1. ## Construct a matrix

v1 = (2 -1 2 1) v2 = (0 -1 1 -3) These vectors are column vectors.

Find a 2x4 matrix A, with linearly independent rows, such that Av1 = 0 and Av2 = 0.

2. A basis of $^{\perp}$ with the usual inner product provides the rows of $A$ .

3. Or, find a, b, c, d, e, f, g, h such that
$\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
and
$\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

That gives you four equations to solve for 8 values. Of course, there will be an infinite number of such matrices just as there are an infinite number of possible bases for the orthogonal space FernandoRevilla suggests.

4. Originally Posted by HallsofIvy
Or, find a, b, c, d, e, f, g, h such that
$\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$
and
$\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

That gives you four equations to solve for 8 values. Of course, there will be an infinite number of such matrices just as there are an infinite number of possible bases for the orthogonal space FernandoRevilla suggests.

You are right, but we only need two equations and four unknowns.

$^{\perp}\equiv\begin{Bmatrix} 2x_1-x_2+2x_3+x_3=0\\{}\quad -x_2+x_3-3x_4=0 \end{matrix}$

has dimension $2$, so two linearly independent solutions of the system are "your" $(a,b,c,d)$ and $(e,f,g,h)$ , that is, we needn't to denote twice the unknowns.

5. Thanks so much for the help!

I was just wondering what does the thing the looks like an upside down T by the (v1, v2) mean?

6. Originally Posted by skittle
I was just wondering what does the thing the looks like an upside down T by the (v1, v2) mean?

The subspace of $\mathbb{R}^4$ whose elements $x=(x_1,x_2,x_3,x_4)$ are orthogonal to $v_1$ and $v_2$. Equivalently:

$\begin{Bmatrix}(x_1,x_2,x_3,x_4)\cdot (2,-1,2,1)=0\\(x_1,x_2,x_3,x_4)\cdot (0,-1,1,-3)=0\end{matrix}\Leftrightarrow\begin{Bmatrix}2x_1-x_2+2x_3+x_4=0\\{}\quad\; \;-x_2+x_3-3x_4=0\end{matrix}$