# Why is invertibility of $A$ necessary for $\langle u,v\rangle:=Au\cdot Av$ to be an inner product?

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.

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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

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 $\langle u,u\rangle_A=0$? $\color{White}{\mathrm{Hint}:Au=0\implies \langle u,u\rangle_A=0.}$
@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
No. What if $Au=0$ but $u\ne0$? –  anon Dec 9 '12 at 7:39