Find a matrix representation of $(fL)(x,y) = L(Ax, Ay)$, where A $\in \mathbb{R}^{3 \times 3}$

Question

I'm interested in the following question:

Let $L((\mathbb{R}^3)^2; \mathbb{R})$ be the vector space of bilinear maps from $\mathbb{R^3}\times \mathbb{R^3}$ into $\mathbb{R}$. Consider the subspace V $\subset$ $L((\mathbb{R}^3)^2; \mathbb{R})$ of maps L that are antisymmetric, that is, they satisfy $$L(x,y) = -L(y, x)$$ for all $x, y \in \mathbb{R}^3$

Let f be the linear map $f: V \to V$, defined by $$(fL)(x, y) = L(Ax, Ay)$$ where $$A =\begin{bmatrix} 1 & 4 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{bmatrix}$$

i.) Choose a basis of V and determine the dimension of V.

ii.) In your chosen basis of V, the linear map F can be represented as a matrix B. Find this matrix B.

I've done the first part. My answer is

$ B = [\vec{b_1}, \vec{b_2}, \vec{b_3}, \vec{b_4}, \vec{b_5}, \vec{b_6}]$, where

$$ \vec{b_1} =\begin{Bmatrix}\begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\end{Bmatrix}$$

$$ \vec{b_2} = \begin{Bmatrix}\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\end{Bmatrix}$$

$$ \vec{b_3} = \begin{Bmatrix}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\end{Bmatrix}$$

$$ \vec{b_4} = \begin{Bmatrix}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}\end{Bmatrix}$$

$$ \vec{b_5} = \begin{Bmatrix}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{bmatrix}\end{Bmatrix}$$

$$ \vec{b_6} = \begin{Bmatrix}\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0 \end{bmatrix}\end{Bmatrix}$$

So, evidently, the dimension of V is 6.

For the second part, I haven't got a clear cut idea of what the method is. I'm thinking I need to substitute each combination of basis vectors, i.e. $(\vec{b1}, \vec{b1})$, $(\vec{b1}, \vec{b2})$ ... $(\vec{b3}, \vec{b3})$, into L, represent those as a linear combination of B, and then use that linear combination as columns of the matrix, but something about that feels wrong. I imagine $L(\vec{v}, \vec{v}) = -L(\vec{v}, \vec{v})=0$ can also come in useful in one form or another.

To be honest, I'm more interested in the general method here rather than the specific numerical answer to this instance, regardless. I'm fairly certain there's some sort of "substitute this, then represent that as a linear combination" technique we need to rinse and repeat a few times, and I'm just looking to get an understanding of that idea.

Any help would be greatly appreciative!

Answer 1

Note that any bilinear form $B$ on $\mathbb R^n\times \mathbb R^n$ is completely determined by its action on $(e_i,e_j)$, $i,j=1,\ldots,n;$ where $\{e_1,\ldots,e_n\}$ is basis for $\mathbb R^n$. This is because given $v = \sum_i v_ie_i$ and $w = \sum_iw_ie_i$ we have $$B(v,w) = B(\sum_iv_ie_i,\sum_jw_je_j) = \sum_{i,j}v_iw_jB(e_i,e_j).$$

Define matrix $\tilde B$ with coefficients $b_{ij} = B(e_i,e_j)$. If you identify vectors $v,w\in\mathbb R^n$ with vector-columns $(v_1\, \ldots \, v_n)^T$ and $(w_1\, \ldots \, w_n)^T$ you can verify that $B(v,w) = v^T\tilde Bw$ where on the RHS you have matrix multiplication. In this way we have established isomorphism between the space of bilinear forms on $\mathbb R^n$ and $M_n(\mathbb R)$, and I will abuse the notation to denote by the same letter both a bilinear form and its corresponding matrix.

You can easily verify that antisymmetric bilinear forms correspond to antisymmetric matrices. Let $E_{ij}$ be the matrix that has $1$ on $(i,j)$-th place and $0$ everywhere else. $\{ E_{ij}-E_{ji}\mid i,j=1,\ldots,n,\ i<j\}$ forms a base for $n\times n$ antisymmetric matrices. In particular, for $n = 3$, $\{ E_{12}-E_{21}, E_{13}-E_{31}, E_{23}-E_{32}\}$ forms basis for $3\times 3$ antisymmetric matrices. You actually wrote those matrices, but for some reason you are double-counting them.

To figure out the map $f$, write $$f(B)(v,w) = B(Av,Aw) = v^TA^TBAw.$$ Therefore, $f(B) = A^TBA$, where, as I said, I abuse the notation to identify forms and corresponding matrices.

You can now easily calculate how $f$ acts on the base vectors $B_1 = E_{12}-E_{21}$, $B_2 = E_{13}-E_{31}$, $B_3 = E_{23}-E_{32}$. For example,

$$f(B_2)\! = \!\left( \begin{array}{ccc} 1 & 0 & 0 \\ 4 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array} \right)\!\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right)\!\left( \begin{array}{ccc} 1 & 4 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array} \right)\! =\! \left( \begin{array}{ccc} 0 & 0 & 3 \\ 0 & 0 & 12 \\ -3 & -12 & 0 \\ \end{array} \right)\! = 3B_2+12B_3.$$

Answer 2

I don't understand your notation for the basis of $V$ but anyway the conclusion is wrong, since the dimension is not $6$ but $3.$ A basis of $V$ is indeed $$(L_{1,2},L_{1,3},L_{2,3})$$ where $$L_{i,j}(e_i,e_j)=1$$ and all others $L_{i,j}(e_k,e_l)=0,$ except of course $L_{i,j}(e_j,e_i)=-1.$ Here, $(e_1,e_2,e_3)$ denotes the canonical basis of $\Bbb R^3.$

You will then easily calculate each of the three $f(L_{i,j})$'s and derive the matrix $B.$ For instance, $$(fL_{1,2})(x,y)=(x_1+4x_2)(2y_2)-(2x_2)(y_1+4y_2)=2L_{1,2}(x,y)$$ hence the 1st column of $B$ is $(2,0,0)^T.$