Construct a matrix

**skittle** · February 28th 2011, 07:57 PM

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.

**FernandoRevilla** · March 1st 2011, 12:10 AM

A basis of $<v_1,v_2>^{\perp}$ with the usual inner product provides the rows of $A$ .

**HallsofIvy** · March 1st 2011, 03:46 AM

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.

**FernandoRevilla** · March 1st 2011, 04:26 AM

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.

$<v_1,v_2>^{\perp}\equiv\begin{Bmatrix} 2x_1-x_2+2x_3+x_3=0\\{}\quad -x_2+x_3-3x_4=0 \end{matrix}$

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.

**skittle** · March 2nd 2011, 07:35 AM

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?

**FernandoRevilla** · March 2nd 2011, 08:19 AM

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:

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

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