Show that $B$ represents an inner product.

Question

Let $V$ be a vector space over $\mathbb{R}$ with $\dim V = n$ Let $B = (b_{ij})$ be an $n \times n$ diagonal matrix which represents a bilinear form on $V$ with respect to a basis of $V$ , $\{v_1,\ldots,v_n\}$. Show that $B$ represents an inner product if and only if $b_{ii}>0$ for all $1 \leq i \leq n$.

I am grateful for any hints on how to solve this.

Answer 1

Hint:

It seems that implicit to this exercise is that the matrix defines positive-definite real inner product space over $V$.

Recall then that by this definition, $\forall\textbf{v}_1,\textbf{v}_2,\textbf{v}_3\in V,\forall c_1,c_2\in\mathbb{R}$, the following should hold:

1. Positive-definiteness: $\langle \textbf{v}_1,\textbf{v}_1\rangle\ge 0$, and equality holds iff $\textbf{v}_1=\textbf{0}$.
2. Symmetry: $\langle \textbf{v}_1,\textbf{v}_2\rangle=\langle \textbf{v}_2,\textbf{v}_1\rangle$.
3. Linearity: $\langle c_1\textbf{v}_1+c_2\textbf{v}_2,\textbf{v}_3\rangle=c_1\langle \textbf{v}_1,\textbf{v}_3\rangle+c_2\langle \textbf{v}_2,\textbf{v}_3\rangle$

Now suppose $B$ represents our potential inner product, where $\langle \textbf{v}_1,\textbf{v}_2\rangle=\textbf{v}_1^TB\textbf{v}_2$.

I recommend you first show that the above is indeed an inner product if $b_{ii}>0$. Then, in the opposite direction, rely on the fact that the definition holds for all $\textbf{v}\in V$, such as ${\boldsymbol\delta}_i$, to force $B$ into having positive entries, given that it is diagonal.