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2 matrices- Same eigenvalues.

by kini.Amith
Tags: eigenvalues, matrices
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kini.Amith
#1
Apr15-11, 08:14 AM
P: 46
given that 2 matrices have the same eigenvalues is it necessary that they be similar? If not, what is the connection between those 2?
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HallsofIvy
#2
Apr15-11, 08:34 AM
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No. Two matrices are similar if and only if they have the same eigenvalues and corresponding eigenvectors. For example, the matrices
[tex]\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}[/tex]
and
[tex]\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}[/tex]
have the same eigenvalues (2 is a double eigenvalue for each) but are not similar. The first has both <1, 0> and <0, 1> as independent eigenvectors corresponding to eigenvalue 2, the second has only <1, 0> and its multiples as eigenvectors.

(If two n by n matrices have the same n distinct eigenvalues, then, because eigenvectors corresponding to distinct eigenvalues are indpependent, they will be similar.)
kini.Amith
#3
Apr15-11, 08:37 AM
P: 46
then what do 2 vectors having the same evalues have in common

Deveno
#4
Apr15-11, 08:37 AM
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2 matrices- Same eigenvalues.

consider the matrix:

[1 0 0]
[0 0 0]
[0 0 0], and then:

[1 0 0]
[0 1 0]
[0 0 0]. are they similar?
kini.Amith
#5
Apr15-11, 08:42 AM
P: 46
i nderstand that they need not be similar, but then what do they have in common?
Deveno
#6
Apr15-11, 09:20 AM
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P: 906
the trouble with finding an "eigenbasis" is that, sometimes you can't. the trouble isn't that an eigenspace (the subspace generated by the eigenvectors corresponding to a particular eigenvalue) can have dimension > 1, but rather than the dimension of such an eigenspace can be less than the algebraic multiplicity of the eigenvalue (this happens when the matrix isn't diagonalizable).

so two matrices can have the same eigenvalues, with the same (algebraic) multiplicites, and yet not be similar.

put another way, in some "nice cases" one can use a diagonal matrix as a "nice form" (similar to) a given matrix, in which case the eigenvalues essentially tell you everything you need to know. but there are what you would call "degenerate cases" where you need to know more to know "which type" of matrix you have. this "something more" is captured by a class of matrices called nilpotent, A = D + N, where D is (similar to) a diagonal matrix, and N is nilpotent.

similarity is just a non-basis way of saying: change the basis. if A is a linear transformation in one basis, PAP-1 is the same transformation in another basis. if the eigenvectors of an nxn matrix are all linearly independent, then we can change A to a matrix that "stretches every dimension by the eigenvalue λi."
(this is A in the basis of eigenvectors).

but we might not get enough eigenvectors. for example C =

[0 1]
[0 0], has eigenvalue 0, with characteristic equation det(C - xI) = 0 of x^2 = 0, so the eigenvalue 0 has algebraic multiplicity 2. but the eigenspace
E0 = {(x,y) : C(x,y) = (0,0)} is span{(1,0)}, which has dimension 1.

compare C to the 0-matrix, which also has the same characteristic equation, but is definitely not similar to C. C is one of those "bad" matrices, the nilpotent kind, that have the same eigenvalues as some other matrix, but aren't similar to them at all.
kini.Amith
#7
Apr15-11, 09:36 AM
P: 46
k. got some idea. will think some more about it.
thnx
HallsofIvy
#8
Apr15-11, 10:51 AM
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Quote Quote by kini.Amith View Post
i nderstand that they need not be similar, but then what do they have in common?
You have asked that repeatedly. Please tell us what you mean by "in common"!
Deveno
#9
Apr15-11, 10:54 AM
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well, they share the same eigenvalues, lol
HallsofIvy
#10
Apr15-11, 10:54 AM
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If two matrices have exactly the same eigenvalues then they can both be written in "normal form" with those same eigenvalues on the diagonal, "0"s everywhere except possibly just above the main diagonal. How many "1"s there will be above the main diagonal
there are depends upon the eigenvectors.
Deveno
#11
Apr15-11, 11:00 AM
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and if you subtract out the diagonal parts, you will be left with nilpotent parts.

so there is a 3-stage comparison (being deliberately vague here, because the decompostion isn't unique, we can "re-arrange" the parts):

same diagonal parts, no nilpotent parts. <--preferred status

same diagonal parts, same nilpotent parts. <--almost as good, "same generalized eigenbasis"

same diagonal parts, different nilpotent parts. <---these matrices are fundamentally "different"
kini.Amith
#12
Apr15-11, 12:24 PM
P: 46
got it. i asked the 'what do they have in common' part repeatedly coz i have read frequently that the eigenvalues tell us many things about a matrix, so i guessed if 2 matrices have the same eigenvalues, we must be able to relate each other in some way.
i'm just learrning this topic, so i have no clear grasp of the concepts. hence the childish nagging questions.
Deveno
#13
Apr15-11, 05:05 PM
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eigenvalues do tell us a lot. if the matrix is diagonalizable, in some sense they tell us everything. and that is very often the case.

the basic idea is this: how can we put a matrix in a form that doesn't lose information, but is easy to work with? equivalently: is there a basis for a vector space V, that allows for easy computation of the linear transformations we are interested in?

and the answer is: sometimes. and when that happens, it's a happy thing.
stringy
#14
Apr21-11, 11:18 AM
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P: 90
Quote Quote by HallsofIvy View Post
No. Two matrices are similar if and only if they have the same eigenvalues and corresponding eigenvectors.
Would you mind clarifying this point? It's well known that a similarity transformation preserves the spectrum, but the eigenvectors?

The matrices

[ 0 1 ]
[ 0 0 ]

and

[ 0 0 ]
[ 1 0 ]

are similar via the permutation matrix

[ 0 1 ]
[ 1 0 ],

but they don't share the same eigenvectors.
wisvuze
#15
Apr21-11, 01:52 PM
P: 367
Quote Quote by stringy View Post
Would you mind clarifying this point? It's well known that a similarity transformation preserves the spectrum, but the eigenvectors?

The matrices

[ 0 1 ]
[ 0 0 ]

and

[ 0 0 ]
[ 1 0 ]

are similar via the permutation matrix

[ 0 1 ]
[ 1 0 ],

but they don't share the same eigenvectors.
The two matrices may not generally share the same eigenvectors, but the relation should be that if v is an eigenvector for matrix A, then Qv should be an eigenvector for matrix B, where Q is the change of basis matrix, so that Av = Q^-1 B Q v
stringy
#16
Apr21-11, 02:40 PM
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P: 90
Yup, I was just curious if HallsofIvy meant something else and just misspoke.

I was thinking perhaps there's another characterization of similar matrices out there that I don't know about.
spg89
#17
Apr22-11, 06:20 AM
P: 7
i think the value of principle diagonal is same in both matrices...


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