Fundamental Theorem of Linear Maps

Fundamental theorem of linear maps

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V,W)$. Then $\text{range }T$ is finite-dimensional and

$$\dim V=\dim \text{null }T+\dim \text{range }T$$

Proof. Let $u_1,\dots ,u_m$ be a basis of $\text{null }T$; thus $\dim \text{null }T=m$. The linearly independent list $u_1,\dots ,u_m$ can be extended to a basis

$$u_1,\dots ,u_m,v_1,\dots ,v_n$$

of $V$ (every linearly independent list extends to a basis). Thus, $\dim V=m+n$.

To complete the proof, we need to show that $\text{range }T$ is finite-dimensional and $\dim \text{range }T=n$. We will do this by proving that $T{v}_1,\dots ,T{v}_n$ is a basis of $\text{range }T$, for which we need to show that it spans $\text{range }T$ and it is linearly independent.

First, we show that $T{v}_1,\dots ,T{v}_n$ spans $\text{range }T$. Let $v\in V$. Because $u_1,\dots ,u_m,v_1,\dots ,v_n$ spans $V$, we can write

$$v=a_1{u}_1+\cdots +a_m{u}_m+b_1{v}_1+\cdots +b_n{v}_n,$$

where the $a$‘s and $b$‘s are in $\mathbb{F}$. Applying $T$ to both sides of this equation, we get

$$Tv=b_{1}T{v}_1+\cdots +b_{n}T{v}_n$$

where terms of the form $T{u}_k$ disappeared; because $u_k$ is in $\text{null }T$, any $T{u}_k=0$. The last equation implies that the list $T{v}_1,\dots ,T{v}_n$ spans $\text{range }T$. In particular, $\text{range }T$ is finite-dimensional.

Finally, we show $T{v}_1,\dots ,T{v}_n$ is linearly independent. Suppose $c_1,\dots ,c_n\in \mathbb{F}$ and

$$c_{1}T{v}_1+\cdots +c_{n}T{v}_n=0$$

Then

$$T(c_1{v}_1+\cdots +c_n{v}_n)=0$$

Hence:

$$c_1{v}_1+\cdots +c_n{v}_n\in \text{null }T$$

Because $u_1,\dots ,u_m$ spans $\text{null }T$, we can write

$$c_1{v}_1+\cdots +c_n{v}_n=d_1{u}_1+\cdots +d_m{u}_m$$

where the $d$‘s are in $\mathbb{F}$. This equation implies that all the $c$‘s and $d$‘s are $0$, because $u_1,\dots ,u_m,v_1,\dots ,v_n$ is linearly independent. Thus, $T{v}_1,\dots ,T{v}_n$ is linearly independent and hence is a basis of $\text{range }T$, as desired.