Finding a two-dimensional span given three vectors.

• February 10th 2011, 10:06 AM
shonucic
Finding a two-dimensional span given three vectors.
Find the values of k so that the vectors [3 -k -1 0}^T, [-1 2 -k -1]^T, [0 -1 3 -k]^T span a two-dimensional space.

My intuition says that I should form a 4x3 matrix out of these vectors with the first vector as the first column, the second vector as the second column, etc.

Then I should row-reduce until I have only 2 leading ones. This will give me the two vectors that span the two dimensional space, but I am having trouble reducing.

Is this the right approach, or am I doing this incorrectly.
Thanks very much for any help on this topic.
• February 10th 2011, 10:21 AM
TheEmptySet
k+1
Quote:

Originally Posted by shonucic
Find the values of k so that the vectors [3 -k -1 0}^T, [-1 2 -k -1]^T, [0 -1 3 -k]^T span a two-dimensional space.

My intuition says that I should form a 4x3 matrix out of these vectors with the first vector as the first column, the second vector as the second column, etc.

Then I should row-reduce until I have only 2 leading ones. This will give me the two vectors that span the two dimensional space, but I am having trouble reducing.

Is this the right approach, or am I doing this incorrectly.
Thanks very much for any help on this topic.

So if you want to span to be 2D then you need the matrix have exactly 2 ones when the matrix is in reduced row form.

$\begin{bmatrix} 3 & -1 & 0 \\ -(k+1) & 2 & -1 \\ 0 & -(k+1) & -(k-3)\end{bmatrix}$

Now $(k+1)R_1+3R_2$ gives the matrix

$\begin{bmatrix} 3 & -1 & 0 \\ 0& -k+5 & -1 \\ 0 & -(k+1) & -(k-3)\end{bmatrix}$

What value must $k$ take to make the last to row dependant?
• February 11th 2011, 04:46 AM
HallsofIvy
Another way of looking at it: the span of a set of vectors has dimension two if there are two independent vectors such that all of the other vectors can be written as a linear combination of those two. Here, that just means that one is a linear combination of the other two.

Let's try
$a\begin{bmatrix}3 \\ -k \\ -1 \\ 0\end{bmatrix}+ b\begin{bmatrix}-1 \\ 2 \\ -k \\ -1\end{bmatrix}= \begin{bmatrix} 0 \\ -1 \\ 3 \\ -k\end{bmatrix}$
$\begin{bmatrix}3a- b \\ -ak+ 2 \\ -a- bk \\ -b\end{bmatrix}= \begin{bmatrix}0 \\ - 1 \\ 3 \\ -k\end{bmatrix}$

so we have four equations, 3a- b= 0, -ak+ 2= -1, a- bk= 3, and -b= -k to solve for a, b, and k. From the last equation, k= b. Then 3a- k= 0 so a= k/3 and $a- bk= a- k^2= 3$ so that $a= k/3= 3- k^2$.
• February 11th 2011, 06:19 AM
shonucic
http://www.mathhelpforum.com/math-he...dbea0df917.png

How did you get that matrix from the initial three vectors?