Fdtxjr

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 $dim(V) =dim(ker L)+dim(operatornameImL)$.

My attempt so far:

Let $(barv_1, ldots, barv_k)$ be the base of the subspace $ker L$. Since $(barv_1, ldots, barv_k)$ is linearly independent, we can add vectors to it to be the vector space $V$ 's base $(barv_1, ldots, barv_k, baru_k+1, ldots,baru_n)$. Now $dim(ker L)=k$ and $dim(V)=n$ and I think if I go on proving that $(L(baru_k+1), ldots, L(baru_n))$ is the base of the subspace $operatornameImL$, I'm going to end up with $dim(operatornameImL)=n-k$, which would prove the theorem.

First I'm going to prove that $(L(baru_k+1), ldots, L(baru_n))$ is a spanning set of the subspace $operatornameImL$. Let $barwin operatornameImL$, which means that there's a $barvin V$ for which $barw =L(barv)$. We know that $(barv_1, ldots, barv_k, baru_k+1, ldots, baru_n)$ is the base of $V$ so $$barv =a_1barv_1 +ldots +a_kbarv_k +a_k+1baru_k+1 +ldots +a_nbaru_n$$ for some $a_1, ldots, a_nin mathbbR$. Now
beginalign
barw &= L(barv) = L(a_1barv_1 +ldots +a_kbarv_k +a_k+1baru_k+1 +ldots +a_nbaru_n) \
&= a_1L(barv_1) +ldots +a_kL(barv_k) +a_k+1L(baru_k+1) +ldots +a_nL(baru_n) \
&= bar0 +ldots +bar0 +a_k+1L(baru_k+1) +ldots +a_nL(baru_n)
endalign
since $L(barv_1), ldots, L(barv_k)in ker L$. Now $operatornameImL =operatornamespan(L(baru_k+1), ldots, L(baru_n))$.

Next I'm trying to prove that $(L(baru_k+1), ldots, L(baru_n))$ is linearly independent. Assume
$$c_k+1L(baru_k+1) +ldots +c_nL(baru_n) =bar0$$
for some $c_k+1baru_k+1, ldots, c_nbaru_nin ker L$. Now
$$L(c_k+1baru_k+1 +ldots +c_nbaru_n) =bar0$$
which means that $c_k+1baru_k+1 +ldots +c_nbaru_nin ker L.$ 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 $c_k+1baru_k+1 +ldots +c_nbaru_nin ker L$, doesn't this mean that $c_k+1baru_k+1 +ldots +c_nbaru_n$ 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.

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
$$L(c_k+1overlineu_k+1+cdots+c_noverlineu_n) = overline0$$
which means that $c_k+1overlineu_k+1+cdots+c_noverlineu_ninoperatornameKer(L)$. 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 $overlineu_1,overlineu_2,dots,overlineu_k$ are a basis for the kernel, there are constants $c_1,c_2,dots,c_k$ so that
$$c_k+1overlineu_k+1+cdots+c_noverlineu_n = -c_1overlineu_1-cdots-c_koverlineu_k$$
$$c_1overlineu_1+cdots+c_koverlineu_k+c_k+1overlineu_k+1+cdots+c_noverlineu_n = 0$$
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 $L(overlineu_k+1),dots,L(overlineu_n)$ are linearly independent.

The proof that $L(overlineu_k+1),dots,L(overlineu_n)$ span the image was good all along, and this completes the the proof of the theorem.