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Basis for a Nul Space 
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#1
Dec1413, 06:50 PM

P: 33

Hey guys so we need to find the basis for
0 1 [itex]\sqrt{2}[/itex] 0 0 0 0 0 0 I know how you do it. But my prof says that one of the basis vectors is (1 0 0) but I don't know how he arrives at this? 


#2
Dec1413, 09:22 PM

P: 81

Since the number of free columns (or number of free variables) equals 2, you will get 2 special solutions for Ax=0, and hence 2 basis vectors. The rest you know how to do: to find 1 of the special solutions, set one of the free variables to 1 and the rest 0, and solve for the pivot variables. Doing this procedure, would give the following basis vectors for the nullspace: (1,0,0) and (0,√2,1). Alternatively, you can still see why the above 2 vectors are a basis. You can easily see that all the solutions of the form x_{2}= √2 x_{3} where x_{3} is any real number will solve the system, along with x_{1}= any real number. Hence, you see that the full solution to Ax=0 is x_{1} (1,0,0) + x_{3} (0,√2,1), and you can easily pick out the basis vectors again. 


#3
Dec1613, 08:49 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,363

Equivalently, the null space of this matrix is the set of all (x, y, z) such that
[tex]\begin{pmatrix}0 & 1 & \sqrt{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}[/tex] [tex]\begin{pmatrix}y+ \sqrt{2}z \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}[/tex] which, as Vahsek said, reduces to the single equation [itex]y+ \sqrt{z}= 0[/itex] or [itex]y= \sqrt{2}z[/itex] (the other two rows being just 0= 0). There is no condition on x so x can be any thing. That is, we can write [itex](x, y, z)= (x, \sqrt{2}z, z)= (x, 0, 0)+ (0, \sqrt{2}z, z)= x(1, 0, 0)+ z(0, \sqrt{2}, 1)[/itex]. A vector is in the null space of this matrix if and only if it is a linear combination of [itex](1, 0, 0)[/itex] and [itex](0, \sqrt{2}, 1)[/itex], exactly as Vahsek said. 


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