find nondiagonal matrix whose eigenvalues and eigenvectors are given

Printable View

April 23rd 2010, 04:33 PM
superdude

[RESOLVED]find nondiagonal matrix whose eigenvalues and eigenvectors are given

Find a 2x2 nondiagonal matrix whose eigenvalues are 2 and -3 and associated eigenvectors are
$\left[ {\begin{array}{c} -1 \\ 2 \\ \end{array} } \right] $ and $ \left[ {\begin{array}{c} 1 \\ 1 \\ \end{array} } \right]$ respectivley.

The answer key has:
Let $D=\left[ {\begin{array}{cc} 2 & 0 \\ 0 & -3 \\ \end{array} } \right] $ and $P=\left[ {\begin{array}{cc} -1 & 1\\ 2 & 1 \\ \end{array} } \right] $
Then $P^{-1}AP=D$ so $A =PDP^{-1}=\frac{1}{3} \left[ {\begin{array}{cc} -4 & -5 \\ -10 & 1 \\ \end{array} } \right] $

I don't see the intuition behind the first part, how do they know to come up with D and P by slamming together the eigenvectors and eigen values?
April 24th 2010, 01:37 AM
tonio

Quote:

Originally Posted by superdude

Find a 2x2 nondiagonal matrix whose eigenvalues are 2 and -3 and associated eigenvectors are
$\left[ {\begin{array}{c} -1 \\ 2 \\ \end{array} } \right] $ and $ \left[ {\begin{array}{c} 1 \\ 1 \\ \end{array} } \right]$ respectivley.

The answer key has:
Let $D=\left[ {\begin{array}{cc} 2 & 0 \\ 0 & -3 \\ \end{array} } \right] $ and $P=\left[ {\begin{array}{cc} -1 & 1\\ 2 & 1 \\ \end{array} } \right] $
Then $P^{-1}AP=D$ so $A =PDP^{-1}=\frac{1}{3} \left[ {\begin{array}{cc} -4 & -5 \\ -10 & 1 \\ \end{array} } \right] $

I don't see the behind the first part, how do they know to come up with D and P by slamming together the eigenvectors and eigen values?

We know that if $A$ is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix $P$ , which its

columns are eigenvectors of $A$ , s.t. $P^{-1}AP=D$ is a diagonal matrix whose diagonal elements are precisely the eigenvalues of $A$ (all this is classic theory, not

something new). Well, they just did all the process backwards...(Wink)

Tonio
April 24th 2010, 03:39 PM
superdude

Quote:

Originally Posted by tonio

We know that if $A$ is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix $P$ , which its

columns are eigenvectors of $A$

Tonio

How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be $ D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right] $
Is it because we start with the smallest values of $\lambda$ and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?
April 24th 2010, 06:22 PM
tonio

Quote:

Originally Posted by superdude

How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be $ D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right] $

This matrix isn't diagonal so it cannot be in our case , and if the first eigenvalue appears in D then the first column of the invertible matrix P will be an eigenvector corresponding to this eigenvalue...

Tonio

Is it because we start with the smallest values of $\lambda$ and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?

.
April 25th 2010, 03:49 AM
HallsofIvy

Quote:

Originally Posted by superdude

How do you know which eigenvectors are which column? Does it matter? So in my example if I found the eigenvectors how would I know it wouldn't be $ D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right] $
Is it because we start with the smallest values of $\lambda$ and work our way up and the eigenvectors found get put into the matrix D left from right? If this is true, then what happens if 2 eigenvectors are associated with one eigenvalue (is this even possible)?

You are confusing "eigenvalues" with "eigenvectors". D is the diagonal matrix having the eigenvalues on the diagonal. Swapping eigenvectors would give you $P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix}$ . If you kept the same the same D matrix, you would get a matrix having the same eigenvalues and eigenvectors but with the eigenvector now corresponding to the othere eigenvalue.

If you swap both, say $P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix}$ and $D= \begin{bmatrix}-3 & 0 \\ 0 & 2\end{bmatrix}$ then you would get another matrix having the same eigenvalues and the same eigenvectors corresponding to those eigenvalues as the first matrix.

Choose D to be the diagonal matrix having the eigenvalues on the diagonal and P to be the matrix having the corresponding eigenvectors as columns- the first column is an eigenvector for the eigenvalue in the first column, the second column of P corresponds to the eigenvalue in the second column of D, etc.