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I have a final exam tomorrow, am sure a 3x3 eigen value problem like the one below is there. But I find it very hard to find eigen values without zeros in the matrix

Show me how you do it quickly so that I can apply it tomorrow; thanks

enter image description here

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  • $\begingroup$ There isn't a nice algorithm for doing it as far as I'm aware other than the usual way of computing determinants. $\endgroup$ – Cameron Williams Aug 9 '14 at 3:46
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    $\begingroup$ The numbers in your matrix are kind of large for hand calculation, so why do you think it is a good representative example of what could be on the exam? Is it taken from a real exam review or is it from elsewhere? Have you heard of the characteristic polynomial of a square matrix? $\endgroup$ – KCd Aug 9 '14 at 3:46
  • $\begingroup$ Use the condition: $\det(\lambda I-A)=0$. Good luck! $\endgroup$ – idiot Aug 9 '14 at 3:48
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    $\begingroup$ Just use the linearity of the determinant to reduce the polynomial to get something more easy to handle. $\endgroup$ – Jorge Ramos Aug 9 '14 at 3:51
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You could (specially if you have a calculator at hand to crunch numbers) do this:

  1. Get the trace (or sum of the diagonal elements). It is $32-25-2=5$
  2. Compute the determinant. It is $-8$
  3. Compute the square of the matrix and get the trace of it. It is $21$

You now know that:

$$\lambda_{1} \lambda_{2} \lambda_{3} = -8 \space\space\space\space\space(1)$$ $$\lambda_{1} + \lambda_{2} + \lambda_{3} = 5 \space\space\space\space\space(2)$$ $$\lambda_{1}^{2} + \lambda_{2}^{2} + \lambda_{3}^{2} = 21 \space\space\space\space\space(3)$$

Where $\lambda_{1}, \lambda_{2}, \lambda_{3}$ are the eigenvalues to work out. Now, let's say you feel lucky and want to assume that all the eigenvalues are integer. Then, from equation $(3)$ you know the largest one could be only $3$ or $4$ in absolute value, in which case the second largest would have to be $\pm 2$ and then $\pm 1$. If the largest were $3$, the second would have to be $3$ and the third couldn't be an integer. Now, from $(1)$ you know either one is negative or all are. Now, at once you see from $(2)$ that $4$ got to be positive, and the last choice of whether $1$ or $2$ is negative is solved by mentally trying, to find that the answer is indeed $-1$, $2$, $4$.

Yes, we assumed the answer to be made of integers, but, from experience, this is a good assumption to make to give it a first go in exams questions, as it's usually the case!

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I would not exactly call this quick, but a few tricks could simplify the steps:

  1. We have $\det(A)=\det(A^T)$
  2. If you add/subtract multiples of one row to another the determinant remains unchanged
  3. (1)+(2) means that you can add/subtract multiples of one column to another "for free" too
  4. You may want to modify the signs of a row or column which will change the sign of the determinant, but since we are trying to determine when $\det(A-\lambda I)=0$ we do not mind whether the sign has changed or not
  5. We do not need it for this example, but you could also easily scale a row or column by any factor $r\neq 0$ since $r\cdot\det(A-\lambda I)=0\iff\det(A-\lambda I)=0$. In fact (4) follows directly from this using $r=\pm 1$

so with all this in place we have $$ \begin{align} \begin{vmatrix}32-\lambda&-12&-2\\66&-25-\lambda&-4\\54&-21&-2-\lambda\end{vmatrix}&=\begin{vmatrix}32-\lambda&12&2\\66&25+\lambda&4\\54&21&2+\lambda\end{vmatrix}\\ \quad\\ &=\begin{vmatrix}32-\lambda&0&2\\66&1+\lambda&4\\54&9-6\lambda&2+\lambda\end{vmatrix} \end{align} $$ where the first equality was produced changing signs of the last two columns and the last equality is due to having subtracted $6$ times the last column from the middle column. Thus we got rid of annoying negative signs and are down to having to compute only two minors: $$ \begin{vmatrix}1+\lambda&4\\9-6\lambda&2+\lambda\end{vmatrix}=\lambda^2+27\lambda-34 $$ and $$ \begin{vmatrix}66&1+\lambda\\54&9-6\lambda\end{vmatrix}=540-450 \lambda $$ to finally arrive at $$ \begin{align} \det(A-\lambda I)&=(32-\lambda)(\lambda^2+27\lambda-34)+2(540-450 \lambda)\\ &=-\lambda^3+5\lambda^2-2\lambda-8 \end{align} $$ This is still tedious work! BTW the zeros for the above expression are $\lambda=-1,2,4$.

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  • $\begingroup$ @GitGud: Hi! Sorry I just got the impression that noone seemed to care for my answer and waited for something more relevant to come along. I will undelete it again, no problem! Basically deleting answers is not such a good idea unless they are entirely irrelevant, after all. $\endgroup$ – String Aug 11 '14 at 1:28
  • $\begingroup$ I understand that. I often do the same. You're free to keep it deleted if you want. I can read it anyway. I'll do it later. $\endgroup$ – Git Gud Aug 11 '14 at 1:30

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