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I have a simple matrix:

$$ \left[ \begin{array}{ccc} 3&2&3\\ 0&0&2\\ 0&2&0 \end{array} \right] $$

And once I try to find the characteristic polynomial:

$$ \left[ \begin{array}{ccc} \lambda-3&-2&-3\\ 0&\lambda&-2\\ 0&-2&\lambda \end{array} \right] $$

I'm left with:

$$ \lambda^2(\lambda-3) $$

However online calculators suggest a charecteristic polynomial of:

$$ −\lambda^3+3\lambda^2+4\lambda−12 $$

What am I doing wrong?

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  • $\begingroup$ How did you get $\lambda^2(\lambda-3)$? -2(-2) isn't 0 $\endgroup$ – randomgirl Apr 16 '16 at 3:19
  • $\begingroup$ Multiplying down the diagonal. $\endgroup$ – Tommyixi Apr 16 '16 at 3:20
  • $\begingroup$ err... multiplying down the diagonal? WHY? Do you know how to compute a determinant? $\endgroup$ – mathguy Apr 16 '16 at 3:20
  • $\begingroup$ Oh... Right, so that only works with an upper or lower triangluar matrix, correct? $\endgroup$ – Tommyixi Apr 16 '16 at 3:21
  • $\begingroup$ ^Right, multiplying down the diagonal works because the zeroes cancel other stuff out. you are really finding the determinant $\endgroup$ – Elliot G Apr 16 '16 at 4:09
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You made a simple mistake in computing the determinant. To see the computation,

$$c_A(\lambda) = (\lambda -3)(\lambda^2 - 4) + 2(0) + 3(0) = \lambda^3 - 3\lambda^2 -4\lambda + 12$$ Which differs from wolfram alpha's answer by a sign, so the eigenvalues (roots) are still the same. Your computation coincides with that of an upper triangular matrix.

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