Basis for ROW(A), COL(A) and NUL of a square matrix

[Susanne217]

deleted

No one answered

 

[vela]

Looks fine up until the null space. Ax=0 yields the equations

$$x_1+(1+i)x_2+x_4 = 0$$
$$x_3+5ix_4 = 0$$

You have four unknowns and only two equations, so you can solve for two of the variables in terms of the other two.

$$x_1 = -(1+i)x_2-x_4$$
$$x_3 = -5ix_4$$

In vector form, this would be

$$\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = x_2\begin{pmatrix}{-1-i\\1\\0\\0}\end{pmatrix} + x_4\begin{pmatrix}-1\\0\\-5i\\1\end{pmatrix}$$

The two constant vectors are a basis of the null space.

 

[Susanne217]

Sorry I deleted it. I thought no would answer and got frustrated. But thanks.