using general form of characteristic polynomial to determine equation for 3*3 matrix

Question

If $A$ is a $n$ x $n$ matrix and $f(\lambda)=det(A-\lambda I_n)$ then the characteristic polynomial is given as

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

For a 3x3 matrix I dont know how they arrived at the following with the $1/2$ term

$-\lambda^{3}+tr(A) \lambda^{2}+ \frac{1}{2}(tr(A)^2-tr(A^2)) \lambda +det(A)$

I thought it was just a matter of substituting n=3....

But when you substitute $n=3$ what do you write for the $\dots$ term?! This problem tells you the answer to that question. — ancient mathematician, Apr 29 '21 at 14:45
Ah ok... but still I cannot identify a pattern for this $...$ term... — shoggananna, Apr 29 '21 at 14:49
$-f(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$ so the coefficient you want is the sum of the $\lambda_i$ two at a time, which you can express as half the sum of the $\lambda_i$ all squared minus the sum of the squares of the $\lambda_i$. But there's a solution now. — ancient mathematician, Apr 29 '21 at 14:53

score 2 · Accepted Answer · answered Apr 29 '21 at 14:46

If you use this formula for a $3 \times 3$ matrix, then the coeffocient of $ \lambda$ will be $d_1d_2 +d_2d_3+d_3d_1,$ which is is equal to the given expression. Here $d_1, d_2, d_3$ are the diaginal element of the given $3 \times 3$ matrix, say $A.$

Also note that $d_1d_2 +d_2d_3+d_3d_1 = \frac{(d_1+d_2+d_3)^2- (d_1^2 +d_2^2 +d_3^2)}{2}.$

using general form of characteristic polynomial to determine equation for 3*3 matrix

1 Answers1