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

[vish_maths]

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 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 ?

 

[micromass]

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]

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

 

[vish_maths]

The answer given states that s = dim [ Null ( T - k I )^{dim V} ]
where k is the corresponding eigen value.

 

[vish_maths]

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]

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]

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]

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

 

[vish_maths]

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]

Found it

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 \subset null (T - λ I )¹\subset ... \subset 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 \supset range (T - λ I )¹\supset ... \supset 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.