Derive an expression to find how many times an eigen value is repeated


by vish_maths
Tags: derive, eigen, expression, repeated, times
vish_maths
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#1
Feb20-13, 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 T0 * Null T1 *.....*Null Tdim V = Null Tdim 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 ?
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micromass
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#2
Feb21-13, 10:47 AM
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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...
micromass
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#3
Feb21-13, 10:47 AM
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The only thing I can think of is that the eigenvalue comes up s times on the diagonal. Maybe they mean that?

vish_maths
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#4
Feb21-13, 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.
vish_maths
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#5
Feb21-13, 11:12 AM
P: 47
Quote Quote by micromass View Post
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...
An expression means in this context a formula to find number of times an eigen value is repeated in an upper triangular matrix.
micromass
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#6
Feb21-13, 11:15 AM
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Quote Quote by vish_maths View Post
The answer given states that s = dim [ Null ( T - k I )dim V ]
So you need to prove that formula for s? How is this thread different from your previous thread then?
vish_maths
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#7
Feb21-13, 11:19 AM
P: 47
Quote Quote by micromass View Post
So you need to prove that formula for s? How is this thread different from your previous thread then?
I thought i ended up confusing a lot of things over there in that thread. So, i wrote afresh, This is what i meant actually.
micromass
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#8
Feb21-13, 11:36 AM
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And why does Theorem 8.10 not answer the question for you?? I think it basically says and proves what you want.
vish_maths
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#9
Feb21-13, 11:43 AM
P: 47
Quote Quote by micromass View Post
And why does Theorem 8.10 not answer the question for you?? I think it basically says and proves what you want.
Hi

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 :(
vish_maths
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#10
Mar13-13, 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 ) ] ≥ n-r

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 ) = n-r .

If we try to look at the safest boundary conditions :

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 .

=> maximum value of m from statement (2) = r ---------------- (3)
=> dim range (T - λ I )r = n-r
=> 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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