Is Your Vector Part of the Matrix's Solution Space?

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To determine if a vector x is in the solution space of a matrix M, one must check if Mx equals a specific vector b. In this case, the discussion focuses on the null space, where b is zero, leading to the equation Mx=0. The example vectors v1 and v2 were tested, and it was concluded that v1 is not in the null space since Mv1 does not equal zero. There is also concern about the phrasing of the question, as it may imply a broader context involving a non-zero vector b. Clarification on the problem's wording is necessary for accurate interpretation.
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Homework Statement


a)Let M be a m*n matrix and x be a n*1 coordinate vector. How can you check whether or not x is in the solution space of M?

[0 1 1 1 0]
M=[1 1 0 0 1]
[0 1 1 0 1]
[1 0 1 0 0]

b)To decide whether or not the following are in the solution space of M
i) v1=[1 0 0 1 1]T ii) v2=[1 0 0 1 1]T
*T means the transpose of the matrices

Any help is appreciated


Homework Equations



{x\inℝn:Ax=0}

The Attempt at a Solution



I started by solving the homogeneous linear equation:
M*v1=0

[0 1 1 1 0]
[1 1 0 0 1]*[0 1 0 1 1]T = 0
[0 1 1 0 1]
[1 0 1 0 0]

[2 2 2 0]T ≠ 0

∴ v1 is not in the solution space of M

Am i doing the right here??
 
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What, exactly, was the wording of the question? To ask about the "solution space" of just a matrix, M, makes no sense. We talk about the "solution space" of an equation like "Mx= b" and the answer depends on b as well as M. The solution space of "Mx= 0", with b specfically equal to 0, is the "null space" or "kernel" of matrix M.

Yes, to determine whether a given vector, x, is in the solution space of Mx= b, simply multiply M and the given x and see if the result is equal to b. Assuming that the problem is really asking whether the given x is in the "null space", so that b= 0, it is immediately clear that the top row times x does not give 0 and so x is not in the null space.

But I am still concerned about the wording of the question. If it really said "solution space", it is possible that there is some non-zero "b", perhaps given in a previous part of the problem, that you missed.
 
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