Consider $A =\left( \begin{array}{ccc} -1 & 2 & 2\\ 2 & 2 & -1\\ 2 & -1 & 2\\ \end{array} \right)$. Find the eigenvalues of $A$.

So I know the characteristic polynomial is:

$$f_A(\lambda) = (-\lambda)^n+(trA)(-\lambda)^{n-1}+...+\det A$$

I found the $\det A = 27$, so the characteristic polynomial of $A$ is:


However, the textbook I'm using doesn't give any method for finding the value of $c$. I have the solution to the problem, and the value of $c$ is in fact $9$, but is there any method to numerically solve for it?

If we have a $4\times 4$ matrix, we'll end up with a characteristic polynomial of the form $$\lambda^4-tr A(\lambda)^3+c_1\lambda^2-c_2\lambda+\det A$$

Similarly, is there a method to solve for $c_1,c_2$ in this case?

  • $\begingroup$ Finding this determinant $\det(A-\lambda I)$ is your best bet in general. There are shortcuts one can take sometimes. For instance in this case the rows always add up to $3$, so $3$ is an eigenvalue. You've already found $\det(A)=27$, so looking at the trace the other 'two' eigenvalues follow easily. Finding $\det(A)=27$ isn't that much easier than finding the characteristic polynomial though. $\endgroup$ – Git Gud Nov 26 '14 at 9:54
  • $\begingroup$ Another way for finding $c$ in the $3\times 3$ case is to use Cayley-Hamilton, thus the characteristic polynomial evaluated at the matrix $A$ is the zero matrix. $\endgroup$ – Dimitar Ho Nov 26 '14 at 10:18

Developping $$ f_A(\lambda) =\begin{vmatrix} -1-\lambda & 2 & 2\\ 2 & 2-\lambda & -1\\ 2 & -1 & 2-\lambda\\ \end{vmatrix} $$ gives as coefficient of $-\lambda$ the number $\left|\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right| + \left|\begin{smallmatrix}-1&2\\2&2\end{smallmatrix}\right| + \left|\begin{smallmatrix}-1&2\\2&2\end{smallmatrix}\right|=3-6-6=-9$ (this is your $c$, note the sign; the coefficeint of $\lambda$ then is $+9$ of course). Actually your matrix is so full of coefficients $2$ that it is not so clear what is going on. Better is to say the coefficient of $-\lambda$ in $$ \begin{vmatrix} a-\lambda & b & c\\ d & e-\lambda & f\\ p & q & r-\lambda\\ \end{vmatrix} $$ is the number $\left|\begin{smallmatrix}e&f\\q&r\end{smallmatrix}\right| + \left|\begin{smallmatrix}a&c\\p&r\end{smallmatrix}\right| + \left|\begin{smallmatrix}a&b\\d&e\end{smallmatrix}\right|$.

  • $\begingroup$ Is there a generalization for this? Can we find the two coefficients in the $4\times 4$ case? $\endgroup$ – Jacob Nov 26 '14 at 9:46
  • $\begingroup$ *developing (I will delete this comment later) $\endgroup$ – Git Gud Nov 26 '14 at 9:48
  • $\begingroup$ Yes there is a generalisation, just develop the determinant. The result is a fairly big mess, but can be described as: the coefficient of $(-\lambda)^{n-k}$ is the sum of all $k\times k$ minors of $A$ taken on the same set of row indices as column indices. There are $\binom nk$ such $k$-subsets of indices, which is a lot and explains the mess. $\endgroup$ – Marc van Leeuwen Nov 26 '14 at 9:50

Hint: The definition of the characteristic polynomial is

$$f_A(\lambda) = \det(A-\lambda I).$$

  • $\begingroup$ I'm trying to find the characteristic polynomial without explicitly finding the determinant of the matrix $A-\lambda I$. The book I'm using has the formula above, but it doesn't show how we can solve for $c$. $\endgroup$ – Jacob Nov 26 '14 at 9:39
  • $\begingroup$ @Jacob But since $f_A$ IS EXACTLY the determinant of the matrix $A-\lambda I$, it is impossible to find one without finding the other. The fact that the first element is $(-1)^n$, the second $tr(A)$ and the last $\det A$ are all consequences of the definition of $f_A$. Your book probably uses that definition too, and there is no easy way of calculating the factor next to $\lambda^{n-2}$. $\endgroup$ – 5xum Nov 26 '14 at 9:43

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