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

In summary: You are correct, the theorem 8.10 does not answer the question. It states that 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.
  • #1
vish_maths
61
1
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 ?
 
Last edited:
Physics news on Phys.org
  • #2


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


The only thing I can think of is that the eigenvalue comes up s times on the diagonal. Maybe they mean that?
 
  • #4


The answer given states that s = dim [ Null ( T - k I )dim V ]
where k is the corresponding eigen value.
 
  • #5


micromass said:
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.
 
  • #6


vish_maths said:
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?
 
  • #7


micromass said:
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.
 
  • #8


And why does Theorem 8.10 not answer the question for you?? I think it basically says and proves what you want.
 
  • #9


micromass said:
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 :(
 
  • #10
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 [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.
 

1. What is an eigen value?

An eigen value is a scalar value that represents the amount of stretch or compression of a vector in a linear transformation. It is a fundamental concept in linear algebra and is used in various applications such as physics, engineering, and computer graphics.

2. Why is it important to find repeated eigen values?

Finding repeated eigen values is important because it helps in understanding the geometric structure of a linear transformation and can reveal important information about the corresponding matrix, such as its rank and characteristic polynomial.

3. Can you explain how to derive an expression to find the number of repeated eigen values?

To derive an expression for finding repeated eigen values, we can use the concept of algebraic multiplicity, which is the number of times an eigen value appears as a root of the characteristic polynomial. By finding the algebraic multiplicity of each eigen value, we can determine how many times it is repeated.

4. Is there a formula for calculating the algebraic multiplicity of an eigen value?

Yes, the algebraic multiplicity of an eigen value can be calculated by finding the degree of its corresponding factor in the characteristic polynomial.

5. How does finding repeated eigen values help in solving systems of linear equations?

Finding repeated eigen values can help in solving systems of linear equations because it allows us to determine the dimension of the eigenspace associated with each eigen value. This information can then be used to find the general solution to a system of linear equations.

Similar threads

  • Linear and Abstract Algebra
Replies
3
Views
965
  • Linear and Abstract Algebra
Replies
9
Views
2K
Replies
4
Views
968
  • Linear and Abstract Algebra
Replies
10
Views
3K
  • Linear and Abstract Algebra
Replies
6
Views
5K
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
207
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
Back
Top