What is the nullspace of a 3x3 complex matrix?

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SUMMARY

The nullspace of the complex matrix C = (1, -1, 1; 2, 0, 1+i; 0, 1+i, -1) is determined by reducing it to row echelon form, resulting in (1, -1, 1; 0, 1, -0.5+0.5i; 0, 0, 0). The basis for the nullspace is given by the vector (-0.5-0.5i, 0.5-0.5i, 1), which spans the subspace defined by the equations x + (0.5 + 0.5i)z = 0 and y + (-0.5 + 0.5i)z = 0. The use of tools such as Mathematica and Matlab confirmed the presence of a zero row in the reduced row echelon form, validating the findings.

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math2010
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Homework Statement


I have the 3x3 matrix C=(1,-1,1; 2,0,1+i; 0,1+i,-1) and I want to find its nullspace (a set of vectors that span that subspace).


The Attempt at a Solution


So first I have reduced the matrix to row echelon form and I got 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?

By "i" I mean imaginary since this is a complex matrix.
 
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I ended up with a different row-reduced matrix, with no rows of zeroes.
 


Are you sure? Because I used Mathematica to check the reduced row echelon form of this matrix, and it seems the rref has a row of zeros!

Also, does the set containing (-1,0.5+0.5i,1) and (1,0,0) span the subspace?
 
Last edited:


I tried row-reducing it again using Matlab and I still got a zero row:

1 0 0.5 + 0.5i
0 1 -0.5 + 0.5i
0 0 0
 


That reduced matrix corresponds to equations

x+(0.5+0.5 i)z = 0
y+(-0.5+0.5 i)z = 0

Solving for the other variables in terms of z, you get a solution of

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

The vector multiplying the z on the RHS is a basis of the nullspace.
 


math2010 said:
I tried row-reducing it again using Matlab and I still got a zero row:

1 0 0.5 + 0.5i
0 1 -0.5 + 0.5i
0 0 0
I agree with your result now.
 

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