Basis for Null Space: Finding Basis Vectors | Explained and Solved

[Dgray101]

Hey guys so we need to find the basis for

0 1 $\sqrt{2}$
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?

 

[Vahsek]

Hey guys so we need to find the basis for

0 1 $\sqrt{2}$
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]

Equivalently, the null space of this matrix is the set of all (x, y, z) such that
$$\begin{pmatrix}0 & 1 & \sqrt{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$

$$\begin{pmatrix}y+ \sqrt{2}z \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

which, as Vahsek said, reduces to the single equation $y+ \sqrt{z}= 0$ or $y= -\sqrt{2}z$ (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 $(x, y, z)= (x, -\sqrt{2}z, z)= (x, 0, 0)+ (0, -\sqrt{2}z, z)= x(1, 0, 0)+ z(0, -\sqrt{2}, 1)$. A vector is in the null space of this matrix if and only if it is a linear combination of $(1, 0, 0)$ and $(0, -\sqrt{2}, 1)$, exactly as Vahsek said.