# Eigenvalues

• Aug 29th 2011, 01:36 AM
Arron
Eigenvalues
I am trying to determine the eigenvalues of A

A =
(6 0 2)
(0 7 0)
(2 0 3)
Which is a 3x3 matrix.

So far I have.

The characteristic equation is det (A-lambda I)=0;

|6-lam 0 2|
|0 7-Lam 0| = 0
|2 0 3-Lam|

I expand the determinant, and obtain

(6-Lam)
|7-lam 0|
|0 3-lam|
-0+
|0 6-Lam| = 0
|2 0 |

Not sure how to simplyfy and factorise this expession.
Can anyone help?
• Aug 29th 2011, 01:53 AM
Prove It
Re: Eigenvalues
Quote:

Originally Posted by Arron
I am trying to determine the eigenvalues of A

A =
(6 0 2)
(0 7 0)
(2 0 3)
Which is a 3x3 matrix.

So far I have.

The characteristic equation is det (A-lambda I)=0;

|6-lam 0 2|
|0 7-Lam 0| = 0
|2 0 3-Lam|

I expand the determinant, and obtain

(6-Lam)
|7-lam 0|
|0 3-lam|
-0+
|0 6-Lam| = 0
|2 0 |

Can anyone help?

You need to solve $\displaystyle \displaystyle |\mathbf{A} - \lambda \mathbf{I}| = 0$ for $\displaystyle \displaystyle \lambda$.

\displaystyle \displaystyle \begin{align*} \mathbf{A} - \lambda \mathbf{I} &= \left[\begin{matrix}6 & 0 & 2 \\ 0 & 7 & 0 \\ 2 & 0 & 3\end{matrix}\right] - \left[\begin{matrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix}\right] \\ &= \left[\begin{matrix} 6 - \lambda & 0 & 2 \\ 0 & 7 - \lambda & 0 \\ 2 & 0 & 3 - \lambda \end{matrix}\right] \\ |\mathbf{A} - \lambda \mathbf{I}| &= \left|\begin{matrix}6 - \lambda & 0 & 2 \\ 0 & 7 - \lambda & 0 \\ 2 & 0 & 3 - \lambda\end{matrix}\right| \\ &= (6 - \lambda)\left|\begin{matrix} 7 - \lambda & 0 \\ 0 & 3 - \lambda \end{matrix}\right| - 0 \left|\begin{matrix} 0 & 0 \\ 2 & 3-\lambda \end{matrix}\right| + 2\left|\begin{matrix}0 & 7-\lambda \\ 2 & 0 \end{matrix}\right| \\ &= (6 - \lambda)(7 - \lambda)(3 - \lambda) - 2\cdot 2(7 - \lambda) \\ &= (7-\lambda)\left[(6-\lambda)(3-\lambda) - 4\right] \\ &= (7-\lambda)\left(18 -6\lambda -3\lambda + \lambda ^2 - 4\right) \\ &= (7-\lambda)(\lambda ^2 - 9\lambda + 14)\end{align*}

So now solve $\displaystyle \displaystyle (7-\lambda)(\lambda - 7)(\lambda - 2) = 0$ for $\displaystyle \displaystyle \lambda$ :)
• Aug 29th 2011, 03:22 AM
HallsofIvy
Re: Eigenvalues
Quote:

Originally Posted by Arron
I am trying to determine the eigenvalues of A

A =
(6 0 2)
(0 7 0)
(2 0 3)
Which is a 3x3 matrix.

So far I have.

The characteristic equation is det (A-lambda I)=0;

|6-lam 0 2|
|0 7-Lam 0| = 0
|2 0 3-Lam|

I expand the determinant, and obtain

(6-Lam)
|7-lam 0|
|0 3-lam|
-0+
|0 6-Lam| = 0
|2 0 |

Not sure how to simplyfy and factorise this expession.
Can anyone help?

Expanding on the middle column (or middle row), because of those "0"s, give you
$\displaystyle (7- \lambda)\left|\begin{array}{cc}6-\lambda & 2 \\ 2 & 3- \lambda\end{array}\right|$
$\displaystyle = (7- \lambda)(\lambda^2- 9\lambda+ 18- 4)= (7- \lambda)(\lambda^2- 9\lambda+ 14)$
and that last quadratic factors easily.
• Aug 29th 2011, 04:07 AM
Arron
Re: Eigenvalues
Thanks, so the eigenvalues are;

Lambda=7, lambda=7 and lambda=2.