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I have to find a characteristic polynomial for the follwoing matrix:

$$A=\left( \begin{matrix}0 & -1 & 1\\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{matrix}\right)$$

My goal to find the roots of the characteristic polynomial.

The characteristic polynomial is given by solving $det(\lambda I -A)$, and then we get:

$$det(\lambda I -A)=det\left( \begin{matrix}t & 1 & -1\\ 1 & t & -2 \\ -1 & -2 & t \end{matrix}\right)$$

The final polynomial I get is $$P_A=t^3-6t+4$$

While there is a way to solve this polynomial I believe there is a way to make this determinant more "comfortable" to solve.

Can I add the third column to the second column?

$$det(\lambda I -A)=det\left( \begin{matrix}t & 0 & -1\\ 1 & t-2 & -2 \\ -1 & t-2 & t \end{matrix}\right)$$

And Then, can I do the following (and why?)

$$ det(\lambda I -A)=(t-2)*det\left( \begin{matrix}t & 0 & -1\\ 1 & 1 & -2 \\ -1 & 1 & t \end{matrix}\right)$$ What kind of elementary operations, or other arithmetic ones I can do, to make the determinant easier to solve?

Thanks,

Alan

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    $\begingroup$ Your phrasing seems very confused. Am I correct in assuming that what you want to do is find the roots of the characteristic polynomial? $\endgroup$ – Jack M Nov 2 '15 at 9:30
  • $\begingroup$ Yes, thank you. I will edit. $\endgroup$ – Alan Nov 2 '15 at 9:35
  • $\begingroup$ Did you try finding the characteristic equation after column operation as suggested in the question? $\endgroup$ – hjpotter92 Nov 2 '15 at 9:51
  • $\begingroup$ I have, I'll be greatful if you can look at my edit. A common factor was taken out of the matrix and I can't see why this can be done. $\endgroup$ – Alan Nov 2 '15 at 9:53
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    $\begingroup$ You have the row 1, col 2 and the row 2, col 1 elements interchanged in your second-to-last matrix. But yes, the operations you are doing are legal. $\endgroup$ – Will Orrick Nov 2 '15 at 10:02
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Personally, I prefer this method $$|A-\lambda I|=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ -1 & -\lambda & 2 \\ 1 & 2 & -\lambda \\ \end{array}\right|$$ $$=-\lambda\left| \begin{array}{rr} -\lambda & 2\\ 2 & -\lambda \\ \end{array}\right| +\left| \begin{array}{rr} -1 & 2\\ 1 & -\lambda \\ \end{array}\right| +\left| \begin{array}{rr} -1 & -\lambda\\ 1 & 2 \\ \end{array}\right|$$ $$=-\lambda^3+6\lambda-4 =(2-\lambda)(\lambda^2+2\lambda-2)$$ $$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$ You can also triangularize the matrix first. This makes the determinant trivial to compute $$|A-\lambda I|=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ -1 & -\lambda & 2 \\ 1 & 2 & -\lambda \\ \end{array}\right|$$ $$=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ 0 & 2-\lambda & 2-\lambda \\ 1 & 2 & -\lambda \\ \end{array}\right|$$ $$=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ 0 & 2-\lambda & 2-\lambda \\ \lambda & 2\lambda & -\lambda^2 \\ \end{array}\right|$$ $$=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ 0 & 2-\lambda & 2-\lambda \\ 0 & 2\lambda-1 & 1-\lambda^2 \\ \end{array}\right|$$ $$=\left| \begin{array}{rrr} -\lambda & -1 & 1\\ 0 & 2-\lambda & 2-\lambda \\ 0 & 0 & -\lambda^2-2\lambda+2\\ \end{array}\right|$$ $$=\left| \begin{array}{rrr} 1 & \lambda^{-1} & -\lambda^{-1}\\ 0 & 2-\lambda & 2-\lambda \\ 0 & 0 & \lambda^2+2\lambda-2\\ \end{array}\right|$$ $$=(2-\lambda)(\lambda^2+2\lambda-2)$$ $$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$ In any case, I suggest you look up the properties of the determinant.

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For 2 by 2 matrices and for 3 by 3, there is a fairly satisfactory cookbook method that you can do the same way every time, in order to find the characteristic polynomial (the eigenvalues are the roots of that).

For a 2 by 2, call the trace $\sigma_1$ and the determinant $\sigma_2.$ Then the characteristic polynomial is $$ \color{blue}{\lambda^2 - \sigma_1 \lambda + \sigma_2} \; . $$ Note the alternating signs.

For a 3 by 3, call the trace $\sigma_1$ and the determinant $\sigma_3.$ I will get to the new $\sigma_2$ in a minute. Then the characteristic polynomial is $$ \color{blue}{ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3} \; . $$

Note the alternating signs.

Alright, suppose we call the matrix $A$ and entries lower case $a_{ij}.$ Then $\sigma_2$ is the sum of the three 2 by 2 determinants, sometimes called principal minors, where the two diagonal entries of each submatrix are chosen from the diagonal elements of $A,$ and each 2 by 2 submatrix makes a little square in $A.$ In symbols $$ \sigma_2 = \det \left( \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right) + \det \left( \begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} \right) + \det \left( \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right) $$

For your matrix, $$ \sigma_2 = \det \left( \begin{array}{cc} 0 & -1 \\ -1 & 0 \end{array} \right) + \det \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right) + \det \left( \begin{array}{cc} 0 & 2 \\ 2 & 0 \end{array} \right) = -1 -1 - 4 = -6. $$ We also have trace $\sigma_1 = 0$ and $\sigma_2 = -6$ and determinant $\sigma_3 = -4.$ The characteristic polynomial is

$$ \lambda^3 - \sigma_1 \lambda^2 + \sigma_2 \lambda - \sigma_3 = \lambda^3 -6 \lambda + 4 . $$

I recommend confirming this in symbols, find the recipe for 2 by 2 $$ \det \left( \begin{array}{cc} \lambda - a_{11} & - a_{12} \\ - a_{21} & \lambda - a_{22} \end{array} \right) $$ and carefully gather the coefficients of $\lambda^2,$ also $\lambda,$ also constant terms (no $\lambda$).

Same task for 3 by 3 $$ \det \left( \begin{array}{rrr} \lambda - a_{11} & - a_{12} & - a_{13} \\ - a_{21} & \lambda - a_{22} & - a_{23} \\ - a_{31} & - a_{32} & \lambda - a_{33} \\ \end{array} \right) $$ which will be messier, of course. But then you will be sure that it works, by your own efforts.

I remember saying this after class to an individual student who was having trouble, just as an alternative, he said "Why didn't you tell us?" That was years ago. Certainly there is some benefit to this, in terms of making fewer errors.

On confirming things, if you are able to solve for the eigenvalues (one of them is $2$) then it will be possible to find an eigenvector $v$ as you were taught. The final check is to go back to the original matrix $A$ and confirm that $Av = 2v.$ The other eigenvalues were $-1 + \sqrt 3$ and $-1 - \sqrt 3$ Worth doing all the eigenvectors for practice.

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