How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:

$\displaystyle

\begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}

$

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- Apr 17th 2009, 12:00 PMjennifer1004polynomial, eigenvalues and eigenvectors
How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:

$\displaystyle

\begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}

$ - Apr 17th 2009, 12:10 PMrunning-gag
Hi

t is an eigenvalue of matrix A iff there exists a non-zero vector X such that AX = tX or (A-tI)X = 0 where I is the identity matrix (1 on the diagonal ; 0 everywhere else)

This means that the determinant of A-tI must be equal to 0

Set the determinant of $\displaystyle

\begin{pmatrix}2-t & -2 & 3\\0 & 3-t & -2\\0 & -1 & 2-t\end{pmatrix}

$ to 0 and solve for t to find the eigenvalues - Apr 17th 2009, 01:06 PMrubix
Hi there jennifer1004

for easiness let us denote given matrices as A.

A = $\displaystyle \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}

$

**first to find characteristic polynomial:**

**det ( A - xID)**

1) subtract x from diagonal (the one going from left to right) of the matrix A; leave rest as it is.

$\displaystyle \begin{pmatrix}2-x & -2 & 3\\0 & 3-x & -2\\0 & -1 & 2-x\end{pmatrix}

$

2) take the determinant of matrices obtained from # 1

det A = [ (2-x) (3-x) (2-x) ] - [ (-1) (-2) (2-x) ]

3) set the result you get from # 2 equal to zero. This is your characteristic function.

**secondly, to find e-values:**

4) solve for x in your characteristic function. These value(s) of x is your e-value(s).

**lastly, to find e-vectors:**5) If you obtained more than one e-value then you'll have to do follow following step more than once - one e-value at a time. Say, you had two e-values namely p and q. Go over p and then over q. I'll do it for p.

(A - xID) V = 0 ; where ID is the identify matrix and V is the e-vector that you are looking for.

6) go back to #1 and plug p in there. You'll get

$\displaystyle \begin{pmatrix}2-p & -2 & 3\\0 & 3-p & -2\\0 & -1 & 2-p\end{pmatrix}

$

{remember p is a number...it is a e-value}

7) find A - xID

$\displaystyle \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix} - p \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix} = $

8) use the formula (A - xID) V = 0 and solve

(A - xID) $\displaystyle \begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix} $

hope that helps! - Apr 18th 2009, 04:51 AMTwighi
Easiest way to solve the determinant is probably using a cofactor expansion across column 1.

$\displaystyle (2-\lambda)((3-\lambda)(2-\lambda) -2) $

Now solve the quadratic equation and you have your eigenvalues.

And as rubix already explained, you find your eigenvectors by finding a basis for the nullspace of the matrix $\displaystyle A-\lambda_{1}I $ , where $\displaystyle \lambda_{j} \, j=1,2,3 $ are the eigenvalues. - Apr 20th 2009, 07:17 AMjennifer1004
Thank you for your explanations. I will read them over more carefully very shortly. I am trying to focus and absorb so many different types of info right now that it is so hard for me to just understand everything. It is a little overwhelming, but thank you for all the time you spent helping me.

- Apr 21st 2009, 12:26 PMjennifer1004
I got the polynomial of $\displaystyle x^{3}-7x^{2}+14x-8$

Eigenvalues of 2, 4, 1

Eigenvectors <-1,1,1> for 1, <1,0,0> for 2 and <7/2, -2, 1> for 4. I'm not sure if these are even correct but would these be my final eigenvectors, or am I supposed to multiply them by the corresponding eigenvalue. For example, <1, 0, 0> for 2 would be <2, 0, 0> instead as my final eigenvector? - Apr 21st 2009, 04:06 PMjennifer1004
I think I got this now. I believe I have to multiply the eigenvectors by the corresponding eigenvalue. My final eigenvectors should be:

for 2 <2 0 0>

for 4 <14 -8 4>

for 1 <-1 1 1>

?