Is the eigenvalue of this matrix really non-real ? If so How can I calculate this by hand?

I'm trying to find the Eigenvalues of the following $2\times 2$ matrix : $$\begin{bmatrix}-2 & -7\\ 1 & 2\end{bmatrix}$$

I've been getting mixed results: by hand I've tried calculating the eigenvalue and I ended up with $-\sqrt{11}$ and $\sqrt{11}$. But when I used maple to help me calculate the eigenvalues it gives me the following non-real numbers: $i\sqrt{3}$ $-i\sqrt{3}$

From what I've checked the answer Maple is giving me is correct and the one I keep getting by hand is incorrect. I would like some help understanding how I can get the same result manually if possible. Thanks for your time.

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I assume that by "I," you meant $i$, the imaginary unit. I have edited the question, feel free to change it if this is not what you intended. –  Stahl Jul 8 '13 at 18:44
You want to solve the eigenvalue equation $$\left|\begin{array}{cc} -2-\lambda & -7 \\ 1 & 2-\lambda\end{array}\right| = 0.$$ The expression you get is $-(2+\lambda)(2-\lambda)+7 = 0.$ From which you get $4-\lambda^2 = 7$ and thus $\lambda^2 = -3$. –  Cameron Williams Jul 8 '13 at 18:44
You mistaken a sign, check your computations. Instead of $x^2+7-4=0$ you got $x^2-7-4=0$... –  N. S. Jul 8 '13 at 18:46
A quick way to check would be to compute the square of the matrix, and see what you get. You get $-3I$, and not $11I$... –  Thomas Andrews Jul 8 '13 at 19:00

To find the eigenvalues of a square matrix $M$, you need to solve the equation $\det(M-\lambda I)=0$, where $I$ is the square identity matrix. In your case, you have:

$$\det\left(\left[\begin{array}{cc} -2 & -7 \\ 1 & 2 \end{array}\right]-\lambda\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\right) = 0$$ $$\det\left(\left[\begin{array}{cc} -2 -\lambda& -7 \\ 1 & 2-\lambda \end{array}\right]\right) = 0$$ $$(-2-\lambda)(2-\lambda)-(1)(-7)=0$$ $$\lambda^2+3=0$$ It follows that $\lambda^2 = -3$, i.e. $\lambda =\pm\operatorname{i}\!\sqrt{3}$.

Notice that even-by-even matrices often have no real eigenvalues. In the case or the plane, any sort of rotation will stop vector being kept on the same line. Where as, in the three dimensional case, even rotations have a fixed axis (think of the earth turning about its axis.)

Graphically, you can see this. Try plotting quadratics, cubics, quartics, quintics, etc. You'll see that every cubic and quintic will meet the $x$-axis, i.e. every degree three or five characteristic polynomial has at least one real root, i.e. every 3-by-3 or 5-by-5 matrix has a real eigenvalue. There are lots of quadratics and quartics that miss the $x$-axis, i.e. there are lots of 2-by-2 and 4-by-4 matrices with no real roots.

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It's pretty simple to find the eigenvalues of a $2\times 2$ matrix, since if we denote by $\lambda_1$ and $\lambda_2$ the eigenvalues of the given matrix $A$ then we have $$\lambda_1+\lambda_2=\mathrm{tr}(A)=0\quad\text{and}\quad \lambda_1\lambda_2=\det(A)=3$$ hence $-(\lambda_1)^2=3$ and then the eigenvalues are non real.

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