# Complex Eigen Space

• Apr 14th 2010, 05:57 PM
dwsmith
Complex Eigen Space
$\begin{vmatrix}
1-\lambda & 1\\
-2 & 3-\lambda
\end{vmatrix}$

$\lambda=2\pm \mathbf{i}$

When $\lambda=2-\mathbf{i}$, $\begin{bmatrix}
-1+\mathbf{i} & 1\\
-2 & 1+\mathbf{i}
\end{bmatrix}\Rightarrow\begin{bmatrix}
1 & \frac{-1}{2}-\frac{1}{2}\mathbf{i}\\
0 & 0
\end{bmatrix}$
.

$x_1=\bigg(\frac{1}{2}-\frac{1}{2}\mathbf{i}\bigg)x_2$
$x_2$

$x_2*\begin{bmatrix}
1-\mathbf{i}\\
2
\end{bmatrix}$

$x_2*\begin{bmatrix}
1-\mathbf{i}\\
1
\end{bmatrix}$

Why am I having a discrepancy?
• Apr 14th 2010, 08:27 PM
tonio
Quote:

Originally Posted by dwsmith
$\begin{vmatrix}
1-\lambda & 1\\
-2 & 3-\lambda
\end{vmatrix}$

$\lambda=2\pm \mathbf{i}$

When $\lambda=2-\mathbf{i}$, $\begin{bmatrix}
-1+\mathbf{i} & 1\\
-2 & 1+\mathbf{i}
\end{bmatrix}\Rightarrow\begin{bmatrix}
1 & \frac{-1}{2}-\frac{1}{2}\mathbf{i}\\
0 & 0
\end{bmatrix}$
.

$x_1=\bigg(\frac{1}{2}-\frac{1}{2}\mathbf{i}\bigg)x_2$
$x_2$

Here is where you went wrong: it must be $x_1=\left(\frac{1}{2}+\frac{1}{2}\,i\right)x_2\iff (1-i)x_1=x_2$ , and thus an eigenvector corresponding to the above eigenvalue is, for example, $\binom{1+i}{2}$

Tonio

$x_2*\begin{bmatrix}
1-\mathbf{i}\\
2
\end{bmatrix}$

What the above and the below mean, anyway??

$x_2*\begin{bmatrix}
1-\mathbf{i}\\
1
\end{bmatrix}$

Why am I having a discrepancy?

.
• Apr 14th 2010, 08:29 PM
dwsmith
What does what mean?

• Apr 14th 2010, 08:40 PM
tonio
Quote:

Originally Posted by dwsmith
What does what mean?

Oh, my answer is correct with the info you gave. No doubt about that. You check what you wrote after substituing 2-i and etc.: the first line in that matrix was $1\,\,\,-\frac{1}{2}-\frac{1}{2}i$ , so solving this gives what I wrote. You forgot to change the sign of the imaginary part.

Tonio
• Apr 14th 2010, 08:41 PM
dwsmith
That was I was original doing until all my complex eigen spaces were wrong.
• Apr 14th 2010, 08:41 PM
tonio
Quote:

Originally Posted by dwsmith
What does what mean?

What does $x_2*\binom{1-i}{2}$ mean...

Tonio
• Apr 14th 2010, 08:42 PM
tonio
Quote:

Originally Posted by dwsmith
That was I was original doing until all my complex eigen spaces were wrong.

I have no idea what you mean by this.

Tonio
• Apr 14th 2010, 08:43 PM
dwsmith
That is the span of of the eigen space. Any number times the column vector is in the space.
• Apr 14th 2010, 08:44 PM
dwsmith
Quote:

Originally Posted by tonio
You forgot to change the sign of the imaginary part.

Tonio

I was changing the sign on the imaginary part at first until all my answer were wrong.
• Apr 14th 2010, 08:55 PM
tonio
Quote:

Originally Posted by dwsmith
I was changing the sign on the imaginary part at first until all my answer were wrong.

I still can' understand what you mean...but never mind: you wrote http://www.mathhelpforum.com/math-he...d7c6753d-1.gif , which I suppose means $\det(A-\lambda I)$ to find the eigenvalues of a matrix

which is then $A=\begin{pmatrix}1&1\\\!\!\!-2&3\end{pmatrix}$. You say your book says that $\binom{1-i}{1}$ and all its scalar multiples are eigenvalues of A corresponding to $\lambda = 2-1$. This is not so:

$\begin{pmatrix}1&1\\\!\!\!-2&3\end{pmatrix}\begin{pmatrix}1-i\\1\end{pmatrix}=$ $\begin{pmatrix}2-i\\1+2i\end{pmatrix}\neq (2-i)\begin{pmatrix}1-i\\1\end{pmatrix}$ , as you can easily check.

So either you miscopied some of the given terms, or you looked at the answer to the wrong question, or your book is dead wrong.

You may now want to check that the vector I gave you in my first post really is an eigenvector of the above matrix A corresponding to $\lambda = 2-i$ ...

Tonio
• Apr 14th 2010, 09:03 PM
dwsmith
Trust me I understand what you are saying because I did the problem the other way original.

However, I doubled checked the answers and I didn't have a copying error and this isn't just a one time occurrence of how they handling imaginary eigenvectors and spaces.

I could understand a book having a mistake on one problem but on all complex problems would be hard to fathom as well.

Therefore, I am stuck in a quandary since it is hard to have a consistency of the same error.
• Apr 14th 2010, 09:06 PM
tonio
Quote:

Originally Posted by dwsmith
Trust me I understand what you are saying because I did the other way original.

However, I doubled checked the answers and I didn't have a copying error and this isn't just a one time occurrence of how they are imaginary eigenvectors and spaces.

I could understand a book having a mistake on one problem but on all complex problems would be hard to fathom as well.

Therefore, I am stuck in a quandary since it is hard to have a consistency of the same error.

I can't tell about all the problems but in this particular one, with the info you provided, no doubt it is wrong.

May I ask what book is that, and page of exercise?

Tonio
• Apr 14th 2010, 09:09 PM
dwsmith
Linear Algebra with Applications (Ed. 8) by Steven J. Leon.

pg 294, 1e