Spans

Zocken · #1 $Old$ November 5th, 2009, 01:27 AM

Find all values a for which $\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ is in the span of $v_1,v_2,v_3$ . Also find the values of a for which $v_1,v_2,v_3$ are linearly independent.

$v_1=\begin{bmatrix}1 \\ a \\ a\end{bmatrix}$ $v_2=\begin{bmatrix}a \\ 1 \\ a\end{bmatrix}$ $v_3=\begin{bmatrix}a \\ a \\ 1\end{bmatrix}$

When I try to do rref I get all zeros in the bottom...

HallsofIvy · #2 $Old$ November 5th, 2009, 05:42 AM

Quote:

Originally Posted by Zocken $View Post$

Find all values a for which $\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}$ is in the span of $v_1,v_2,v_3$ . Also find the values of a for which $v_1,v_2,v_3$ are linearly independent.

$v_1=\begin{bmatrix}1 \\ a \\ a\end{bmatrix}$ $v_2=\begin{bmatrix}a \\ 1 \\ a\end{bmatrix}$ $v_3=\begin{bmatrix}a \\ a \\ 1\end{bmatrix}$

When I try to do rref I get all zeros in the bottom...

For the first problem you want all a such that x[1 a a]+ y[a 1 a]+ z[a a 1]= [1 1 1] for some numbers x, y, z. That gives the three equations x+ ay+az= 1, ax+y+az= 1, and ax+ay+ z= 1. For what values of a does that have a solution?

For the second problem you want to look at x[1 a a]+ y[a 1 a]+ z[a a 1]= [0 0 0]. An obvious solution is x=y= z= 0. For what values of a is that the only solution?

Are you saying you get all zeros in the last row of the augmented matrix, for all a?

Zocken · #3 $Old$ November 5th, 2009, 06:03 AM

yes, when I augment the matrix with 1 1 1 I get the bottom with zeros. Is there something I am missing?