How to find $f$ for a symmetric bilinear form?

Question

Let's say we have the symmetric matrix:$$A = \left(\begin{array}{cc} 1&2 \\ 2&0 \end{array}\right)$$

How do I find the symmetric bilinear form of this $A$?

Answer 1

The bilinear form $B$ that this matrix represents is given by $$B(u,v) = u^TAv = \sum_i \sum_j A_{i,j}u_iv_j$$

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EDIT:

To give a little more information, a bilinear form is a function from a vector space $V$ to its field of scalars $K$, $B:V \times V \rightarrow K$ which is linear in each argument:

• $B(u_1+u_2,v) = B(u_1,v) + B(u_2,v)$

• $B(u,v_1+v_2) = B(u,v_1) + B(u,v_2)$

• $\alpha B(u,v) = B(\alpha u,v) + B(u,\alpha v), \alpha \in K$

Given a basis of $V$, $\{e_1,e_2,...,e_n\}$, a bilinear form can always be represented by a matrix by examining how it acts on the basis vectors:

• define $B(e_i,e_j) = A_{i,j}$

$$B(u,v) = B(\sum_i u_ie_i,\sum_j v_je_j )$$ $$ = \sum_i u_i B(e_i,\sum_j v_je_j)$$ $$ = \sum_i \sum_j u_i v_j B(e_i,e_j)$$ $$ = \sum_i \sum_j u_i v_j A_{i,j}$$ $$ = u^TAv$$

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Examples:

The determinant of a $2 \times 2 $ matrix is a bilinear form over its column vectors (in general it is an $n$-linear form over the columns of an $n \times n$ matrix).

The first fundamental form, which is like a sort of generalized dot product that lets you measure angles and distances on curved surfaces, is also a bilinear form.

Answer 2

Bilinear forms are exactly those linear maps taking two vectors to a scalars, so this form of $A$ maps to exactly one of those linear maps, as denoted by @Eigenchris.

For further 'meaning', consider the bilinear form $u^TAv$. $Av$ is just some new vector $v'$, so we have $u^Tv'$, which is just the dot product between $u$ and $v'$. So first $v$ gets mapped to some different vector by a linear transformation, then we take the dot product between that vector and $u$.