# Algebraic multiplicity of an eigenvalue

I am reading Sheldon Axler's Linear Algebra done right. On pg. 168, Theorem 8.10 states the following:

The number of times an eigenvalue $k$ appears on the diagonal of an $n\times n$ upper triangular matrix is equal to $\dim \operatorname{null} ( T - k I )^n$.

The proof to the theorem is given through induction, which I sadly find as non intuitive and unsatisfactory.

How can I prove the above result without induction? I have spent around a month pondering over it and feel sad about not being able to prove it. Can anyone please give me a direction?

I know the following results: ( If V*W implies V is a proper subset of W and V*=W implies V is a subset of W ) , then given for an eigen value k

Null(T-kI) * Null (T-kI)^(2} * .....* Null ( T - K I)^(dim V - 1) = Null ( T - K I)^(dim V )

The book until this theorem hasn't touched Jordan form nor has it introduced determinants nor the concept of characteristic polynomial.

How do I go about it from here. Any help would be deeply appreciated.

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For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Mar 10 '13 at 7:52
okay, I was able to edit it. I am still looking the pages to make it better readable. –  VHP Mar 10 '13 at 8:16
Well, not really a teacher. We have a math club where we discuss alternate proofs from given books. So, there one of the tasks was to find an alternate proof. –  VHP Mar 10 '13 at 9:45
@YACP We try to discuss everything in general. So, going by our previous discussions : i tried it with different matrices and found that the dimension of range of T actually becomes n-r. But, how do i prove it mathematically. In short, the whole problem lies in analyzing the behavior of higher powers of T. Do we have any definite result leading to dimension of range of higher powers of T ? I just know that the null space of higher power of T contains the null space of the lower powers of T and this goes until the power = dimension of vector space V. Possible to deduce as a proof from here ? –  VHP Mar 11 '13 at 5:24
@YACP .. I am so sorry. I am new to the forum and I was accessing the site from my phone. Even I was wondering how the answer got deleted. It was purely accidental –  VHP Mar 11 '13 at 9:26

if an eigen value λ is repeated r times on the main diagonal of M(T) [ where M(T) denotes the matrix associated with the linear mapping T ] , then M(T - λ I ) has r zeroes on the main diagonal.

Speaking of higher powers of M(T - λ I ) ( say kth power ) , notice that under any circumstance, the non zero diagonal element of M(T - λ I ) would simply be their kth power for M(T - λ I ). -------------------- (A)

since, the eigen vectors for non distinct eigen values may/may not be linearly independent

$\implies$ dim [ null (T - λ I ) ] ≤ r $\implies$ dim [ range (T - λ I ) ] ≥ n-r

Now, we know that :

null (T - λ I )0 ⊂ null (T - λ I )1⊂ ......... ⊂ null (T - λ I )m =null (T - λ I )(m+1)=...... = null (T - λ I )(dim V) =..

$\implies$ 0 < dim null (T - λ I ) < ... < dim null (T - λ I )m = dim[ null (T - λ I )(m+1) ] = .... = dim null (T - λ I )(dim V) = ....

...................... (1)

We also know that range (T - λ I )0 ⊃ range (T - λ I )1⊃ .......... ⊃ range (T - λ I )m = range (T - λ I )(m+1) =... = range (T - λ I )(dim V) = ...

$\implies$ n > dim range (T - λ I )1 > ..... > dim range (T - λ I )m = dim range (T - λ I )(m+1)=...... dim range (T - λ I )(dim V)

........................ (2)

after carefully analysing the statement (A) , it states that the minimum dimension of range of any power of ( T - λ I ) = n-r .

If we try to look at the safest boundary conditions ( when the dimension gets reduced in just steps of 1 in (2) ):

max [ dim range (T - λ I ) ] = n-1

We already know that max [ dim [range (T - λ I )^m ] ] = n-r and not less than that .

$\implies$ maximum value of m from statement (2) = r ---------------- (3) $\implies$ dim range (T - λ I )r = n-r $\implies$ dim null (T - λ I )r = r

$\implies$from (1) : dim null (T - λ I )(dim V) = r .

Hence, there you have the expression for the algebraic multiplicity of an eigen value.

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@YACP, i have posted it :) . –  VHP Mar 15 '13 at 4:03
@YACP I understand you have no interest in this question. However, thank you very much. One of the points in your answer led me to the proof :) . Please don't consider helping as a mistake. I am sure with the new knowledge i will help many other curious minds . Regarding the Latex usage, i am getting better. I am very new , so, hopefully, i will be good in some time. Hope to see you around . Good day ! Cheers –  VHP Mar 16 '13 at 13:02