_2\in K.\) \(\,\) Thus we have proved that \(\ \text{Ker}\,F < V.\)

Now let \(\,\ w_1,\,w_2\in\text{Im}\,F.\ \ \) Then \(\ \ w_1=F(v_1),\ \,w_2=F(v_2)\,,\ \,\) \(\ \exists\ v_1,\ \exists\ v_2\in V.\)

Then for \(\ \forall\ \ b_1,b_2\in K:\ \) \(\ b_1\,w_1+\,b_2\,w_2\ =\ b_1\,F(v_1)\,+\,b_2\,F(v_2)\ =\ F(b_1\,v_1+\,b_2\,v_2)\in\text{Im}\,F.\ \) This proves that \(\ \ \text{Im}\,F < W.\quad\bullet\)

The \(\,\text{Ker}\,F\ \) and \(\,\text{Im}\,F\ \) being vector (sub)spaces, the following two definitions are sensible:

\[\begin{split}\begin{array}{rclcl} \text{nul}\ F & :\,= & \dim\,\text{Ker}\,F & \qquad\quad & \text{nullity of the homomorphism}\ F \\ \text{rk}\ F & :\,= & \dim\,\text{Im}\,F & \qquad\quad & \text{rank of the homomorphism}\ F \end{array}\end{split}\]

Theorem 7. \(\ \) The Rank-Nullity Theorem. \(\\\)

If \(\,F\ \) is a linear transformation from the vector space \(\,V\ \) to the vector space \(\,W,\ \) then

(1)\[\text{nul}\ F\ +\ \text{rk}\ F\ =\ \dim\,V\,.\]

Proof. \(\,\) Assume that \(\ \text{nul}\,F = k,\ \) while \(\ U = \{u_1,\,u_2,\,\dots,\,u_k\}\ \) is a basis of \(\ \text{Ker}\,F.\ \) \(\\\) Since \(\,\text{Ker}\,F < V,\ \,\) the set \(\,U\,\) can be extended to the basis \(\,B\,\) of the whole space \(\,V:\)

\[B\ \,=\ \,U\,\cup\,Y\ \,=\ \, \{\,u_1,\,u_2,\,\dots,\,u_k,\ y_1,\,y_2,\,\dots,\,y_r\,\}\,.\]

Within this notation \(\ \dim\,V=\,k+r,\ \) where \(\,k = \text{nul}\,F.\ \) To prove the thesis (1), it is enough to show that \(\, r = \text{rk}\ F.\ \) For that purpose we shall demonstrate that the image of the set \(\,Y:\)

(2)\[F(Y)\ :\,=\ \{\,Fy_1,\,Fy_2,\,\dots,\,Fy_r\,\}\]

is a basis of \(\,\text{Im}\,F.\ \,\) This shall be accomplished in two steps.

1.) To show that the set (2) is linearly independent, we assume that

\[c_1\,Fy_1\,+\;c_2\,Fy_2\,+\;\ldots\;+\;c_r\,Fy_r\ =\ \theta_W\,,\qquad c_1,\,c_2,\,\ldots,\,c_r\,\in\,K.\]

Due to linear independence of vectors in the basis \(\,B,\,\) we obtain

\[ \begin{align}\begin{aligned}F\left(c_1\,y_1\,+\;c_2\,y_2\,+\;\ldots\;+\;c_r\,y_r\right)\ =\ \theta_W\,,\\c_1\,y_1\,+\;c_2\,y_2\,+\;\ldots\;+\;c_r\,y_r\ \in\ \text{Ker}\,F\,,\\c_1\,y_1\,+\;c_2\,y_2\,+\;\ldots\;+\;c_r\,y_r\ =\ -\ b_1\,u_1\,-\;b_2\,u_2\,+\;\ldots\;-\ b_k\,u_k\,,\\b_1\,u_1\,+\;b_2\,u_2\,+\;\ldots\;+\;b_k\,u_k\ +\ c_1\,y_1\,+\;c_2\,y_2\,+\;\ldots\;+\;c_r\,y_r\ =\ \theta_V\,,\\b_1=\,b_2=\;\ldots\;=\;b_k\,=\;c_1=\,c_2=\;\ldots\;=\;c_r\ =\ 0\,.\end{aligned}\end{align} \]

Thus we have evidenced the implication

\[c_1\,Fy_1\,+\;c_2\,Fy_2\,+\;\ldots\;+\;c_r\,Fy_r\ =\ \theta_W \quad\Rightarrow\quad c_1\,=\,c_2\,=\,\ldots\,=\,c_r\ =\ 0\,,\]

which expresses the linear independence of the set \(\ F(Y)\).

2.) To show that the set \(\ F(Y)\ \) spans \(\,\text{Im}\,F,\ \) we consider an arbitrary vector \(\,w\in \text{Im}\,F:\)

\[w\ =\ F(v),\quad v=b_1\,u_1+\;\ldots\;+\;b_k\,u_k\;+\ c_1\,y_1+\;\ldots\;+\;c_r\,y_r\,.\]

Taking into account that \(\ \,Fu_i=\theta_W,\ \,i=1,2,\dots,k,\ \,\) we get

\begin{eqnarray*} w & = & F\,(b_1\,u_1+\;\ldots\;+\;b_k\,u_k\;+\ c_1\,y_1+\;\ldots\;+\;c_r\,y_r) \\ & = & b_1\,Fu_1+\;\ldots\;+\;b_k\,Fu_k\;+\ c_1\,Fy_1+\;\ldots\;+\;c_r\,Fy_r \\ & = & c_1\,Fy_1+\;\ldots\;+\;c_r\,Fy_r\,\in\,L(Fy_1,\dots,Fy_r)\,. \end{eqnarray*}

Thus \(\ \text{Im}\,F\subset L(Fy_1,\dots,Fy_r).\ \) On the other hand, \(\ \text{Im}\,F\supset L(Fy_1,\dots,Fy_r),\ \) \(\\\) whereby \(\ \text{Im}\,F=L(Fy_1,\dots,Fy_r)=L\left(F(Y)\right).\)

As a linearly independent spanning set, \(\ F(Y)\ \) is a basis of \(\ \text{Im}\,F,\ \ \) and \(\ r\,=\,\text{rk}\ F.\) \(\ \ \bullet\)

The following Criterion for the Isomorphism of Vector Spaces is based on the above-mentioned Theorem 7.:

Theorem 8. \(\\\)

Any two finite-dimensional vector spaces over a field \(\,K\ \) are isomorphic \(\\\) if, and only if, they are of the same dimension:

\[V\,\simeq\,W\qquad\Leftrightarrow\qquad\dim\,V\,=\ \dim\,W\,.\]

Proof.

\(\Rightarrow\ :\ \) We assume that the spaces \(\ V\ \) and \(\ W\ \) are isomorphic: \(\ V\simeq W,\ \,\) that is \(\ \text{Iso}(V,W)\neq\emptyset.\)

Let \(\ F\ \) be an isomorphism of the space \(\,V\,\) onto the space \(\,W.\ \) The map \(\ F\ \) being bijective, \(\\\) the preimage of each element of \(\,W\,\) is a set containing exactly one element of \(\ V.\ \) \(\\\) In particular, the zero vector \(\ \theta_W\ \) is the image of the zero vector \(\ \theta_V\ \) only, \(\,\) meaning that \(\ \text{nul}\,F=\,\dim\,\text{Ker}\,F=0.\)

But \(\ F\ \) is also surjective: \(\ F(V)=\text{Im}\,F=W,\ \,\) wherefrom \(\ \,\text{rk}\ F=\dim\,\text{Im}\,F=\,\dim\,W.\ \)

Using Theorem 7., \(\,\) we obtain

\[\dim\,V\ =\ \,\text{nul}\,F\ +\ \,\text{rk}\,F\ =\ 0\ +\ \dim\,W\ =\ \dim\,W.\]

\(\Leftarrow\ :\ \) Let \(\ \,\dim\,V=\,\dim\,W=n.\)

Then every basis of \(\,V,\,\) as well as every basis of \(\,W,\,\) has \(\,n\ \) elements. Suppose that the set \(\ B = \{\,v_1,v_2,\,\dots,\,v_n\,\}\ \) is a basis of the space \(\,V,\ \) whereas \(\ C = \{\,w_1,w_2,\,\dots,\,w_n\,\}\ \) is a basis of the space \(\,W.\ \) On the grounds of the Corollary of Theorem 5., we define the linear transformation \(\,F:\,V\rightarrow W\ \) by giving its values on vectors in \(\ B:\)

(3)\[F(v_i)\ :\,=\ w_i\,,\qquad i=1,2,\dots,n.\]

Then the image of an arbitrary vector \(\ \,v = \displaystyle\sum_{i\,=\,1}^n\ a_i\,v_i \in V\ \,\) is given by

(4)\[F\left(\,\sum_{i\,=\,1}^n\ a_i\,v_i\right)\ \ =\ \ \sum_{i\,=\,1}^n\ a_i\,Fv_i\ \ =\ \ \sum_{i\,=\,1}^n\ a_i\,w_i\,.\]

If \(\,w=F(v),\ \) where \(\ v\in V,\ w\in W,\ \) then the coordinates of \(\,w\,\) in the basis \(\,C\,\) are equal to the respective coordinates of \(\,v\,\) in the basis \(\,B.\)

Given a basis, the vectors are in a one-to-one correspondence with the families of their coordinates. Consequently, the linear transformation \(\,F\,\) defined by Eqs. (3) or (4) is a bijection, hence an isomorphism. So \(\,\text{Iso}(V,W)\neq\emptyset,\ \) and \(\,V\simeq W.\) \(\ \ \bullet\)

Isomorphic vector spaces may consist of objects of different kinds. Having the same structure, they are however mathematically equivalent.

Corollary.

All \(\,n\)-dimensional vector spaces over a field \(\,K\ \) are isomorphic to the space \(\,K^n.\)