Eigenvalues and Hermitian Matrices
Date: 01/03/2006 at 03:32:30 From: Sunder Subject: Linear Algebra: Proof Let A be a symmetric matrix (such a matrix appears when considering multivariate distributions). Show that the eigenvalues of A are real. Surfing the net, I found out that it's about Hermitian matrices. Is there a simpler approach? I don't understand it well enough to prove it. Your guidance is much appreciated.
Date: 01/03/2006 at 04:51:25 From: Doctor Luis Subject: Re: Linear Algebra: Proof Hi Sunder, Let's take a real symmetric matrix A. The eigenvalue equation is: Ax = ax where the eigenvalue a is a root of the characteristic polynomial p(a) = det(A - aI) and x is just the corresponding eigenvector of a. The important part is that x is not 0 (the zero vector). Well, anyway. Let's calculate the following inner product (here, x_i* is the complex conjugate of x_i): <x,Ax> = sum_i x_i* (Ax)_i = sum_i x_i* (sum_j A_ij x_j) = sum_i sum_j x_i* A_ij x_j That's the inner product expanded out, which we'll use later. But for now, note that since x is an eigenvector, we know that Ax = ax. We can use this fact to conclude: <x,Ax> = <x,ax> = sum_i x_i* (ax)_i = sum_i x_i* a x_i = a sum_i x_i* x_i = a (sum_i |x_i|^2) Note that sum_i |x_i|^2 is always positive since x is nonzero. We'll use this fact later, too. Next, we should find the following inner product (again, y* means complex conjugate of y): <Ax,x> = sum_i (Ax)_i* x_i = sum_i (sum_j A_ij x_j)* x_i = sum_i (sum_j A_ij* x_j*) x_i = sum_i sum_j x_i A_ij* x_j* But now, we can use the fact that A^t = A and that A is real. In particular, that A_ij* = A_ij, and A_ji = A_ij. <Ax,x> = sum_i sum_j x_i A_ij x_j* = sum_i sum_j x_i A_ji x_j* = sum_j sum_i x_j* A_ji x_i = sum_I sum_J x_I* A_IJ x_J (renaming j->I, i->J) = sum_i sum_j x_i* A_ij x_j (dummy variables J->j, I->i) = <x,Ax> So, because A is real and symmetric, we have A = A^t and <Ax,x> = <x,Ax>. Now, take the eigenvalue equation again: Ax = ax Now, take the transpose and then complex conjugate: (Ax)^t = (ax)^t x^t A^t = a x^t x^t A = a x^t (since A^t = A) (x^t A)* = (a x^t)* (x*)^t A* = a* (x*)^t (x*)^t A = a* (x*)^t (since A* = A) Now, just multiply both sides by x, (on the right), (x*)^t A x = a* (x*)^t x sum_i (x*)_i (Ax)_i = a* sum_i (x*)_i x_i sum_i x_i* (sum_j A_ij x_j) = a* sum_i x_i* x_i sum_i sum_j x_i* A_ij x_j = a* (sum_i |x_i|^2) or <Ax,x> = a* (sum_i |x_i|^2) But, we already found that <x,Ax> = a (sum_i |x_i|^2), and that <Ax,x> = <x,Ax>. Therefore, 0 = <Ax,x> - <x,Ax> = a* (sum_i |x_i|^2) - a (sum_i |x_i|^2) 0 = (a* - a) (sum_i |x_i|^2) Since sum_i |x_i| > 0, we can divide this last equation by it, which gives us 0 = a* - a or a = a* Since a is any eigenvalue of A, we have proven that the complex conjugate of a is a itself. This can only happen if a is real, which concludes the proof. Note that we spent most of the time doing inner product math in the long-winded explanation given above. All we really wanted to say was that <x,Ax> = <A'x,x>, where A' is the adjoint matrix to A (adjoint for matrices means transpose and complex conjugation). A matrix which is its own adjoint, i.e. A = A', is called self-adjoint or Hermitian. That's all it means. Clearly, a real Hermitian matrix is just a symmetric matrix. Do you see why? Now, the short proof. Consider the inner product <u,v> = sum_i u_i* v_i and let A be a Hermitian matrix. Let x be an eigenvector of A with eigenvalue a. Then, <x,Ax> = <x,ax> = a <x,x> and <Ax,x> = <ax,x> = a* <x,x> (can you explain this last step?) Lastly, note that <x,Ax> = <A'x,x> (adjoint matrix) = <Ax,x> (since A is self-adjoint) Thus, 0 = <Ax,x> - <x,Ax> = a* <x,x> - a <x,x> 0 = (a* - a) <x,x> 0 = a* - a (we can divide by <x,x> since it's nonzero) a = a* Therefore, any eigenvalue a of a Hermitian matrix A is real. See how much simpler the Hermitian proof is? Well, let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/
Date: 01/04/2006 at 01:00:42 From: Sunder Subject: Thank you (Linear Algebra: Proof) Dear Dr. Luis, Thanks for your prompt reply. You asked good questions to prod my understanding.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum