- #1

vish_maths

- 61

- 1

## Homework Statement

Prove without induction that Multiplicity of an eigen value , k = dim[ Null(T - k I)^( dim V) ]

## Homework Equations

[(T - k I)^dim V ] v =0

[Thoughts]

i understand that normal eigen vectors with same eigen values may not be

linearly independent.

[(T - k I)^dim V ] v =0

then, the fact that k = dim[ Null(T - k I)^dim V ]

somehow gives an intuition that in this case, the Eigen vectors with

the same Eigen value k are linearly independent ?

This is confusing to me.

## The Attempt at a Solution

If i can know, that for [(T - k I)^dim V ] v =0 , the solutions are

linearly independent, then the desired result can be proved.

OR if i prove that the solutions to the above equation are eigen vectors which form a basis, then i have the solution.

What could be a direction ?