Eigenvalues of a matrix B= f(A) given eigenvalues of A

[L-x]

Homework Statement

Find the eigenvalues/vectors of A. (I can do this bit :P, A is a 3x3 matrix)
What are the eigenvalues and eigenvectors of the matrix B = exp(3A) + 5I, where I is
the identity matrix?

Homework Equations

The Attempt at a Solution

I have (correctly) found that A has eigenvectors and corresponding eigenvalues such that
A|1>=|1>, A|2>=2|2>, A|4>=4|4>.
|1>=(1,1,1), |2>=(1,-1,0), |4>=(1,1,-2) though i don't actually think you need this.

as B can be expanded in a power series of A, and A commutes with the identity (obiously)

[B,A]=0

=> |1>, |2>, |4> are eigenvectors of B

I am not confident in the reasoning that follows, does it seem correct?

exp(3A)|j>=e^(3j)|j>
5I|j>=5|j>
j=1,2,4
=> the eigenvectors of B are |1>, |2>, |4> with respective eigenvalues (e^3 +5), (e^6 +5), (e^12 +5)

Does this seem correct? Thanks in advance for your help.

 

[micromass]

Hi L-x!

Seems ok!

 

[Ray Vickson]

There is a theorem in matrix analysis that for an nxn matrix A having n distinct eigenvalues (lambda_1, ..., lambda_n), and for any analytic function f(x) = sum_{n=0}^infinity c_n *x^n, the function f(A) =def= c_0*Identity + sum_{n=1}^infinity c_n * A^n has the form
f(A) = sum_{j=1}^{n} E_j * f(lamba_j), where E_1,...,E_n are nxn matrices that do not depend on the form of f(.) . Furthermore, if v1, v2,..., vn are eigenvectors for the n eigenvalues, we have E_i*v*j = 0 for i =/= j. So, in your case, taking f(x) = x^0 we have Identity = 1^0*E1 + 2^0*E2 + 3^0*E3 = E1+E2+E3, taking f(x) = x gives A = E1 + 2E2 + 3E3, and taking f(x) = x^2 gives A^2 = E1 + 4*E2 + 9*E3. So, you can determine E1, E2 and E3. Then, of course, for f(x) = exp(3x) + 5 we have f(A) = E1*(exp(3)+5) + E2*(exp(6)+5) + E3*(exp(9)+5). Thus, f(A)*v1 = (exp(3)+5)*v1, etc. (Note that you do not need to a actually find E1, E2 and E3 to answer the question.)

RGV

 

[L-x]

There is a theorem in matrix analysis that for an nxn matrix A having n distinct eigenvalues (lambda_1, ..., lambda_n), and for any analytic function f(x) = sum_{n=0}^infinity c_n *x^n, the function f(A) =def= c_0*Identity + sum_{n=1}^infinity c_n * A^n has the form
f(A) = sum_{j=1}^{n} E_j * f(lamba_j), where E_1,...,E_n are nxn matrices that do not depend on the form of f(.) . Furthermore, if v1, v2,..., vn are eigenvectors for the n eigenvalues, we have E_i*v*j = 0 for i =/= j. So, in your case, taking f(x) = x^0 we have Identity = 1^0*E1 + 2^0*E2 + 3^0*E3 = E1+E2+E3, taking f(x) = x gives A = E1 + 2E2 + 3E3, and taking f(x) = x^2 gives A^2 = E1 + 4*E2 + 9*E3. So, you can determine E1, E2 and E3. Then, of course, for f(x) = exp(3x) + 5 we have f(A) = E1*(exp(3)+5) + E2*(exp(6)+5) + E3*(exp(9)+5). Thus, f(A)*v1 = (exp(3)+5)*v1, etc. (Note that you do not need to a actually find E1, E2 and E3 to answer the question.)

RGV

This looks interesting but I'm having quite a lot of trouble following your notation. Do you know the name of the theorem so i could look it up perhaps?

Thanks for the help.

 

[Ray Vickson]

This looks interesting but I'm having quite a lot of trouble following your notation. Do you know the name of the theorem so i could look it up perhaps?

Thanks for the help.

The old books “Matrices”, by Gantmacher, or the book “Theory of Matrices”, by P. Lancaster (Academic Press, 1969) have these results. I am sure there are numerous more modern treatments, but those two are the ones I have on my bookshelf.

Anyway, below I will give a short proof for the case of a 3 x 3 matrix A having 3 distinct, non-zero real eigenvalues r1, r2 and r3. The same form of proof goes through for a general nxn matrix with n distinct, real, nonzero eigenvalues. The basic result is the same, but with a somewhat different proof, if some of the eigenvalues are complex or some are zero. Distinctness remains important; without it we need to invoke Jordan Canonical forms, and f(A) can involve not only f(r) for eigenvalue r, but also f’(r). f’’’(r), etc., depending on the multiplicity of r and the form of its Jordan block(s).

So, let ui and vi be the left and right-eigenvectors of A for eigenvalue ri, chosen so that ui*vi = 1. Note that for rj different from ri we have ui*vj = 0; see, eg,
http://answers.yahoo.com/question/index?qid=20100729070957AAlQuHF . Note that the vi form a basis for the whole space. Now let Ei = vi*ui for all i; we have that Ei is a 3x3 matrix because it is of the form column x row in that order. The orthogonality results imply that Ei*Ei = Ei and Ei*Ej = 0 if i is different from j. Letting B = sum ri*Ei, we have that B*vi = ri*vi for all i, hence (B-A)*vi = 0 for all vi, hence (B-A)w = 0 for all vectors w, because all such w have the form sum ci*vi . Thus, B = A, so we have shown that A = sum ri*Ei. Now look at C = E1+E2+E3. We have C*vi = vi for all i, so C*w = w for all vectors w. That is, C = Identity matrix I. Now we are almost done.

Look at any polynomial p(x) = c0 + c1*x + c2*x^2 + … + cm*x^m, and _define_ p(A) = c0*I + c1*A + c2*A^2 + … + cm*A^m. We have A^2 = sum ri^2 * Ei*Ei + sum_{i < j}ri*rj Ei*Ej + sum_{j < i} rj*ri*Ej*Ei = sum ri^2 * Ei + 0 + 0. Similarly, A^3 = sum ri^3*Ei, etc. Finally, we have I = sum Ei = sum ri^0 * Ei (this is where having all ri nonzero comes in!), so we have that p(A) = c0*A^0 + c1*A^1 + … + cm*A^m is of the form p(A) = sum p(ri)*Ei. If f(x) is an analytic function whose radius of convergence includes the largest |eigenvalue|, then the same type of argument goes through, by taking some limits. The result is f(A) = sum f(ri)*Ei.

This stuff is typically used in solving constant-coefficient couples ODEs, because if x is a vector and A is a constant matrix, the DE system dx/dt = Ax has solution x(t) = exp(A*t)*x(0), so we need to know how to compute matrix exponentials.

RGV

 

[conquest]

How about just changing to the basis given by the eigenvectors (they should be linearly independent since there are 3 distinct eigenvalues). then in this basis A is just diagonal so the exponent is just the diagonal matrix with the exponents on the diagonal (so the diagonal entries exponentiated on the diagonal).