# Nullspace of Complex Matrix

• Feb 11th 2010, 01:15 PM
iain3915
Nullspace of Complex Matrix
Hi. I have the 3*3 matrix C=(1,-1,1; 2,0,1+i; 0,1+i,-1) and I must find an orthonormal basis for it's nullspace.(Headbang)

So first I have reduced the matric to row echelon form and I have this matrix
(1,-1,1; 0,1,-0.5+0.5i; 0,0,0)

How do I read off from this the nullspace of this matrix? What is a basis for this nullspace?

My plan once I have that is to use gram schmitt to create an orthonormal basis.

Any help much appreciated!
• Feb 11th 2010, 03:43 PM
Roam
Quote:

Originally Posted by iain3915
Hi. I have the 3*3 matrix C=(1,-1,1; 2,0,1+i; 0,1+i,-1) and I must find an orthonormal basis for it's nullspace.(Headbang)

So first I have reduced the matric to row echelon form and I have this matrix
(1,-1,1; 0,1,-0.5+0.5i; 0,0,0)

How do I read off from this the nullspace of this matrix? What is a basis for this nullspace?

My plan once I have that is to use gram schmitt to create an orthonormal basis.

Any help much appreciated!

You have to start by producing a set of vectors that span the subspace. The nullspace of your matrix is the solution space of "Ax=0", so the problem boils down to solving this linear system.

$A= \begin{bmatrix}1&-1&1 \\ 2&0&1+i \\ 0&1+i&-1\end{bmatrix}
\begin{bmatrix}x \\ y \\ z\end{bmatrix}$

Then you must apply the Gram-Schmit process that you mentioned to the vectors you found. If there are more than one vector, you must first construct the orthogonal basis vectors. If there is only one vector you just go ahead and normalize it to obtain the orthonormal basis vector. For example if you have $w_1= (-1,1,0)$, then

$
q_1 = \frac{w_1}{\| w_1 \|} = (\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0)
$

Try it.
• Feb 11th 2010, 03:52 PM
iain3915
Thanks Roam,

My problem is producing the set of vectors that span the subspace. (Silly, I know!)

Does the set containing (-1,0.5+0.5i,1) and (1,0,0) span the subspace?
• Feb 11th 2010, 09:02 PM
Roam
Quote:

Originally Posted by iain3915
Does the set containing (-1,0.5+0.5i,1) and (1,0,0) span the subspace?

No, that's not right at all! That reduced matrix corresponds to equations "x+(0.5+0.5 i)z = 0" and "y+(-0.5+0.5 i)z = 0". So if you solve for the other variables in terms of z, you get a solution of

$\begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}-0.5-0.5i \\ 0.5-0.5i \\ 1\end{bmatrix} z$

And that's a basis of the nullspace.