# Determining whether a function is an inner product

• Mr Davis 97
In summary: Injectivity" is a property of functions. "Invertibility" is a property of matrices. Finally, "inner product space" is a property of vector spaces.In summary, the conversation discusses whether a function with certain properties can be considered an inner product on a specific vector space. The participants go through a few different approaches to prove or disprove this, eventually concluding that the function cannot be an inner product due to the size of the vector space and the properties of the function.
Mr Davis 97

## Homework Statement

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##

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

fresh_42 said:
Are you sure you got it right? I would start to think about counterexamples. E.g. what if ##v \in \ker(T) - \{0\}##?
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?

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?

fresh_42 said:
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?
I'm not sure about your question, since I don't know whether I am right or wrong yet

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?

Mr Davis 97 said:
I'm not sure about your question, since I don't know whether I am right or wrong yet
I think you've successfully proved that it cannot be an inner product.

fresh_42 said:
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?
##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?

Mr Davis 97 said:
##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?

Yes, that's right.

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##:
Mr Davis 97 said:
##v = 0## [does not] follow[s from ##T(V)=0##] 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?
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\}##.

Mr Davis 97

## 1. What is an inner product?

An inner product is a mathematical operation that takes two vectors as input and produces a scalar value as output. It is often denoted by <x,y> and is used to measure the angle between two vectors, as well as the length of a vector.

## 2. How do you determine if a function is an inner product?

To determine if a function is an inner product, it must satisfy four properties: linearity in the first argument, conjugate symmetry, positive definiteness, and homogeneity in the second argument. These properties can be checked by plugging in values and verifying that the results hold.

## 3. What is linearity in the first argument?

Linearity in the first argument means that the function is linear with respect to the first vector in the inner product. This property is satisfied if the function follows the distributive property and the scalar multiple property.

## 4. What is conjugate symmetry in an inner product?

Conjugate symmetry, also known as Hermitian symmetry, means that the inner product of two vectors is equal to the complex conjugate of the inner product of the same two vectors in reverse order. In other words, <x,y> = <y,x*>, where * denotes complex conjugation.

## 5. Why is positive definiteness important for an inner product?

Positive definiteness ensures that the inner product of a vector with itself is always a positive real number. This property is necessary for the inner product to measure the length of a vector, as the length must be a positive value. It also helps to determine if two vectors are orthogonal (perpendicular) to each other.

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