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

Find a 2x4 matrixA, with linearly independent rows, such thatAv1 =0andAv2 =0.

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- Feb 28th 2011, 07:57 PMskittleConstruct a matrix
**v**1 = (2 -1 2 1)**v**2 = (0 -1 1 -3) These vectors are column vectors.

Find a 2x4 matrix*A*, with linearly independent rows, such that*A***v**1 =**0**and*A***v**2 =**0**. - Mar 1st 2011, 12:10 AMFernandoRevilla
A basis of $\displaystyle <v_1,v_2>^{\perp}$ with the usual inner product provides the rows of $\displaystyle A$ .

- Mar 1st 2011, 03:46 AMHallsofIvy
Or, find a, b, c, d, e, f, g, h such that

$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}2 \\ -1 \\ 2 \\ 1\end{bmatrix}= \begin{bmatrix} 2a- b+ 2c+ d \\ 2e- f+ 2g+ h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

and

$\displaystyle \begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}0 \\ -1 \\ 1 \\ -3\end{bmatrix}= \begin{bmatrix} -b+ c- 3d \\ -f+ g- 3h\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}$

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 AMFernandoRevilla

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

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

has dimension $\displaystyle 2$, so two linearly independent solutions of the system are "your" $\displaystyle (a,b,c,d)$ and $\displaystyle (e,f,g,h)$ , that is, we needn't to denote twice the unknowns. - Mar 2nd 2011, 07:35 AMskittle
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 AMFernandoRevilla

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

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