28
$\begingroup$

I read through similar questions, but I couldn't find an answer to this:

How do you determine the symmetric matrix A if you know:

$\lambda_1 = 1, \ eigenvector_1 = \pmatrix{1& 0&-1}^T;$

$\lambda_2 = -2, \ eigenvector_2 = \pmatrix{1& 1& 1}^T;$

$\lambda_3 = 2, \ eigenvector_3 = \pmatrix{-1& 2& -1}^T;$

I tried to solve it as an equation system for each line, but it didn't work somehow.

I tried to find the inverse of the eigenvectors, but it brought a wrong matrix.

Do you know how to solve it?

Thanks!

$\endgroup$
1

3 Answers 3

43
$\begingroup$

Writing the matrix down in the basis defined by the eigenvalues is trivial. It's just $$ M=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{array} \right). $$ Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely: $$ S = \left( \begin{array}{ccc} 1 & 1 & -1 \\ 0 & 1 & 2 \\ -1 & 1 & -1 \\ \end{array} \right). $$ This is just the matrix whose columns are the eigenvectors. We can change to the standard coordinate bases by computing $SMS^{-1}$. We get $$ SMS^{-1} = \frac{1}{6}\left( \begin{array}{ccc} 1 & -8 & -5 \\ -8 & 4 & -8 \\ -5 & -8 & 1 \\ \end{array} \right). $$ You can check that this matrix has the desired eigensystem. For example, $$ \frac{1}{6}\left( \begin{array}{ccc} 1 & -8 & -5 \\ -8 & 4 & -8 \\ -5 & -8 & 1 \\ \end{array} \right) \left( \begin{array}{c} -1 \\ 2 \\ -1 \end{array} \right) = \left( \begin{array}{c} -2 \\ 4 \\ -2 \end{array} \right). $$

$\endgroup$
5
  • $\begingroup$ sorry, I don't get it. Shouldn't be the result Ax = λx for all the three pairs of values and vectors? Like for example for the first pair: result-matrix * u1 = 1 * u1? $\endgroup$ Jan 26, 2015 at 0:54
  • 1
    $\begingroup$ @user3435407 Yep - I guess I don't get what you don't get. I added one check. $\endgroup$ Jan 26, 2015 at 1:28
  • 6
    $\begingroup$ People: If you were searching for the answer of this question too, now I found a good live demonstration of the above mentioned solution here: youtube.com/watch?v=HWnCv4iHCDc $\endgroup$ Jan 26, 2015 at 3:40
  • $\begingroup$ I don't understand: how was M found in this example (beside a trivial thing given those precise values)? $\endgroup$ Dec 30, 2021 at 23:04
  • $\begingroup$ @totalMongot M was obtained by putting the eigenvalues in the diagonal and setting everything else to 0. $\endgroup$ Jul 8, 2023 at 23:20
16
$\begingroup$

call the eigenvectors $u_1, u_2$ and $u_3$ the eigenvectors corresponding to the eigenvalues $1, -2, $ and $2.$ then $$A = 1\dfrac{u_1u_1^T}{u_1^Tu_1} - 2\dfrac{u_2u_2^T}{u_2^Tu_2} + 2\dfrac{u_3u_3^T}{u_3^Tu_3}$$

you can verify this by computing $Au_1, \cdots$. this expression for $A$ is called the spectral decomposition of a symmetric matrix.

$\endgroup$
1
  • $\begingroup$ ok, thanks, I need to count it through first, I'll let you know when I managed it! $\endgroup$ Jan 26, 2015 at 0:11
11
$\begingroup$

An $n\times n$ matrix with $n$ independent eigenvectors can be expressed as $A=PDP^{-1}$, where $D$ is the diagonal matrix $\operatorname{diag}(\lambda_1\:\lambda_2\:\cdots\lambda_n)$ and $P$ is the matrix $(\vec{v}_1\:|\:\vec{v}_2\:|\cdots|\:\vec{v}_n)$ where $v_i$ is the corresponding eigenvector to $\lambda_i$.

$$D=\begin{pmatrix}1&0&0\\0&-2&0\\0&0&2\end{pmatrix}$$ $$P=\begin{pmatrix}1&0&-1\\1&1&1\\-1&2&-1\end{pmatrix}$$

$\endgroup$
2
  • $\begingroup$ Does this approach have a name? $\endgroup$ Aug 26, 2021 at 14:50
  • $\begingroup$ I think this is colled diagonalization of a matrix, found the answer after asking the question $\endgroup$ Aug 26, 2021 at 15:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .