# How to find the eigen vector for the given matrix?

Given matrix is $$A = \begin{bmatrix} 2 & 5\\ 4 & -2 \end{bmatrix}$$ How to find the eigen-vector? I'm stuck here $$\begin{bmatrix} 2 - 4i & 5\\ 4 & -2 -4i \end{bmatrix} \times \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$ Problem is here.
Problem Set Part I Problems

EDIT:: $$A = \begin{bmatrix} 2 & -5\\ 4 & -2 \end{bmatrix}$$

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These eigenvalues are real numbers and thus should not contain the imaginary unit $i$. –  ja72 Aug 21 '12 at 18:25
all right thanks for the hint :)) –  hasExams Aug 21 '12 at 18:44

"the eigen-vector" is vague. You're looking for an eigenvector associated with an eigenvalue. The eigenvalues of the given matrix are: $\pm 4i.$ So solve for $v_1$ and $v_2$ in $$\pmatrix{ 2 & -5\\ 4 & -2} v_1 = 4i v_1 \\ \pmatrix{ 2 & -5\\ 4 & -2} v_2 = -4i v_2$$

Edit (from the comments): Let the eigenvalues be $x_1, x_2.$ It's known (see your textbook) that $$x_1 + x_2 = \text{Trace}(A) = 0. \\ x_1 x_2 = \text{Det}(A) = 16.$$ Solve for $x_1, x_2$ to get the eigenvalues $\pm 4i.$

Edit (from the comments): Do you know about elimination? You will need $1$ step to solve $$\begin{bmatrix} 2 - 4i & -5\\ 4 & -2 -4i \end{bmatrix} \times \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

Multiply the first row by $\dfrac{−(2+4i)}{5}$ and add it the second row. Solve for $y$ then substitute back in to solve for $x.$

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huh .... i thought the eigen value is $\pm 4i$. Could you check the original problem?? –  hasExams Aug 21 '12 at 18:30
@testuser: Please show how you arrive at your eigenvalue. –  ja72 Aug 21 '12 at 18:31
@ja72 sorry typo :(( –  hasExams Aug 21 '12 at 18:34
all right ... thanks!! could you edit your answer so that it matches my Q a bit?? –  hasExams Aug 21 '12 at 18:50

The eigenvalues are the solutions to $\det(A-\lambda I)=0$. Since $$A-\lambda I=\begin{bmatrix}2-\lambda&-5\\4&-2-\lambda\end{bmatrix}\;,$$ $\det(A-\lambda I)=(2-\lambda)(-2-\lambda)+20=\lambda^2+16$, and $\lambda=\pm4i$. You should now take each of these eigenvalues and solve the system $A\vec x=\lambda\vec x$, i.e., $(A-\lambda I)\vec x=\vec 0$. For example, with $\lambda=4i$ you have to solve

$$\begin{bmatrix}2-4i&5\\4&-2-4i\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\;.$$

It’s not as messy as it looks: just multiply the first row by $2$ and the second by $1-2i$, and you get the augmented matrix

$$\begin{bmatrix}4-8i&-10&0\\4-8i&-10&0\end{bmatrix}\;,$$

since $(1-2i)(-2-4i)=-2(1-2i)(1+2i)=-2(1-4i^2)=-10$.

Added: Since it may be the manipulations with complex numbers that are causing the difficulty, I’ll go ahead and finish the reduction. The single non-zero row reduces to $\begin{bmatrix}2-4i&-5&0\end{bmatrix}$, so to finish the reduction you need to calculate $$\frac{-5}{2-4i}=\frac{-5}{2-4i}\cdot\frac{2+4i}{2+4i}=\frac{-10-20i}{4-16i^2}=\frac{-10-20i}{20}=-\frac12-i\;.$$ This gives you the eigenvector $$\begin{bmatrix}\frac12+i\\1\end{bmatrix}\;,$$ and indeed

$$\begin{bmatrix}2&-5\\4&-2\end{bmatrix}\begin{bmatrix}\frac12+i\\1\end{bmatrix}=\begin{bmatrix}-4+2i\\4i\end{bmatrix}=4i\begin{bmatrix}\frac12+i\\1\end{bmatrix}\;.$$

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I'm terribly sorry for typo ... could you check again please )( –  hasExams Aug 21 '12 at 18:35
@testuser: I’ve updated it; the computation is actually very similar. –  Brian M. Scott Aug 21 '12 at 18:40

You have two equations with two unknowns with the r.h.s. equals zero. The only non-trivial solution is found by solving for one variable in terms of the other, in any of the two equations. The results is identical.

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I'm terribly sorry for typo ... could you check again please )( –  hasExams Aug 21 '12 at 18:38

An eigenvector is a vector that solves the equation $(A-\lambda I)v = 0$, where $0$ represents the zero vector.

So first you must solve for your eigenvalues. It appears as though you attempted that, but I'm not sure that your results are correct. The characteristic polynomial is $$\det \begin{pmatrix} 2-\lambda & -5 \\ 4 & -2-\lambda \end{pmatrix} = (2-\lambda)(-2-\lambda)-(-5)\cdot 4 =-4+2\lambda-2\lambda+\lambda^2+20.$$

Set this equal to zero, and solve for $\lambda$.

Thus, one eigenvector will correspond to the solution to

$$\begin{pmatrix}2-\lambda & -5 \\ 4 & -2-\lambda\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}.$$

This can be solved easily through Gaussian elimination, but the answer might be a little ugly. The first Gaussian elimination step is to find a multiple $k$ such that when you subtract $k$ times the first row from the second row, you end up with a zero in the (2,1) entry:

$$\begin{pmatrix}2-\lambda & -5 \\ 4-k(2-\lambda) & -2-\lambda-k\cdot (-5)\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0-k\cdot (0)\end{pmatrix} \Longrightarrow$$ $$\begin{pmatrix}2-\lambda & -5 \\ 0 & -2-\lambda-k\cdot (-5)\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}.$$

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I'm terribly sorry for typo ... could you check again please )( –  hasExams Aug 21 '12 at 18:37
Sure, but now that I have shown you the steps, you should do the computations. –  Arkamis Aug 21 '12 at 18:39