Thread: Finding a two-dimensional span given three vectors.

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.



Now gives the matrix



What value must take to make the last to row dependant?

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



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 so that .

http://www.mathhelpforum.com/math-he...dbea0df917.png

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