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For a bilinear form on $ \mathbb R^2$ the matrix of the bilinear form, A, with respect to the standard basis is $a_{ij}= \langle e_i, e_j\rangle$, for $ i =1,2$, j = $1,2$. Then for any two vectors $x,y$ one can write $\langle x,y\rangle = x^tAy$.

i don't understand the matrix of a bilinear form in say $ \mathbb R^{2*2}$. If I take a bilinear form $\langle A,B\rangle = trace(AB)$ with respect to the standard basis $e_{ij}$ I get a $4*4$ matrix, C, of the form. Then the product $\langle A,B\rangle$ = $A^tCB$ isn't defined since is consists of a $2*2$ matrix multiplied either side of a $4*4$. At least this is my incorrect intuition. What is the correct way to view the matrix of a form on a space of matrices?

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Well, I have used this form quite a bit, and never felt the need to think about its matrix, so one answer might be: don't think about it. – Chris Godsil Nov 6 '12 at 0:00
I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Nov 6 '12 at 8:05
up vote 1 down vote accepted

Let $b:V\times V\rightarrow K$ be a bilinear form on a vector space $V$ of finite dimension over a field $K$. Then the matrix $B\in K^{n\times n}$ of $b$ with respect to some basis $v_1,\ldots ,v_n$ of $V$ has the entries $b_{ij}:=b(v_i,v_j)$ and satisfies the equation


where $x,y\in K^n$ are the coordinate vectors of $v,w$ with respect to the given basis. That is $v=x_1v_1+x_2v_2+\ldots +x_nv_n$, $x=(x_1,\ldots ,x_n)$ and similar for $y$.

In your case $V=\mathbb{R}^{2\times 2}$ and as a basis you took the matrices $E_{ij}$ having only an entry $1$ at the position $(i,j)$ in the matrix. Thus instead of the equation $\langle A,B\rangle=A^tCB$ that you don't understand and that is indeed meaningless you have the equation

$\langle A,B\rangle =a^tCb$

where $a\in R^4$ is the coordinate vector of $A$ with respect to the basis $(E_{ij})$. That is depending on the way you order the base elements the vector $a$ has the matrix coefficients $a_{ij}$ as entries, in the corresponding ordering.

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