# Construct a matrix

• Feb 28th 2011, 07:57 PM
skittle
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.
• Mar 1st 2011, 12:10 AM
FernandoRevilla
A basis of $^{\perp}$ with the usual inner product provides the rows of $A$ .
• Mar 1st 2011, 03:46 AM
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.
• Mar 1st 2011, 04:26 AM
FernandoRevilla
Quote:

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.
• Mar 2nd 2011, 07:35 AM
skittle
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?
• Mar 2nd 2011, 08:19 AM
FernandoRevilla
Quote:

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:

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