# Another question of finding eigenvalues with parameters

Let

$$A=\left(\begin{matrix}a&b&c\\a-s&b+s&c\\a-t&b&c+t\end{matrix}\right)$$

Find A's eigenvalues.

So, I did some row operations without changing the value of A's determinant, so that I got

$$A'=\left(\begin{matrix}a&b&c\\-s&0&0\\-t&0&t\end{matrix}\right)$$

But I still can't get a simple way for finding the eigenvalues. Are there more row operations I can do?

Thanks in advance for any assistance!

• Doing row operations generally changes the characteristic polynomial, though it leaves the determinant (which is up to a sign just one coefficient of the cahracterisitic polynomial) unchanged. So the method you applied is not valid, and will give a wrong result. – Marc van Leeuwen Nov 26 '13 at 13:19
• So what's the correct method? Is there a method in which I can simplify the work trying to find the characteristic polynomial? – err Nov 26 '13 at 13:24
• You can do row and/or column operations on the matrix $XI_3-A$, whose determinant is by definition the characteristic polynomial. This is more laborious than doing operations on $A$. Alternatively you can conjugate $A$ by simple matrices, which amounts to specific combinations of row and column operations; the latter respects the characteristic polynomial and does not involve $X$, but takes quite a bit of practice to apply correctly, and it is hard to produce specific simplifications using it. – Marc van Leeuwen Nov 26 '13 at 13:34
• Also, where did $b$ go on the way from $A$ to $A'$? – Henning Makholm Nov 26 '13 at 13:38

According to Maple, the characteristic polynomial of $A$ is \begin{align} x^3 &{}+(-c-a-t-b-s)x^2 \\&{}+ (-a^2+2as+st+at+sc+tc+ab+bt)x \\&{}+ (tac-asc-2ast+scb-abt+a^2t-tsc-tcb),\end{align} and it does not factor. This means the problem is pretty hopeless. The constant term is also clearly not $-\det(A')$, so either you mistyped the matrix $A$ in the first place, or you made errors in the row operations leading to $A'$ (but which would not lead to solving this problem anyway if done correctly, just to get the right constant term of the characteristic polynomial).
Added. With the now modified matrix, the simplest thing is to conjugate by $I_3-E_{1,2}-E_{1,3}$ (with $E_{i,j}$ an elementary matrix; conjugations means left-multiply by this matrix and right-multiply by the inverse $I_3+E_{1,2}+E_{1,3}$); this transforms $A$ into the similar matrix $$A'=\begin{pmatrix}a+b+c&b&c\\0&s&0\\0&0&t\end{pmatrix},$$ which being triangular clearly has eigenvalues $a+b+c$, $s$, and $t$, and so has$~A$.