0
$\begingroup$

Let $b : \Bbb{R}^2 \times \Bbb{R}^2 \to \Bbb{R}$ be the bilinear form defined by $$b((x_1,x_2),(y_1,y_2)) = x_1 y_1 - 2x_1 y_2 + x_2 y_1 + 3x_2 y_2.$$ Find the $2 \times 2$ matrix $B$ of $b$ relative to the basis $U=\{u_1,u_2\} = \{(0,1),(1,1)\}$

From what I understand, if $B=[b_{ij}]$ is the matrix representation, then $b_{ij} = b(u_i,u_j)$. If this is so, then I should get $$b_{11} = b((0,1),(0,1)) = 0 \cdot 0 - 2 \cdot 0 \cdot 1 + 1 \cdot 0 + 3 \cdot 1 \cdot 1 =3,$$ $$b_{12} = b((0,1),(1,1)) = 4,$$ $$b_{21} = b((1,1),(0,1)) = 1,$$ and $$b_{22} = b((1,1),(1,1)) = 3$$

However, the answer is $B = \begin{bmatrix} 0 & 4 \\ -1 & 3 \\ \end{bmatrix}$. Where did I go wrong?

$\endgroup$

2 Answers 2

2
$\begingroup$

If a vector has coordinates $[a,b]$ in $U$, then the vector is $a(0,1)+b(1,1)=(b,a+b)$.

Let's compute $b$ in coordinates in $U$. I will be using $[\cdot,\cdot]$ to denote coordinates in $U$ and $(\cdot,\cdot)$ to denote the vectors themselves or, since they look the same, coordinates in $\{(1,0),(0,1)\}$

\begin{align}b([a_1,a_2],[b_1,b_2])&=b((a_2,a_1+a_2),(b_2,b_1+b_2))\\ &=a_2b_2-2a_2(b_1+b_2)+(a_1+a_2)b_2+3(a_1+a_2)(b_1+b_2)\\ &=3a_1b_1+4a_1b_2+2a_2b_1+3a_2b_2\\ &=a_1(3b_1+4b_2)+a_2(2b_1+3b_2)\\ &=\left[\begin{matrix}a_1&a_2\end{matrix}\right]\left[\begin{matrix}3b_1+4b_2\\2b_1+3b_2\end{matrix}\right]\\ &=\left[\begin{matrix}a_1&a_2\end{matrix}\right]\left(\begin{matrix}3&4\\2&3\end{matrix}\right)\left[\begin{matrix}b_1\\b_2\end{matrix}\right] \end{align}

$\endgroup$
2
  • $\begingroup$ I am sorry but I don't really understand what you are aiming at. $\endgroup$
    – user193319
    Aug 3, 2017 at 16:45
  • $\begingroup$ @user193319 I computed the matrix of the bilinear form in the basis $U$. That matrix is, by definition, the matrix such that by doing matrix multiplication with the coordinates of the input in that basis, gives you the value of the form. $\endgroup$
    – Peyton
    Aug 3, 2017 at 22:38
0
$\begingroup$

Are you sure you've written the bilinear form correctly? If instead it is $$b((x_1,\boldsymbol{y_1}),(\boldsymbol{x_2},y_2))=x_1y_1−2x_1y_2+x_2y_1+3x_2y_2,$$ then you will get the answer you're looking for.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .