# Solving for complex eigen vectors

The given matrix is $$A= \begin{pmatrix} 3 & -2\\ 4 & -1\\ \end{pmatrix}$$

The roots are found as $1\pm2i$. The article then uses the eigen value $1+2i$ and rewrites the system as

$$(2-2i)x-2y=0$$ $$4x-(2+2i)y=0$$

Then it goes on saying:

In fact the two equations are identical since $(2+2i)(2-2i)=8$.So the system reduces to one equation $(1-i)x-y=0$

I didnt get this.

• The system of linear equations stems from inputting the value of the eigenvalue in the equation $\;\det (tI-A)\;$ , which makes the matrix $\;\lambda I-A\;$ singular. Thus, that homogeneous system has a non-trivial solution and thus one of the equations must be a linear combination of the other one(s) ... – Timbuc May 10 '15 at 19:43
• For any $2 \times 2$ matrix $A$, if it is singular, then you must be able to write one column (row) in terms of the other. – copper.hat May 10 '15 at 19:44
• Ok I got this: $2[(1-i)x-y]=(2-2i)x-2y$ and $[(1-i)x-y](2+2i) = 4x-(2+2i)y$, but still didnt get from where $(2+2i)(2-2i)=8$ has come. Must be doing something really stupid here. – Maha May 10 '15 at 19:54

Hint:

from the system: $$\begin{cases} (2-2i)x-2y=0\\ 4x-(2+2i)y=0 \end{cases}$$ multipling the first equation by $4$ and the second by $(2-2i)$, you have:

$$\begin{cases} 4(2-2i)x-8y=0\\ 4(2-2i)x-(2+2i)(2-2i)y=0 \end{cases}$$

$$\begin{cases} 4(2-2i)x-8y=0\\ 4(2-2i)x-8y=0 \end{cases}$$

I think that the remark about $(2+2i)(2-2i)=8$ is completely useless.

If $\lambda$ is an eigenvalue of the matrix $A$, then, by definition, $A-\lambda I$ has rank $<n$ (where $n$ is the order of the matrix $A$).

Thus the matrix of the system you're looking at has rank $<2$. Since the rank is at least $1$, you conclude that one of the equations is redundant, without any need to check. Just choose a non trivial one, in this case either of the two.

In case $n>3$ one can't predict which of the equations will be redundant, because the only information is that the rank is less than $n$, but it could be anything from $0$ to $n-1$, depending on the matrix and the eigenvalue.

• OK if that is useless then how the reduction to $(1-i)x-y$ is done? – Maha May 10 '15 at 20:12
• @Mahesha999 Just take the first equation and remove the common factor $2$. – egreg May 10 '15 at 20:13
• hey damn thanks...just want to know if there is any fact behind: If $\lambda$ is an eigenvalue of the matrix $A$, then, by definition, $A-\lambda I$ has rank $<n$ (where $n$ is the order of the matrix $A$). I am missing something simple intuitive stuff? – Maha May 11 '15 at 10:29
• @Mahesha999 There is $v\ne0$ with $Av=\lambda v$, which is the same as saying that $v\ne0$ is in the null space of $A-\lambda I$; having non zero null space and rank $<n$ are equivalent conditions for an $n\times n$ matrix. Thus one can think to the definition of eigenvalue as “$A-\lambda I$ is not invertible”. – egreg May 11 '15 at 11:35