- #1
vish_maths
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- 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 ?