Are Matrices with the Same Eigenvalues Always Similar?

In summary, two matrices are similar if and only if they have the same eigenvalues and corresponding eigenvectors. However, even if two matrices have the same eigenvalues, they may not necessarily be similar if their corresponding eigenvectors are not linearly independent. This is because similarity is a way of changing the basis of a linear transformation, and if there are not enough linearly independent eigenvectors, the transformation cannot be fully captured in a new basis. Therefore, two matrices with the same eigenvalues may have different "parts" that make them fundamentally different, such as nilpotent parts.
  • #1
kini.Amith
83
0
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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  • #2
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.)
 
  • #3
then what do 2 vectors having the same evalues have in common
 
  • #4
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?
 
  • #5
i nderstand that they need not be similar, but then what do they have in common?
 
  • #6
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.
 
  • #7
k. got some idea. will think some more about it.
thnx
 
  • #8
kini.Amith said:
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"!
 
  • #9
well, they share the same eigenvalues, lol
 
  • #10
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.
 
  • #11
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"
 
  • #12
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.
 
  • #13
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.
 
  • #14
HallsofIvy said:
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.
 
  • #15
stringy said:
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
 
  • #16
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.
 
  • #17
i think the value of principle diagonal is same in both matrices...
 

1. What are eigenvalues?

Eigenvalues are a mathematical concept used in linear algebra to describe the properties of a matrix. They are defined as the scalar values that, when multiplied by a vector, result in the same vector. In other words, they represent the scaling factor of a vector when it is transformed by a matrix.

2. How do you determine if two matrices have the same eigenvalues?

To determine if two matrices have the same eigenvalues, you can calculate the determinant of both matrices. If the determinant is the same for both matrices, then they have the same eigenvalues. Additionally, you can also compare the characteristic polynomial of the matrices, as they will have the same roots if the matrices have the same eigenvalues.

3. Why is it important for two matrices to have the same eigenvalues?

Having the same eigenvalues means that the matrices have similar properties and can be transformed in a similar way. This is useful in various applications, such as in solving systems of linear equations, finding the principal components of a dataset, and in understanding the behavior of dynamical systems.

4. Can two matrices with the same eigenvalues have different eigenvectors?

Yes, it is possible for two matrices to have the same eigenvalues but different eigenvectors. This means that although the matrices have the same scaling properties, they may transform vectors in different directions. This is because the eigenvectors are not unique and can be scaled or rotated.

5. How does the concept of eigenvalues relate to the diagonalization of matrices?

The diagonalization of a matrix involves finding a diagonal matrix that is similar to the original matrix. In this process, the eigenvalues of the original matrix are the diagonal elements of the diagonal matrix. Therefore, two matrices with the same eigenvalues are easier to diagonalize and can be transformed into the same diagonal matrix.

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