Can a Matrix Be Expressed as the Sum of a Diagonalizable and a Nilpotent Matrix?

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The discussion focuses on expressing a given matrix A as the sum of a diagonalizable matrix and a nilpotent matrix. The eigenvalues of the matrix A are identified as 1, -1, and 3, with corresponding eigenvectors provided. A proposed nilpotent matrix B is presented, which has a specific structure that confirms its nilpotency. The conversation also raises the question of whether the diagonalizable and nilpotent matrices commute. Overall, the thread seeks clarity on the steps involved in this matrix decomposition process.
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1. write the given matrix A as the sum of a diagonalizable matrix and a nilpotent matrix.

A =

7, 3, 3, 2
0, 1, 2,-4
-8,-4,-5,0
2, 1, 1, 3

Homework Equations



3. The eigenvalues are 1, -1, 3 and the associated eigenvalues are

(1,-2,0,0)
(0,1,-1,0)
(1,-2,0,1), respectively.

Some steps in solving this question/ types of these questions would be amazing. Please explain in layman's terms, because I'm a little over my head :)
 
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Using your notation,

B=

(0,3,3,2)
(0,0,2,-4)
(0,0,0,0)
(0,0,0,0)

Then B is nilpotent. Likewise for the bottom half. But do they commute?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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