# Proof about Matrix representation in Linear transformation.

If dimV = m , dimU = n , F : V -> U : linear dim(ImF)=r then there exists bases of V and U s.t matrix representation of F has the form $$\pmatrix{I_r & 0\\0 & 0}$$ (where Ir is r-square identity matrix)

step(i) :

I found that basis of kerF has m-r elts that is, {W1,W2,...,W(m-r)} and dimension of V is m.

step(ii) :

So I want to set basis of V as {V1,V2,...,V(r),W1,W2,...,W(m-r)}

for this process, I learned some theorem about this, but I can't remember it. I want to know what theorem need from step(i) to step(ii)

Let $V$ be a vector space of dimension $n$. If $S = \{v_1,\dots,v_k\}$ is a linearly independent set in $V$, then $S$ can be extended to a basis of $V$. That is, there exist vectors $v_{k + 1}, \dots, v_n$ such that $\{v_1,\dots,v_n\}$ is a basis of $V$.
• That's not necessary here. Of course, $\{F(v_1),\dots,F(v_r), F(w_1),\dots,F(w_{m-r})\}$ is a spanning set for $Im(F)$. Since $F(v_1) = \cdots = F(v_r) = 0$, it suffices to show that $F(w_r), \dots, F(w_{m-r})$ are linearly independent. To show that these vectors are linearly independent, note that if they were not linearly independent, then we could necessarily find another element of $\ker(F)$ from the span of $w_r,\dots,w_{m-r}$. – Omnomnomnom May 1 '17 at 17:29