Register to reply

Basis for a Nul Space

by Dgray101
Tags: basis, space
Share this thread:
Dgray101
#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?
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
Vahsek
#2
Dec14-13, 09:22 PM
P: 84
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
#3
Dec16-13, 08:49 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552
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.


Register to reply

Related Discussions
Linear Algebra: Basis vs basis of row space vs basis of column space Calculus & Beyond Homework 8
Finding basis of a column space/row space Linear & Abstract Algebra 1
Basis for tangent space and cotangent space Differential Geometry 23
Fourier of Basis Points (Basis in Reciprocal space) (Convolution Theorem) Atomic, Solid State, Comp. Physics 0
Basis for null space, row space, dimension Calculus & Beyond Homework 1