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Linear algebra help please - Jordan form of matrices?
    #15338956 -

Can someone please help me understand the concepts behind the Jordan form of matrices?

My book gives one or two examples of Jordan matrices but it is very vague on what the definition of a Jordan form is or what properties define it.

From what I read, J = (M^-1)(A)(M) but the book is very vague on what the matrix M is.  It basically says M is any matrix that comes as close as possible to diagonalizing A.  The few examples of Jordan matrices that the book provides show matrices that have eigenvalues on the diagonals, and 1's above the diagonals.  So I know the Jordan form is somewhat related to diagonalization but I'm not entirely sure how, or exactly over which eigenvalues I should put the ones.

Can someone please help me out?


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Re: Linear algebra help please - Jordan form of matrices? [Re: Celestial Traveler]
    #15338982 -

the answer is 42


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Re: Linear algebra help please - Jordan form of matrices? [Re: Salomon]
    #15339001 -

Salomon said:
the answer is 42



That's the grade I'll get on my assignment if I don't get help :confused:


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Re: Linear algebra help please - Jordan form of matrices? [Re: Celestial Traveler]
    #15339035 -

For example can someone find the Jordan form of this matrix?

[1 -2]
[0 -1]


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Re: Linear algebra help please - Jordan form of matrices? [Re: Celestial Traveler]
    #15339076 -

Celestial Traveler said:
For example can someone find the Jordan form of this matrix?

[1 -2]
[0 -1]




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Re: Linear algebra help please - Jordan form of matrices? [Re: ModularMind]
    #15339078 -

:facepalm:


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Re: Linear algebra help please - Jordan form of matrices? [Re: Celestial Traveler]
    #15339084 -

In the finite dimensional case, σ(T) = {λi} is a finite discrete set in the complex plane. Let ei be the function that is 1 in some open neighborhood of λi and 0 elsewhere. By property 3 of the functional calculus, the operator

is a projection. Moreoever, let νi be the index of λi and

The spectral mapping theorem tells us

has spectrum {0}. By property 1, f(T) can be directly computed in the Jordan form, and by inspection, we see that the operator f(T)ei(T) is the zero matrix.
By property 3, f(T) ei(T) = ei(T) f(T). So ei(T) is precisely the projection onto the subspace



Notice that the expression of f(T) is a finite sum because, on each neighborhood of λi, we have chosen the Taylor series expansion of f centered at λi.


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