<s>[INST] Calculate the determinant of $A-5I$ Question
Let $ A = 
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
 [/INST] The calculation of $\lambda_4 \lambda_5$ (which in this case is equivalent to the cubic term of the characteristic polynomial) is indeed given by the sum of principal $2\times2$ subdeterminants
$$\lambda_4 \lambda_5 = \sum_{i=2}^5 \sum_{j=1}^{i-1} \left|\begin{matrix}a_{ii}&a_{ij}\\a_{ji}&a_{jj}\end{matrix}\right|.$$
In this case we actually have the easy formula $a_{ij} = 5i+j-5$, so the above subdeterminant simplifies to $-5(i-j)^2$.  We can of course evaluate the double sum algebraically, but it's just as easy to regroup the sum according to the value of $i-j$:
$$\sum_{1\le j < i \le 5} (i-j)^2 = 4\cdot 1^2 + 3\cdot 2^2 + 2\cdot 3^2 + 1\cdot4^2 = 4 + 12 + 18 + 16 = 50.$$
Thus $\lambda_4 \lambda_5 = -5(50) = -250$ and $\det(A-5I) = (-5)^3 (-250-5\cdot 65+25) = 68750$.
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