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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!

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    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 question2012-12-09

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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 $\langle u,u\rangle_A=0$? $\color{White}{\mathrm{Hint}:Au=0\implies \langle u,u\rangle_A=0.}$

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    @user43956 The iff property of inner products does not say anything about *two* vectors independently being involved. For instance, we can have $u$ and $v$ in $\Bbb R^n$ such that $\langle u,v\rangle=0$ with the usual Euclidean inner product (they merely have to be orthogonal). The conclusion is, $A$ must be invertible, because otherwise there is a nonzero $u$ for which $Au=0$, in which case we have $\langle u,u\rangle_A=\langle Au,Au\rangle=\langle0,0\rangle=0$ even though $u\ne0$, contrary to the noted property of inner products.2012-12-09