Question

Is there a quick way to find the index of nilpotency of a matrix? Like A=[0 2 1; 0 0 3; 0 0 0]?

Answer 1

without having to multiply the matrix by itself until we get all zeroes

Answer 2

for a 3x3 this shouldnt be very difficult

Answer 3

Yeah I know but I have exam tomorrow so just worried I might get a 4x4 or the power for the matrix might be high

Answer 4

By index I assume the number of times it needs to be composed to destroy it?

Answer 5

If thats the case then you just look at the lowest diagonal

Answer 6

What do you mean by lowest diagonal?

Answer 7

What do you mean by index?

Answer 8

The smallest n such that A^n=0?

Answer 9

like for this example when the matrix A is A^3 it's all zeroes so nilpotency index is 3

Answer 10

Just look at the first diagonal with entries, and see how many diagonals including that one are above it.

Then A^(that number + 1)

Answer 11

oh so if there's only 1 entry that's not a diagonal?

Answer 12

Suppose only the corner entry has a value. So that is only one diagonal, so its A^(1+1) or A^2=0

Answer 13

Oh okay that make sense so then you'd know the index of a 3x3 is always 3

Answer 14

No! it depends on the lowest diagonal with values.

Answer 15

It could be either 3 or 2

Answer 16

If only the corner is filled in a 3x3 matrix it would be 2

Answer 17

okay I just want to make sure we're on the same page.. A=[ 0 0 0; 1 0 0; 8 1 0] it's 3 because of the 0 0 0 and 1 1 ?

Answer 18

but if it's like all zeroes then nilpotency index can only be 2 max

Answer 19

Index = 2
\[\left[\begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\]
Index = 3
\[\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{matrix}\right]\]

Answer 20

Okay that's very clear that's a lot!