- Consider a bilinear function $f : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ that is symmetric, i.e. $$ f(A_1, A_2) = f(A_2, A_1) \quad\forall A_1, A_2 \in \mathbb{R}^n $$ and satisfies $$ f(I_i; I_j) = 0 \text{ if } i < j \text{ and } f(I_i; I_i) = 1 \text{ for each } i $$ Prove that such a function is unique in each dimension n. What is this function anyway?
Sorry about the sloppy formatting. Basically, I'm a bit confused because this function looks like an inner product, but I know there are several different inner products that satisfy the axioms for a euclidean space. Doesn't that mean that f is not a unique function? Can anyone help me out here?
I will assume that $\{I_i\}$ denotes the standard basis for $\mathbb{R}^n$. For any $x, y \in \mathbb{R}^n$, write $$ x = \sum \alpha_i I_i, \text{ and } y = \sum \beta_j I_j $$ Since $f$ is symmetric, and satisfies $f(I_i, I_j) = 0$ if $i< j$, it satisfies $f(I_i, I_j) = 0$ for $i\neq j$. Since it is bilinear, you get $$ f(x,y) = \sum \alpha_i \beta_i $$ In other words, $f$ is the usual inner product.