show that if N is a nilpotent nxn matrix then identity matix+ N is invertible.

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- Sep 22nd 2009, 02:49 PMamm345Matrices- Nilpotent
show that if N is a nilpotent nxn matrix then identity matix+ N is invertible.

- Sep 22nd 2009, 05:42 PMaliceinwonderland
Since N is a nilpotent square matrix, $\displaystyle N^k = 0$ for some positive integer k.

If k is an odd positive number greater than 1 (if k=1, then it is trivial), then the required inverse is

$\displaystyle I - N + N^2 - N^3 + ,,,, + N^{k-1}$.

To verify this,

$\displaystyle (I + N)(I - N + N^2 - N^3 + ,,,, + N^{k-1})$

$\displaystyle = I - N + N^2 + ,,,, + N^{k-1} +N - N^2 + ...- N^{k-1} + N^{k}$

$\displaystyle = I $, since $\displaystyle N^k = 0$.

If k is an even positive number, then the required inverse is

$\displaystyle I - N + N^2 - N^3 +,,,, - N^{k-1}$. You can verify this in the same manner as the above.