Help! Wrong Eigenvalues Using Matrix A

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SUMMARY

The forum discussion centers on the calculation of eigenvalues and eigenvectors for a matrix A using two methods: \(\lambda I - A\) and \(A - \lambda I\). The user encountered discrepancies in the results, specifically obtaining incorrect eigenvectors from the first method while the second method yielded correct results. The discussion highlights the importance of substituting eigenvectors back into the systems of equations derived from the echelon form to verify their validity, ultimately leading to the identification of a mistake in the calculations.

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Yankel
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Hello all,

I have a problem with eigenvalues. I tried finding eigenvalues and eigenvectors of a matrix A. I did once using:

\[\lambda I-A\]

And a second time using:

\[A-\lambda I\]

For the first eigenvalue I got identical eigenvectors in both methods, but for the second eigenvalue, the first method was wrong, while the second was correct. I can't find the problem. I am attaching my solution using

\[\lambda I-A\]

which is wrong. Can you assist ? Thank you !

View attachment 3800
 

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Why don't you substitute the third, problematic, eigenvector, in each of the three systems of equations you got in the process of finding the echelon form?
 
Not sure I follow you...

I would like to know why this eigenvector is problematic
 
You have three systems of linear equations on the second page with matrices
\[
\begin{pmatrix}1&-1&-1\\-1&1&-1\\0&0&2\end{pmatrix}\qquad
\begin{pmatrix}1&1&1\\0&0&-2\\0&0&1\end{pmatrix}\qquad
\begin{pmatrix}1&1&1\\0&0&1\\0&0&0\end{pmatrix}
\]
I suggest substituting $x=-1$, $y=1$, $z=0$ into those systems to see if it is really a solution.
 
I see, it isn't a solution. So you confirmed the fact that I made a mistake, but where is it ?
 
Maybe I was writing in Russian... (Smile)

Evgeny.Makarov said:
Why don't you substitute the third, problematic, eigenvector, in each of the three systems of equations you got in the process of finding the echelon form?
This way you can see which of the system is correct ($x=−1$, $y=1$, $z=0$ is not a solution) and which is wrong (it is a solution), and thus determine the first system that is wrong. Looking at it carefully, you'll see the mistake.

This is the way to find mistakes in other similar situations.
 
Hi,

After first arrow in second page you have modified first row of the matrix.
 
Evegeny, you were writing in English, not Russian, it's my fault for not getting it quicker. Now I understand (finally) your method, it's a good idea, thank you !

Fallen Angel, thanks ! I made a silly mistake, thanks for finding it !
 

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