What Values of k Ensure Linear Independence in R^4 for These Vectors?

Nathew

Homework Statement


Determine all values of the constant k for which the given set of vectors is linearly independent in \mathbb R^4.
{(1, 1, 0, −1), (1, k, 1, 1), (4, 1, k, 1), (−1, 1, 1, k)}

Homework Equations





The Attempt at a Solution



So far I set up a coefficient matrix
<br /> \begin{pmatrix}<br /> 1 &amp; 1 &amp; 4 &amp; -1 \\<br /> 1 &amp; k &amp; 1 &amp; 1 \\<br /> 0 &amp; 1 &amp; k &amp; 1 \\<br /> -1 &amp; 1 &amp; 1 &amp; k<br /> \end{pmatrix}<br />

And tried converting it to REF

<br /> \begin{pmatrix}<br /> 1 &amp; 1 &amp; 4 &amp; -1 \\<br /> 0 &amp; 1 &amp; k &amp; 1 \\<br /> 0 &amp; 0 &amp; (-k^2+k-3) &amp; (3-k) \\<br /> 0 &amp; 0 &amp; (5-2k) &amp; (k-3)<br /> \end{pmatrix}<br />

I'm not sure if I should keep going trying to reduce this to REF to see which values of k will not work, but it just seems too messy.

Am I approaching this the wrong way?
 
Physics news on Phys.org
Unlikely as it may seem, it looks to me like checking the determinant is as easy or easier than row reduction. If you add the last row to each of the first two rows you get a column with 3 zeros leaving you with one 3x3 determinant, which is easy to just expand.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top