# find nondiagonal matrix whose eigenvalues and eigenvectors are given

• Apr 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.

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?
• Apr 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.

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
• Apr 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)?
• Apr 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)?

.
• Apr 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.