Thread: find nondiagonal matrix whose eigenvalues and eigenvectors are given

1. [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
$\displaystyle \left[ {\begin{array}{c} -1 \\ 2 \\ \end{array} } \right]$ and $\displaystyle \left[ {\begin{array}{c} 1 \\ 1 \\ \end{array} } \right]$ respectivley.

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

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?

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

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

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 $\displaystyle A$ is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix $\displaystyle P$ , which its

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

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

Tonio

3. Originally Posted by tonio
We know that if $\displaystyle A$ is a square matrix which eigenvectors form a basis of the given vector space, then there exists an invertible matrix $\displaystyle P$ , which its

columns are eigenvectors of $\displaystyle A$

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 $\displaystyle D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right]$
Is it because we start with the smallest values of $\displaystyle \lambda$ 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)?

4. 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 $\displaystyle D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right]$

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 $\displaystyle \lambda$ 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)?
.

5. 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 $\displaystyle D=\left[ {\begin{array}{cc} 0 & 2 \\ -3 & 0 \\ \end{array} } \right]$
Is it because we start with the smallest values of $\displaystyle \lambda$ 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 $\displaystyle P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix}$. 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 $\displaystyle P= \begin{bmatrix}1 & -1 \\ 1 & 2\end{bmatrix}$ and $\displaystyle D= \begin{bmatrix}-3 & 0 \\ 0 & 2\end{bmatrix}$ 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.

find a 2*2 non diagonal matrix whose eigen values are 2,-3 and associated eigen vectors are?

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