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**iLloyd054****Member**- Registered: 2014-04-01
- Posts: 10

(a) 2,1,0

Is matrix 0,2,0 diagonalisable?

0,0,2

(b) 3,-2,3

A is a matrix 1,2,1

1,3,0

(i) Find the eigenvalues.

(ii) Find P-1 and B such that P-1AP = B where B is a diagonal matrix

*Last edited by iLloyd054 (2014-04-01 17:58:41)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,783

Hi iLloyd054;

a) It should be because it has three distinct eigenvalues. See c) for the actual diagonalization.

b) The eigenvalues of that matrix are 4, 2, -1.

c)

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**iLloyd054****Member**- Registered: 2014-04-01
- Posts: 10

thank you so much bobbym I appriciate that...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,783

Hi;

You are welcome and welcome to the forum.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**eigenguy****Member**- Registered: 2014-03-18
- Posts: 78

(A) No. Perhaps bobbym sees something distinct about each of those 2s, but they all look the same to me. That matrix is already in Jordan normal form, and that superdiagonal 1 tells me that the eigenspace of 2 is going to be 2 dimensional (if there were a second superdiagonal 1, it would only be one dimensional).

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 103,783

No nothing different I was referring to the matrix in part b which was not the question asked in a).

a) is not diagonalizable.

Sorry for the confusion.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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