[Math] Prove $\dim(V)=\dim(\ker L)+\dim(\operatorname{Im}L)$

linear algebraproof-verification

I'm trying to prove the following theorem:

Let $V$ and $U$ be vector spaces and assume $V$ has a finite dimension. Let $L: V \to U$ be a linear transformation. Now $\dim(V) =\dim(\ker L)+\dim(\operatorname{Im}L)$.

My attempt so far:

Let $(\bar{v}_1, \ldots, \bar{v}_k)$ be the base of the subspace $\ker L$. Since $(\bar{v}_1, \ldots, \bar{v}_k)$ is linearly independent, we can add vectors to it to be the vector space $V$ 's base $(\bar{v}_1, \ldots, \bar{v}_k, \bar{u}_{k+1}, \ldots,\bar{u}_n)$. Now $\dim(\ker L)=k$ and $\dim(V)=n$ and I think if I go on proving that $(L(\bar{u}_{k+1}), \ldots, L(\bar{u}_n))$ is the base of the subspace $\operatorname{Im}L$, I'm going to end up with $\dim(\operatorname{Im}L)=n-k$, which would prove the theorem.

First I'm going to prove that $(L(\bar{u}_{k+1}), \ldots, L(\bar{u}_n))$ is a spanning set of the subspace $\operatorname{Im}L$. Let $\bar{w}\in \operatorname{Im}L$, which means that there's a $\bar{v}\in V$ for which $\bar{w} =L(\bar{v})$. We know that $(\bar{v}_1, \ldots, \bar{v}_k, \bar{u}_{k+1}, \ldots, \bar{u}_n)$ is the base of $V$ so $$\bar{v} =a_1\bar{v}_1 +\ldots +a_k\bar{v}_k +a_{k+1}\bar{u}_{k+1} +\ldots +a_n\bar{u}_n$$ for some $a_1, \ldots, a_n\in \mathbb{R}$. Now
\begin{align}
\bar{w} &= L(\bar{v}) = L(a_1\bar{v}_1 +\ldots +a_k\bar{v}_k +a_{k+1}\bar{u}_{k+1} +\ldots +a_n\bar{u}_n) \\
&= a_1L(\bar{v}_1) +\ldots +a_kL(\bar{v}_k) +a_{k+1}L(\bar{u}_{k+1}) +\ldots +a_nL(\bar{u}_n) \\
&= \bar{0} +\ldots +\bar{0} +a_{k+1}L(\bar{u}_{k+1}) +\ldots +a_nL(\bar{u}_n)
\end{align}
since $L(\bar{v}_1), \ldots, L(\bar{v}_k)\in \ker L$. Now $\operatorname{Im}L =\operatorname{span}(L(\bar{u}_{k+1}), \ldots, L(\bar{u}_n))$.

Next I'm trying to prove that $(L(\bar{u}_{k+1}), \ldots, L(\bar{u}_n))$ is linearly independent. Assume
$$c_{k+1}L(\bar{u}_{k+1}) +\ldots +c_nL(\bar{u}_n) =\bar{0}$$
for some $c_{k+1}\bar{u}_{k+1}, \ldots, c_n\bar{u}_n\in \ker L$. Now
$$L(c_{k+1}\bar{u}_{k+1} +\ldots +c_n\bar{u}_n) =\bar{0}$$
which means that $c_{k+1}\bar{u}_{k+1} +\ldots +c_n\bar{u}_n\in \ker L.$ This is the part that I'm getting stuck at and I don't know how I should go on with proving the linear independence. Since I've now figured $c_{k+1}\bar{u}_{k+1} +\ldots +c_n\bar{u}_n\in \ker L$, doesn't this mean that $c_{k+1}\bar{u}_{k+1} +\ldots +c_n\bar{u}_n$ is a linear combination of $\ker L$'s base vectors? I feel like if it is true then I should use that information to move on but I don't know how to.

Best Answer

Since this is old but not properly answered, I'll knock it out. (See also the comment by Callus, which I didn't notice until I was done writing this)

Now $$L(c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n}) = \overline{0}$$ which means that $c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n}\in\operatorname{Ker}(L)$. This is the part that I'm getting stuck at and I don't know how I should go on with proving the linear independence.

Since $\overline{u}_1,\overline{u}_2,\dots,\overline{u}_k$ are a basis for the kernel, there are constants $c_1,c_2,\dots,c_k$ so that $$c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n} = -c_1\overline{u}_1-\cdots-c_k\overline{u}_k$$ $$c_1\overline{u}_1+\cdots+c_k\overline{u}_k+c_{k+1}\overline{u}_{k+1}+\cdots+c_{n}\overline{u}_{n} = 0$$ Now we use the linear independence of the $u_i$ to conclude that all of the $c_i$ are zero. That includes $c_{k+1}$ through $c_n$, so $L(\overline{u}_{k+1}),\dots,L(\overline{u}_{n})$ are linearly independent.

The proof that $L(\overline{u}_{k+1}),\dots,L(\overline{u}_{n})$ span the image was good all along, and this completes the the proof of the theorem.

Related Question