Can some one explain me a easy alternate proof of rank nullity theorem? OK I have just gone through the Gilbert Strang's 'Introduction to Linear algebra'. I felt it as a very nice book. But I am bad at proving things. I have an idea what null space (kernel) and column space (image) etc. Also I have verified rank nullity theorem for small matrices. But I wanted to understand the generalized proof of dimension theorem. So I read the wiki page , but didn't understood it.
Is there any easy proof, some one can explain me?
 A: So in general, suppose we have a linear transformation $\tau:W \rightarrow V$ over some field $\mathbb{F}$ with $\dim(W)=n$ and dim$(V)=m$. 
The tricky thing is that vectors in the kernel are in $W$ and vectors in the image is in $V$. In the proof of the rank nullity theorem we make the "connection" via another theorem stating that the image of a basis spans the image...so if $\beta$ is a basis for $W$, $\tau(\beta)$ spans $\tau(W)$ - if you do not understand this, then it's best to first study that theorem.
Now start the proof by taking a basis $\alpha=\{\alpha_1 \ldots \alpha_k\}$ for the kernel of $\tau$, which is a subspace of $W$. We can always extend this to be a basis for $W$ itself (another theorem)...so let's extend it by adding vectors $\beta=\{\alpha_{k+1},\ldots,\alpha_n\}$ so that $\alpha \cup \beta$ is a basis for $W$. So as mentioned in the previous paragraph, $\tau(\alpha \cup \beta)$ spans $\tau(W)$. But we can in fact do better, since we know $\tau(\alpha_1)=\cdots=\tau(\alpha_k)=0$, we in fact have that $\tau(\beta)$ spans $\tau(W)$. This is the first part of the proof.
What we have to prove now is that $\tau(\beta)$ is a linearly independent set. We start in the usual way, from the definition of linear independence, suppose we have scalars $a_{k+1},\ldots,a_n$ so that \begin{equation} \sum_{i=k+1}^{n}a_i \tau(\alpha_i)=0, \end{equation} then by the linearity of $\tau$:\begin{equation} \tau\left(\sum_{i=k+1}^{n} a_i \alpha_i \right )=0. \end{equation} But this means $\sum_{i=k+1}^{n} a_i \alpha_i $ is in the kernel of $\tau$. Let $v=\sum_{i=k+1}^{n} a_i \alpha_i $. Since $v$ is in the kernel of $\tau$ and $\alpha$ is a basis for this kernel we can find scalars $a_1,\ldots,a_k$ so that $v=\sum_{i=1}^{k} a_i \alpha_i$. And now we use a simple "trick": \begin{equation} v-v= \sum_{i=1}^{k} a_i \alpha_i-\sum_{i=k+1}^{n} a_i \alpha_i=0,\end{equation} and since $\alpha \cup \beta$ is a basis, and hence linearly independent we must have $a_1=\cdots=a_n=0$, which proves that $\tau(\beta)$ is linearly independent.
So taking it all together: $\{\alpha_1,\ldots,\alpha_k\}$ is a basis for the kernel of $\tau$ (by hypothesis), and we have proved $\{\tau(\alpha_{k+1}),\ldots,\tau(\alpha_{n})\}$ is a basis for the image of $\tau$, and we have  $|\alpha_1,\ldots,\alpha_k|+|\tau(\alpha_{k+1}),\ldots,\tau(\alpha_{n})|=k+(n-k)=n$, which is what we wanted...
let me know if anything is unclear, and I will try to expand on whatever specific request you might have.    

Edit - Request to prove first theorem:
Suppose $\beta=\{\beta_1,\ldots,\beta_n\}$ is a basis for $W$. Consider any vector $w \in W$ which can be expressed in terms of the basis $\beta$ as \begin{equation} w=\sum_{i=1}^n b_i \beta_i\end{equation} for some scalars $b_i$, then $\tau(w) \in \tau(W)$ and we have \begin{eqnarray} \tau(w) &=& \tau\left(\sum_{i=1}^n b_i \beta_i \right) \\
&=& \sum_{i=1}^n b_i\tau(\beta_i), \end{eqnarray} by the linearity of $\tau$. This means any vector in $\tau(W)$ can be expressed as a linear combination of the vectors in $\tau(\beta)$, and therefore $\tau(\beta)$ spans $\tau(W)$.
