Pruning a set of matrices (easy)

[sid9221]

http://dl.dropbox.com/u/33103477/prune.png

I am unsure if the the answer is:

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}}$$

or

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 2 & 7 \\ <br /> -4 & 1<br /> \end{pmatrix}}$$

I'm pretty sure it's the second one but am a bit confused about when the algorithm stops.
Can someone confirm the answer ?

 

[Dustinsfl]

http://dl.dropbox.com/u/33103477/prune.png

I am unsure if the the answer is:

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}}$$

or

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 2 & 7 \\ <br /> -4 & 1<br /> \end{pmatrix}}$$

I'm pretty sure it's the second one but am a bit confused about when the algorithm stops.
Can someone confirm the answer ?

I have never encountered the word prune, but by context, I believe it is saying X is linearly dependent and you want to make a set Y (lin ind) from X that spans $M_{2,2}$ as well.

If that is the case, you just need to verify that none of the matrices in your 3 set are linearly dependent to the other two. If that is the case, then the 3 matrix set is correct.

 

[Ray Vickson]

http://dl.dropbox.com/u/33103477/prune.png

I am unsure if the the answer is:

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}}$$

or

$${\begin{pmatrix}<br /> 2 & 1 \\ <br /> 5 & 1<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 3 & -1 \\ <br /> 7 & 4<br /> \end{pmatrix}},<br /> {\begin{pmatrix}<br /> 2 & 7 \\ <br /> -4 & 1<br /> \end{pmatrix}}$$

I'm pretty sure it's the second one but am a bit confused about when the algorithm stops.
Can someone confirm the answer ?

Lay them out as 4-dimensional vectors, then look for a subset that generates the whole span. In other words, start with the vectors
$$\begin{array}{l}v_1 = [2,1,5,1] \\v_2 = [3,-1,7,4] \\ v_3 = [5,-5,11,10] \\<br /> v_4 = [2,7 -4,1]\end{array}$$ Obviously, $v_1, v_2$ are linearly independent.
Are the three vectors $v_1, v_2, v_3$ linearly independent? If not, $v_3$ is in the span of $v_1, v_2.$ If they are linearly independent, throw in $v_4$ and continue the test.

Note: you can do it all in one step, just by doing row operations on the 4x4 matrix with rows v_i. Basically, you are looking for solutions of the 4x4 linear system $c_1 v_1 + c_2 v_2 + c_3 v_3 + c_4 v_4 = 0.$

RGV