Derive an expression to find how many times an eigen value is repeatedby vish_maths Tags: derive, eigen, expression, repeated, times 

#1
Feb2013, 01:42 AM

P: 47

Hello !
I have an upper triangular matrix for an operator T in which an eigen value has been repeated s times in total. Derive an expression for s . My thoughts : ( Let * imply contained in ) then :I know that : (a) Null T^{0} * Null T^{1} *.....*Null T^{dim V} = Null T^{dim V + 1} = ........ (b) Will i have to investigate the effect of higher powers of ( T  k I ) where k is the intended eigen value ?? (c) the book which i am reading : Sheldon Axler's Linear Algebra hasn't introduced Jordan form as of now. Any direction for this will be appreciated. Thanks Can i prove it from these results ? 



#2
Feb2113, 10:47 AM

Mentor
P: 16,518

I have no idea what you're trying to find. What do you mean with "an expression"? Do you have to find some expression?? This is a very vague question...




#3
Feb2113, 10:47 AM

Mentor
P: 16,518

The only thing I can think of is that the eigenvalue comes up s times on the diagonal. Maybe they mean that?




#4
Feb2113, 11:10 AM

P: 47

Derive an expression to find how many times an eigen value is repeated
The answer given states that s = dim [ Null ( T  k I )^{dim V} ]
where k is the corresponding eigen value. 



#5
Feb2113, 11:12 AM

P: 47





#6
Feb2113, 11:15 AM

Mentor
P: 16,518





#7
Feb2113, 11:19 AM

P: 47





#8
Feb2113, 11:36 AM

Mentor
P: 16,518

And why does Theorem 8.10 not answer the question for you?? I think it basically says and proves what you want.




#9
Feb2113, 11:43 AM

P: 47

I found proof by induction unconvincing ; It assumes that the result is already true. ( It does not give an intuition .. ) I really want to derive the expression considering a situation , say, when i never knew what the answer is going to be in which case, probably induction is not going to work. I have thought about it and i think the answer may lie in investigating the behavior of higher powers of ( T  k I ) but i seem to be stuck for more than a week now, which is frustrating :( 



#10
Mar1313, 05:08 AM

P: 47

ok guys, i have finally found a proof. Took me long .I will post it for common good :)
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 => dim [ null (T  λ I ) ] ≤ r => dim [ range (T  λ I ) ] ≥ nr Now, we know that : null (T  λ I )^{0} [itex]\subset[/itex] null (T  λ I )^{1}[itex]\subset[/itex] ......... [itex]\subset[/itex] null (T  λ I )^{m} =null (T  λ I )^{m+1}=...... = null (T  λ I )^{dim V} =.. => 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} [itex]\supset[/itex] range (T  λ I )^{1}[itex]\supset[/itex] .......... [itex]\supset[/itex] range (T  λ I )^{m} = range (T  λ I )^{m+1} =... = range (T  λ I )^{dim V} = ... => 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 ) = nr . If we try to look at the safest boundary conditions : max [ dim range (T  λ I ) ] = n1 We already know that max [ dim [range (T  λ I )^{m} ] ] = nr and not less than that . => maximum value of m from statement (2) = r  (3) => dim range (T  λ I )^{r} = nr => dim null (T  λ I )^{r} = r => 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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