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I have a matrix A: \begin{bmatrix}7/3&-2/3&-2/3\\-2/3&7/3&-2/3\\-2/3&-2/3&7/3\end{bmatrix} which I solve using the characteristic equation like so: https://i.stack.imgur.com/mqFIK.jpg

The problem is, the right eigenvalues are 1 and 3. What am I doing wrong?

Carl
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    is the middle component of the matrix $\color{red}-7/3$? I don't see the negative sign in that image you linked – J. W. Tanner May 03 '20 at 16:23
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    $1$ is not an eigenvalue of the matrix in the question, but if you change the $-7/3$ in the middle to $+7/3$ then $1$ is an eigenvalue. – Andreas Blass May 03 '20 at 16:25
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    Let $\lambda$ be an eigenvalue of $A$. The $c\lambda$ is an eigenvalue of $cA$. – Michael Hoppe May 03 '20 at 16:33
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    @J.W.Tanner Yes, that is correct, I wrote it in wrong. Fixed it! – Carl May 03 '20 at 16:37
  • You pulled out the factor of $1/3$ prematurely: you’ve computed $\det(A-\frac13\lambda I)$ so all of your eigenvalues are off by a factor of three. – amd May 03 '20 at 16:37
  • @AndreasBlass How would I solve it then? Because when I solve it I get eigenvalues 3 and 9 (with the correct middle 7/3) – Carl May 03 '20 at 16:38
  • @amd Oh! But I have learned that if all of the elements are divided by three, I can take them out by multiplying the determinant by that fraction for every column/row. Is that wrong? – Carl May 03 '20 at 16:40
  • Given the corrected version of your question, the quickest way is to use one of the approaches outlined here. – Ben Grossmann May 03 '20 at 16:41
  • @Carl see Michael's comment. If you multiply everything by $3$, then the eigenvalues you get are of the form $3\lambda$ where $\lambda$ is an eigenvalue of $A$. – Ben Grossmann May 03 '20 at 16:44
  • @Omnomnomnom I know that, however I do not want to calculate the eigenvalue of cA, but A. If you look at my solution via the link I provided, what is it I have done wrong in my calculation (trying to find the eigenvalues for A above)? – Carl May 03 '20 at 16:46
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    @Carl On the right side of the first $=$, the expression you have is equal to $(1/3)^3\det(3A - \lambda E) = \det(A - \frac 13 \lambda E)$. What you need is $\det(A - \lambda E)$. – Ben Grossmann May 03 '20 at 16:50

2 Answers2

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You went wrong right off the bat when you expanded $\det(A-\lambda I)$. The main diagonal elements of $A-\lambda I$ are all $\frac73-\lambda$, so when you pulled out a factor of $\frac13$ to simplify your calculations, they should’ve become $7-3\lambda$. What you’ve written down is instead the determinant of $A-\frac\lambda3I$, so all of your eigenvalues are off by a factor of three. To put it a different way, what you’ve done is to compute the eigenvalues of $3A$. It’s OK to scale a matrix in this way to simplify things since if $Av=\lambda v$ then $(cA)v=(c\lambda)v$, but you then have to remember to scale the resulting eigenvalues by $1/c$ to get the eigenvalues of the original matrix.

Note that for this matrix, it’s easy to find eigenvectors and eigenvalues directly, without solving the characteristic equation. It’s of the form $aI+bJ$, where $J$ is the $n\times n$ matrix of all ones and $b\ne0$. $bJ$ is a rank-one matrix with equal row sums, so $0$ is an eigenvalue with multiplicity $n-1$ and the vector of all $1$s is an eigenvector with eigenvalue $nb$. The eigenvalues of the original matrix are therefore $a$ and $a+nb$, respectively. In this case, $a=3$ and $b=-2/3$.

amd
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the eigenvalues of the matrix with all elements equal to $1$ (and 3 by 3) are (0,0,3).

the eigenvalues of the matrix with all elements equal to $-2/3$ (and 3 by 3) are (0,0,-2).

The eigenvalue of that matrix added to $3I,$ which is your actual matrix, are (3,3,1).

Will Jagy
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