[![enter image description here](https://i.stack.imgur.com/NtuyO.png)](https://i.stack.imgur.com/NtuyO.png)

What I understand: By dimension formula $dim V$ $-$ $1$= $dim \langle v \rangle ^ \perp$ since $V$ is the direct sum of $U$ and $U ^ \perp$ if $U$ is a subspace of $V$. dim $\langle v \rangle$ is clearly $1$ 

So it makes sense that the number of basis vectors for $V$ is $1$ more than the number of basis vectors for $\langle v \rangle ^ \perp$

It also makes sense that if $v$ is an eigenvector then then $e\_1$=$v/\|v\|$ is also an eigenvector and also that $\|e\_1\|=1$ so $\langle e\_1,e\_1 \rangle$=1 and $\langle e\_1,e\_i \rangle$=0 for $i\geq 2$ since $e\_i$ were in $v^\perp$ and $e\_1$ is a scalar multiple of $v$. 
So **IF** $\{e\_2,e\_3,..., e\_n\}$ is a basis of then $v^\perp$ then $\{e\_1,e\_2,e\_3,..., e\_n\}$ is a basis for V since an orthonormal set of vectors is also linearly independent.

But WHY can we assume there exists a set $\{e\_2,e\_3,..., e\_n\}$ which is a basis of $v^\perp$. If I can understand this the whole proof will fall together
but 
I am seriously unstuck at the whole line which starts: "So by induction on dim($V$)..." Where is the induction even? Help will be much appreciated.