Finding the eigenvalues

**anna123456** · April 2nd 2010, 07:53 AM

Consider the matrix B =

$<br /> B = \begin{bmatrix} 1&-3\\1&1\end{bmatrix}<br />$

Find the eigenvalues 1 and 2 of B. Write eugenvalues in the form a+bi, where a; b E R.

When I solve for the determinant i get (1-/|(eigenvalue) )+3 and I'm stuck after because when i substitute a +bi instead of the eigenvalue I get a^2 + 2abi +b^2 +4 and it doesn't really help me.

If you know how to solve this I would really appreciate your help.Thanks!

**craig** · April 2nd 2010, 08:19 AM

When it says write eigenvalues in the form $a + bi$ , it means that you are going to get complex numbers as your values.

To find the eigen values we need to solve this following equation, (which I think you have already stated above, kind of.)

$(1- \lambda)(1- \lambda) - 1(-3) = 0$

Expanding out the brackets above and rearranging a little gives you the following equation:

$\lambda^2 - 2\lambda + 4 = 0$ .

Solve this using the quadratic formula to get your eigenvalues.

Hope this helps.

**Defunkt** · April 2nd 2010, 08:20 AM

Originally Posted by anna123456

Consider the matrix B =

$<br /> B = \begin{bmatrix} 1&-3\\1&1\end{bmatrix}<br />$

Find the eigenvalues 1 and 2 of B. Write eugenvalues in the form a+bi, where a; b E R.

When I solve for the determinant i get (1-/|(eigenvalue) )+3 and I'm stuck after because when i substitute a +bi instead of the eigenvalue I get a^2 + 2abi +b^2 +4 and it doesn't really help me.

If you know how to solve this I would really appreciate your help.Thanks!

$det (B-\lambda I) = det \begin{bmatrix} 1-\lambda &-3\\1&1- \lambda \end{bmatrix} = (1-\lambda )^2 + 3 = 0 \Leftrightarrow (1-\lambda )^2 = -3 ...$

Can you solve now?

**anna123456** · April 2nd 2010, 11:03 AM

Originally Posted by Defunkt

$det (B-\lambda I) = det \begin{bmatrix} 1-\lambda &-3\\1&1- \lambda \end{bmatrix} = (1-\lambda )^2 + 3 = 0 \Leftrightarrow (1-\lambda )^2 = -3 ...$

Can you solve now?

The problem is it doesn't have a solution in this form because when something is squared it's always going to be positve or equal to 0.

**Defunkt** · April 2nd 2010, 12:00 PM

Originally Posted by anna123456

The problem is it doesn't have a solution in this form because when something is squared it's always going to be positve or equal to 0.

Over $\mathbb{R}$ , yes, but not over the field of complex numbers $\mathbb{C}$ . The question even says that you are going to have a complex solution.

$(1-\lambda )^2 = -3 \Leftrightarrow 1 - \lambda = \pm \sqrt{-3} = \pm \sqrt{3}i \Leftrightarrow \lambda = 1 \pm \sqrt{3}{i}$

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