
# 6.5 The dimension formula

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The next theorem is the key result of this chapter. It relates the dimension of the kernel and range of a linear map.

Theorem 6.5.1.  Let $$V$$ be a finite-dimensional vector space and $$T:V\to W$$ be a linear map. Then $$\range(T)$$ is a finite-dimensional subspace of $$W$$ and
$\label{eq:dim formula} \dim(V) = \dim(\kernel(T)) + \dim(\range(T)). \tag{6.5.1}$

Proof.

Let $$V$$ be a finite-dimensional vector space and $$T\in \mathcal{L}(V,W)$$. Since $$\kernel(T)$$ is a subspace of $$V$$, we know that $$\kernel(T)$$ has a basis $$(u_1,\ldots, u_m)$$. This implies that $$\dim(\kernel(T))=m$$. By the Basis Extension Theorem, it follows that $$(u_1,\ldots,u_m)$$ can be extended to a basis of $$V$$, say $$(u_1,\ldots,u_m,v_1,\ldots,v_n)$$, so that $$\dim(V)=m+n$$.

The theorem will follow by showing that $$(Tv_1,\ldots, Tv_n)$$ is a basis of $$\range(T)$$ since this would imply that $$\range(T)$$ is finite-dimensional and $$\dim(\range(T))=n$$, proving Equation 6.5.1.

Since $$(u_1,\ldots,u_m,v_1,\ldots,v_n)$$ spans $$V$$, every $$v\in V$$ can be written as a linear combination of these vectors; i.e.,

\begin{equation*}
v = a_1 u_1 + \cdots + a_m u_m + b_1 v_1 + \cdots + b_n v_n,
\end{equation*}
where $$a_i,b_j\in \mathbb{F}$$. Applying $$T$$ to $$v$$, we obtain
\begin{equation*}
Tv = b_1 T v_1 + \cdots + b_n T v_n,
\end{equation*}

where the terms $$Tu_i$$ disappeared since $$u_i\in \kernel(T)$$. This shows that $$(Tv_1,\ldots, Tv_n)$$ indeed spans $$\range(T)$$.

To show that $$(Tv_1,\ldots, Tv_n)$$ is a basis of $$\range(T)$$, it remains to show that this list is linearly independent. Assume that $$c_1,\ldots, c_n \in \mathbb{F}$$ are such that

$c_1 T v_1 + \cdots + c_n T v_n =0.$

By linearity of $$T$$, this implies that

$T(c_1 v_1 + \cdots + c_n v_n) = 0,$

and so $$c_1 v_1 + \cdots + c_n v_n\in \kernel(T)$$. Since $$(u_1,\ldots,u_m)$$ is a basis of $$\kernel(T)$$, there must exist scalars $$d_1,\ldots,d_m\in\mathbb{F}$$ such that

\begin{equation*}
c_1 v_1 + \cdots + c_n v_n = d_1 u_1 + \cdots + d_m u_m.
\end{equation*}

However, by the linear independence of $$(u_1,\ldots, u_m,v_1,\ldots, v_n)$$, this implies that all coefficients $$c_1=\cdots =c_n=d_1=\cdots =d_m=0$$. Thus, $$(Tv_1,\ldots, Tv_n)$$ is linearly independent, and we are done.

Example 6.5.2. Recall that the linear map $$T:\mathbb{R}^2 \to \mathbb{R}^2$$ defined by $$T(x,y)=(x-2y,3x+y)$$ has $$\kernel(T)=\{0\}$$ and $$\range(T)=\mathbb{R}^2$$. It follows that

$\dim(\mathbb{R}^2) = 2 = 0+2 =\dim(\kernel(T)) + \dim(\range(T)).$

Corollary 6.5.3.  Let $$T$$ in $$\mathcal{L}(V,W)$$.

1. If $$\dim(V)>\dim(W)$$, then $$T$$ is not injective.
2. If $$\dim(V)<\dim(W)$$, then $$T$$ is not surjective.

Proof.

By Theorem 6.5.1, we have that

\begin{equation*}
\begin{split}
\dim(\kernel(T)) &= \dim(V) - \dim(\range(T))\\
&\ge \dim(V) - \dim(W)>0.
\end{split}
\end{equation*}
Since $$T$$ is injective if and only if $$\dim(\kernel(T))=0$$, $$T$$ cannot be injective.
Similarly,
\begin{equation*}
\begin{split}
\dim(\range(T)) &= \dim(V) - \dim(\kernel(T))\\
&\le \dim(V)  < \dim(W),
\end{split}
\end{equation*}
and so $$\range(T)$$ cannot be equal to $$W$$. Hence, $$T$$ cannot be surjective.

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