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There seems to be some inconsistencies in my mind that I'm trying to clear up, regarding the null space and the dimension theorem:

This is the problem:

Find a matrix whose null space is spanned by the vectors: $(2, −3, 1, 1, −1), (1, 0, −2, 1, 1), (2, −2, 1, 0, −1), (−8, 3, 1, 1, 1)$ in $\Bbb R^5$

Now, since the null space is four-dimensional, it leads me to believe that the rank of the sought matrix should be $1$, since I'm in $\Bbb R^5$. Basically, the matrix will correspond to a transformation whose image is one-dimensional.

The answer, which is given without any explanation, gives me a 2x5-matrix with a rank of $2$. After calculating the null space of that matrix, I find (just as I expected), a three-dimensional null space. What gives? Is this an error in my literature? Or am I missing some big chunk of theory? Or did I misunderstand the question? How would you go about finding this matrix?

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Consider the four fundamental subspaces. Calling the operator A, you're given Null(A). From this, you can find an A.

There are only 3 linearly independent vectors in that set of vectors specifying the null space (you can verify this quickly by using rank([2,-3,1,1,-1;1,0,-2,1,1;2,-2,1,0,-1;-8,3,1,1,1]) in Octave). So the null space is a 3 dimensional subspace of $\mathbb{R}^5$. Thus by rank-nullity, you should have rank(A)+ null(A) = 5, or rank(A) = 2.

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    $\begingroup$ Aaa, it's sort of a trick question! I just took for granted that the four vectors were linearly independent, didn't even bother checking for that! Then it all makes a lot more sense! Thanks! :) $\endgroup$ – JuliusL33t Aug 9 '14 at 15:19

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