Diagonalising a matrix

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Hi

Can someone please clarify if you are trying to diagonalize a matrix and you find that one of your eigenvalue is a DOUBLE root, does that mean the original matrix is NOT diagonalisable.

Thanks
 

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  • #2
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Hi

Can someone please clarify if you are trying to diagonalize a matrix and you find that one of your eigenvalue is a DOUBLE root, does that mean the original matrix is NOT diagonalisable.

Thanks

No, it doesn't mean that. For example, the unit matrix of order n has the eigenvalue 1 with (algebraic) multiplicity n, and yet it

is diagonalizable...

A matrix is diagonalizable iff there is a basis (of the vector space we're working on) all the elements of which are eigenvectors

of the matrix iff the minimal polynomial of the matrix can be written as the product of DIFFERENT linear factors.

DonAntonio
 
  • #3
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No, it doesn't mean that. For example, the unit matrix of order n has the eigenvalue 1 with (algebraic) multiplicity n, and yet it

is diagonalizable...

A matrix is diagonalizable iff there is a basis (of the vector space we're working on) all the elements of which are eigenvectors

of the matrix iff the minimal polynomial of the matrix can be written as the product of DIFFERENT linear factors.

DonAntonio
Thanks so much for your reply!

So what you are say is that even if I have something like...
(λ+1)^2(λ-5)=0
I still find the eigenvalues and solve the matrix to determine whether it is diagonalisable
right????
 
  • #4
606
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Thanks so much for your reply!

So what you are say is that even if I have something like...
(λ+1)^2(λ-5)=0
I still find the eigenvalues and solve the matrix to determine whether it is diagonalisable
right????

If you meant to ask whether you still have to find the eigenvalues AND THEN their respective eigenvectors and check whether

you have 3 (as I'm assuming that what you wrote is the characteristic polynomial) linear independent eigenvectors, then

the answer is yes: you still have to do that...or, since the minimal and charac. polynomials have the same irreducible

factors, you could check whether [itex]\,(\lambda+1)(\lambda-5)\,[/itex] is the minimal pol. of the matrix, since in this case

it'd be the product of different linear factors and thus the matrix is diagonalizable.

Yet since most questions/applications usually require knowing at least some of the eigenvectors, perhaps the first

method is more useful.

DonAntonio
 
  • #5
HallsofIvy
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Whether or not an n by n matrix is diagonalizable is determined by whether or not it has n independent eigenvectors. IF there are n distinct eigenvalues, then, because eigenvectors corresponding to distinct eigenvalues are independent, it follows that if an n by n matrix has n distinct eigenvalues, it is diagonalizable but if has fewer than n distinct eigenvalues, it may still be diagonalizable. To give a trivial example, the matrix
[tex]\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}[/tex]
is already diagonal (so obviously "diagonalizeable") with the single eigenvalue 2 and the two independent eigenvectors <1, 0> and < 0, 1>.

[tex]\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}[/tex]
has 2 as its only eigenvalue but now has only multiples of <1, 0> as eigenvectors so not two independent eigenvectors. That matrix cannot be diagonalized.
 

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