# Thread: polynomial, eigenvalues and eigenvectors

1. ## polynomial, eigenvalues and eigenvectors

How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
$\displaystyle \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}$

2. Originally Posted by jennifer1004
How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
$\displaystyle \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}$
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

3. Originally Posted by jennifer1004
How do I find the characteristic polynomial, eigenvalues, and eigenvectors of the following matrix:
$\displaystyle \begin{pmatrix}2 & -2 & 3\\0 & 3 & -2\\0 & -1 & 2\end{pmatrix}$
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:

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

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.

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!

4. ## hi

Easiest way to solve the determinant is probably using a cofactor expansion across column 1.

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

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.

5. 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.

6. 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?

7. 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>

?