a symmetric bilinear form has a basis such that it's matrix with respect to it is diagonal

Question

I'm reviewing a proof regarding $f$, a symmetric bilinear form having a basis such that it's matrix with respect to this basis is diagonal.

Here's a summarization:

• For $n=1$ there's nothing to prove. Let's assume that $n>1$.
• Let's also assume $f\ne 0$ (Otherwise, any basis would work).
• By the polar lemma: there's a $u_1 \in V$ such that $c_1 = f(u_1, u_1)$, and $u_1 \ne 0$.
• We denote $V_1 = \text{span}(u_1)$ and $V_2 = \{v\in V\ :\ f(u_1, v) = 0\}$.
• claim: $V = V_1 \oplus V_2$.

Now, at this point the author claims that $\dim V = \dim V_1 + \dim V_2$. Why?
Also, why is it true that $u_1 \ne 0$?

Thanks.

Answer 1

$f$ is a symmetric bilinear form; this mean that

$$f:V\times V\longrightarrow \mathbb{K}$$

is linear in each components. So if you fix the first one as $u_1$ then you have

$$g=f(u_1,\text{ }):V\longrightarrow \mathbb{K}$$ which is linear. By construction $g(u_1)=f(u_1,u_1)=c$ with $u_1\neq 0$. The kernel of $g$ is exactly the space $V_2$ as subspace of $V$ which has dimension $n-1$. On the other hand $V_1\cong \mathbb{K}$. So you can conclude.

Answer 2

First question: the dimension of a direct sum is the sum of the dimensions of the summands.

In this respect, the general formula is $$\dim(V_1+V_2)+\dim V_1\cap V_2=\dim V_1+\dim V_2.$$

Second question: if $f\neq 0$, there exists a vector $u_1$ such that $c_1=f(u_1,u_1)\neq 0$, which obviously implies $u_1\neq 0$.