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
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!
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.
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.
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?