# Determining whether a function is an inner product

1. Nov 21, 2016

### Mr Davis 97

1. The problem statement, all variables and given/known data
T/F: If $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is a linear transformation and $n>m$, then the function $\langle v , w \rangle = T(v) \cdot T(w)$ is an inner product on $\mathbb{R}^n$

2. Relevant equations

3. The attempt at a solution
The first three axioms of the inner product are straightforward. However, I am not sure how to show that $\langle v , v \rangle = 0$ iff $v0$.

Actually, maybe I have something. If $\langle v , v \rangle = 0$ then $T(v) \cdot T(v) = ||T(v)||^2 \implies ||T(v)|| = 0 \implies T(v) = 0$. Thus, v is zero iff the null space of T is only zero. This is only the case when T is invertible. However, n > m, so T can't be invertible. Thus, v doesn't have to be 0, and thus we don't have an inner product space. Is this on the right track?

Last edited: Nov 21, 2016
2. Nov 21, 2016

### Staff: Mentor

Are you sure you got it right? I would start to think about counterexamples. E.g. what if $v \in \ker(T) - \{0\}$?

3. Nov 21, 2016

### Mr Davis 97

I just added to my original post, so I am not sure if your post is after or before that post. Is my reasoning correct?

4. Nov 21, 2016

### Staff: Mentor

You still have a $u$ where there probably should be a $v$. The equation with the norm could be used to show $<v,v> \geq 0$, but what about my question?

5. Nov 21, 2016

### Mr Davis 97

I'm not sure about your question, since I don't know whether I am right or wrong yet

6. Nov 21, 2016

### Staff: Mentor

What happens to vectors in the kernel of $T$? And there are non-zero ones, because $n >m$. What have you to be sure about? You have $\langle v,v \rangle = 0 \Rightarrow T(v)=0$. Does $v=0$ follow?

7. Nov 21, 2016

### PeroK

I think you've successfully proved that it cannot be an inner product.

8. Nov 21, 2016

### Mr Davis 97

$v = 0$ follows because n > m, which means that T is not invertible which means that $Nul(T) \neq \{ 0 \}$. So it can't be an inner product, right?

9. Nov 22, 2016

### PeroK

Yes, that's right.

10. Nov 22, 2016

### Staff: Mentor

The argument was correct: How to find a $v \neq 0$ such that $\langle v,v \rangle = 0$ which contradicts the "iff" in the condition. But then with such a $v$:
If you want to be very exact, then $n > m$ means that $T$ cannot be injective.
Injectivity of linear functions $f$ is equivalent to $Nul(f) = \{0\}$.