Linear Algebra help!
Determine which of the following subsets of \[\mathbb{R} ^{n}\] are in fact subspaces of \[\mathbb{R} ^{n}\] (n>2).
\[\left\{ x|Ax=b, where A _{m*n} \neq0 and b _{m*1}\neq0 \right\}\]
SOMEBODY HELP!!!
saifoo that doesn't show anything..
now>
I asked if: {x| Ax=b, where A is not a 0 matrix and b is not a 0 column} was a subset of R^n
Take R3, you have 8 dimensions, 1 subspace of 0-dim, 3 of 1-dim, 3 of 2-dim and 1 of 3-dim. In general for Rn, total dim = 2^n Maybe that helps?
No my point is.. is this set: S = {x|Ax=b, where A is a non-zero matrix and b is a non-zero column} a subspace of R^n? I have to use the proof, you know the sum must belong in the set and the multiplication by a scalar must belong in the set too..
Square matrices?
No, I wrote A is m by n and column b is m by 1
a bit higher up..like 3 post
Is m > n?
not necessarily.. A is just a matrix with m rows and n columns... b is a column...
m and n are just any numbers of rows and columns..
So in R3, say, A could be a (any number) by 3 matrix?
okay nevermind, I got the answer.. this is how if you are interested: Suppose S = {x|Ax=b, where A is any non-zero matrix and b is any non-zero column}, or S is the solution set to the system Ax=b
Suppose x1 and x2 (x sub 1 and x sub 2) are solutions to Ax=b, then A x1 = b A x2 = b Now let (x1 + x2) = x3
If S is a subspace, then A x3 = b too, but A x1 + A x2 = b + b = A (x1+x2) = A x3 = 2b
2b is not equal to b, unless b=0, therefore this is not a subspace..
I deserve a good answer too xD, but thanks for the effort estudier ..
actually i have one more question..
I will have to chew on that for a bit, I'm not sure that I follow...
Are non-singular matrices subspaces? What this means is, if I add two non-singular matrices of the same size, will the sum be non-singular and if I multiply a nonsingular matrix by a scalar is the result non-singular as well? Size can be anything..
I am more accustomed to treating Rn as a vector space and matrices just as vector representations.
I think I got this now, your b is in Rm. If Ax = b is solved by a particular solution p, then your solution set is all vectors equal to p + v, where v is is a solution of Ax = 0.
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