Fundamental Theorem of Linear Maps

theoremvector-spaces

Intuition

Formal Statement

Consider a finite-dimensional vector space $V$, and $T\in \mathcal{L}(V,W)$. Then $\mathrm{range}T$ is a finite-dimensional subspace of $W$ and,

$$\dim V=\dim (\mathrm{null}T)+\dim (\mathrm{range}T).$$

Corollaries

1. If $V,W$ are finite-dimensional vector spaces with $\dim V>\dim W$, there exist no injective linear maps $T:V\to W$.
2. If $V,W$ are finite-dimensional vector spaces with $\dim V<\dim W$, there exist no surjective linear maps $T:V\to W$.
3. A homogenous system with more variables than equations has non-trivial solutions.
4. A system with more equations than variables has no solution for some choice of constant terms.

Proof

We want to show: 5) $\mathrm{range}T$ is finite-dimensional. 6) $\dim$ add correctly, so we need bases for $V$, $\mathrm{null}T$, $\mathrm{range}T$.

Since $\mathrm{null}T$ is a subspace of a finite-dimensional vector space $V$, $\mathrm{null}T$ is finite-dimensional.Let $\dim \mathrm{null}T=m\le \dim V$, and let $u_1,\dots ,u_m$ be a basis of $\mathrm{null}T$. Thus, $u_1,\dots ,u_m$ is a linearly independent list in $V$, which is finite-dimensional, so $u_1,\dots ,u_m$ can be extended to a basis of $V$:

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

Thus, $\dim V=m+n$. To finish the proof, we claim $T{v}_1,\dots ,T{v}_n$ is a basis of $\mathrm{range}T$:

Let $v\in V$. Then,

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

Apply $T$ to both,

$$Tv=0+\cdots +0+b_{1}T{v}_1+\cdots +b_{n}T{v}_n,$$

So $T{v}_i$ span $\mathrm{range}T$.

Now we show that $T{v}_1,\dots ,T{v}_n$ is linearly independent. Suppose $c_1,\dots ,c_n\in \mathbb{F}$ and $c_{1}T{v}_1+\cdots +c_{n}T{v}_n=0$, so $T(c_1{v}_1+\cdots +c_n{v}_n)=0$, or equivalently $c_1{v}_1+\cdots +c_n{v}_n\in \mathrm{null}T$. Hence $c_1{v}_1+\cdots +c_n{v}_n=d_1{u}_1+\cdots +d_m{u}_m$ since $u_i$‘s are a basis of $\mathrm{null}T$. Since $u$‘s + $v$‘s are a basis of $V_1$, they are linearly independent, so there exist no non-trivial linear combination of $u$‘s and $v$‘s equal to $0$. Therefore, $T{v}_1,\dots ,T{v}_n$ are linearly independent. $\square$