Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reviewing for my final and I encountered a statement that puzzles me. Let $u,v$ be vectors in $\Bbb R^n$ that are expressed in column form, and let $A$ be an invertible $n\times n$ matrix. Then we can express the Euclidean inner product on $\Bbb R^n$ to be $\langle u,v\rangle = Au \cdot Av$.

Why does $A$ have to be an invertible matrix? I had a review question asking if everything but the invertibility property of $A$ held, if the statement was true, and the solutions in the back says that it was false.

Thanks in advance!

share|cite|improve this question
What do you mean by $*$? Is that the dot product? – icurays1 Dec 9 '12 at 7:10
Yes it is, I don't know how to edit the post such that I can change it though! – tamefoxes Dec 9 '12 at 7:19
I'm confused about the language. When you say "The Euclidean inner product", I think $\langle u,v\rangle=u_1v_1+\ldots+u_nv_n$. Take for instance $A=2I$; then $Au\cdot Av=2u_12v_1+\ldots+2u_n2v_n=4\langle u,v\rangle$. So $Au\cdot Av\neq \langle u,v\rangle$. Do you mean we can express any inner product by a matrix $A$...? – icurays1 Dec 9 '12 at 7:36
I think what OP wants is to define the Euclidean product in the usual way, and then more generally define a special bilinear function in terms of $A$ and the usual Euclidean product. – anon Dec 9 '12 at 7:38
if the matrix A is not the identity matrix, then we can call that inner product a weighted inner product. But I think the textbook is just trying to generalize inner products generated by matrices for this specific question – tamefoxes Dec 9 '12 at 7:39
up vote 2 down vote accepted

By definition an inner product must have the property that $\langle u,u\rangle=0$ if and only if $u=0$.

Suppose $\langle\cdot,\cdot\rangle$ is an inner product, $A$ is not invertible, and $\langle u,v\rangle_A:=\langle Au,Av\rangle$ is a bilinear form; can you show that there is a nonzero vector $u$ for which $\langle u,u\rangle_A=0$? $\color{White}{\mathrm{Hint}:Au=0\implies \langle u,u\rangle_A=0.}$

share|cite|improve this answer
we haven't discussed bilinear form in my linear algebra course, is it required to prove that the matrix A has to be invertible? – tamefoxes Dec 9 '12 at 7:28
@user43956 "Bilinear form" just means a function of two vector arguments that is linear in each argument. An inner product is essentially a bilinear form with the added condition that $\langle u,u\rangle=0$ if and only if $u=0$ (plus conjugate-symmetry). You do not need to know the term "bilinear form" for this proof; I just wanted to have a noun at hand to slap onto $\langle u,v\rangle_A$ (otherwise what type of thing would we refer to it as?). – anon Dec 9 '12 at 7:32
Alright, so going back to your question... wouldn't only the zero vector hold for the bilinear form that we are discussing? – tamefoxes Dec 9 '12 at 7:37
No. What if $Au=0$ but $u\ne0$? – anon Dec 9 '12 at 7:39
so u would be contained in the null space of A? – tamefoxes Dec 9 '12 at 7:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.