Basis for a Nul Space


by Dgray101
Tags: basis, space
Dgray101
Dgray101 is offline
#1
Dec14-13, 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?
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Vahsek
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#2
Dec14-13, 09:22 PM
P: 81
Quote Quote by Dgray101 View Post
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?
The way to find a basis for the nullspace is to identify all the free variables (which correspond to the free columns of the matrix): x1 and x3 are the free variables while x2 is a pivot variable.

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
x2= -√2 x3 where x3 is any real number
will solve the system, along with
x1= any real number.

Hence, you see that the full solution to Ax=0 is
x1 (1,0,0) + x3 (0,-√2,1),
and you can easily pick out the basis vectors again.
HallsofIvy
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#3
Dec16-13, 08:49 AM
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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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