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

or

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

or

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

or

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