Question

Let f be the bilinear form on R² defined by

f[(x₁, x₂), (y₁, y₂)]= 3x₁y₁-2x₁y₂+4x₂y₁-x₂y₂

then find the matrix A of f in the basis {u₁(1, 1)=u₂(1, 2)}

Answer 1

$f: \Reals^2*\Reals^2\to\Reals$is given by$f( (\begin{matrix} x_1 \\ x_2 \end{matrix}), (\begin{matrix} y_1 \\ y_2 \end{matrix}))=3x_1y_1-2x_1y_2+4x_2y_1-x_2y_2$

Let us write the matrix of f in the standard basis.

$f(e_1,e_1)=3,\space f(e_1,e_2)=-2, \space f(e_2,e_1)=4, \space f(e_2,e_2)=-1$

hence the matrix in the standard basis is

$A=( \begin{matrix} 3 & -2 \\ 4 & -1 \end{matrix})$

Now for $B = \lbrace b_1, . . . , b_n \rbrace$ and $U = \lbrace u_1, . . . , u_n\rbrace$ are two bases for $V$ . We may write one basis in terms of the other: $u_i= \sum_{j=1}^{n}\lambda_{j,i}b_j.$ Where coefficients $\lambda_{j,i}$ form the transition matrix $M$ from $B$ to $C$. To found matrix $A$ in new basis $\lbrack A\rbrack_u$ used change of basis formula : $\lbrack A\rbrack_u=M^T \lbrack A\rbrack_b M$

In this case we want to write this matrix i basis

$\lparen \begin{matrix} 1\\ 1 \end{matrix})$ ,$(\begin{matrix} 1\\ 2 \end{matrix})$

The transition matrix is :

$M=( \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})$

it’s transpose is the same. The matrix of f in the new basis is

$\lbrack A\rbrack_u=( \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix}) ( \begin{matrix} 3 & -2 \\ 4 & -1 \end{matrix}) ( \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})= ( \begin{matrix} 4 & 1 \\ 7 & 3 \end{matrix})$