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Can we find out what the eigenvalues of a sum of two initial matrices are, by performing an operation on the eigenvalues & eigenvectors of the initial matrices?

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From eigenvalues alone no. If you have eigenvalues and a full set of eigenvectors, you know the matrix. If there are not enough eigenevctors, then the answer will again be no. –  Robin Chapman Sep 11 '10 at 10:19
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Well, you would also need to know what eigenvalues correspond to what eigenvectors ... –  Ricky Demer Sep 11 '10 at 10:20
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The simple answer is "not simpler than diagonalizing the new matrix". The long answer is: "yes, but it depends on the problem at hand, and you do not provide enough information to even guess a good solution for your problem. Let me just say: There is eigenvalue perturbation theory. So if you know an eigenvalue of $A$ with eigenvector $\psi$. Then you approximately know an eigenvalue of $A + t B$ if $t$ is small enough (depending on various things)". –  Helge Sep 11 '10 at 12:59
    
And one should also specify what "find" means. –  Helge Sep 11 '10 at 12:59
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There are some deeply nontrivial bounds on the eigenvalues of $A+B$ for hermitian $A$ and $B$: see terrytao.wordpress.com/2010/01/12/… –  darij grinberg Sep 11 '10 at 16:27
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up vote 34 down vote accepted

I will interpret this to be the question `What do you need to know beyond the eigenvalues of $A$ and of $B$ to get the eigenvalues for $A+B$ from those of $A$ and $B$?' It's a good question. Here's one way to conceptualize it:

Let's modify it to talk of the roots of the characteristic polynomial, rather than just eigenvalues. Over the complexes, for most but not all matrices, these are the same. Knowing the roots is equivalent to knowing the characteristic polynomials.

You can put the characterteristic polynomial of $A$ into homogeneous form as the determinant of $\lambda_1 I + \lambda_2 A$ --- this translates easily back and forth to the usual form. With two matrices, there is a polynomial in one more variable: the determinant of $\lambda_1 I + \lambda_2 A + \lambda_3 B$.

The characteristic polynomial for $A+B$ is obtained by specializing this 3-variable polynomial. I believe this can be any homogeneous polynomial P( x,y,z) of degree n in 3 variables subject to the condition P(1,0,0)=(the determinant of the identity) = 1, but I don't know for sure: perhaps someone can elucidate this point.

Update: Dave Speyer elucidated (see his comment below the fold):

Yes, this is true. If the curve is smooth, then the space of such representations is $J \setminus \Theta$, where $J$ is the Jacobian of the curve and $\Theta$ is the $\Theta$ divisor. See Vinnikov, "Complete description of determinantal representations of smooth irreducible curves", Linear Algebra Appl. 125 (1989), 103--140 – David Speyer

The question, what additional information do you need besides eigenvalues of $A$ and $B$ to get the eigenvalues of $A + B$, translates into a question about homogeneous polynomials. There is a triangle's worth of coefficients. You are given the coefficients on two of the sides of the triangle. What you want is the sum of coefficients along lines parallel to the third side. There are many ways you might get this information, depending on context. For example, the characteristic polynomial of $A B^{-1}$ corresponds to the 3rd edge of the triangle: in dimension 2, that gives the missing coefficient (all you actually need is Trace$(AB^{-1})$), while in higher dimensions, it's not enough.

To illustrate, here's a plot for a pair of pseudo-random matrices of dimension 9 whose coefficients were chosen uniformly and independently in the interval $[0,1]$. This is the slice $\lambda_1 = 1$ of the homogeneous form of the characteristic polynomiial, so on the two axes, its roots are the negative reciprocals of characteristic roots for $A$ and $B$, and on the diagonal, for $A+B$. The characteristic polynomial of $A B^{-1}$ determines the asymptotic behavior at infinity, in this slice. You can of course only see the real eigenvalues in this real picture, but you can see how they move around as you vary the slice through the origin. Under varying circumstance, you will have more a priori information. If $n=2$, the curves are conics. If the matrices are symmetric, they will intersect each line through the origin $n$ in $n$ points --- etc.

alt text

Addendum: Out of curiosity, I drew some pictures in the symmetric case, again for dimension 9. Here is a typical example, for a pair of pseudo-random 9-dimensional symmetric matrices. Observe the eigenvalues meeting and crossing as lines through the origin sweep around. [The buttons are from TabView in Mathematica, and they don't work in this static image.]

alt text

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"singular values = roots of the characteristic polynomial". I DON'T think this is the standard meaning of the word "singular values". –  darij grinberg Sep 11 '10 at 18:17
    
Thanks, I edited to eliminate this term -- I was confused about the terminology. –  Bill Thurston Sep 11 '10 at 18:36
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+1 for abstract art. –  darij grinberg Sep 14 '10 at 20:47
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"I believe this can be any homogeneous polynomial P( x,y,z) of degree n in 3 variables subject to the condition P(1,0,0)=(the determinant of the identity) = 1." Yes, this is true. If the curve $P=0$ is smooth, then the space of such representations is $J \setminus \Theta$, where $J$ is the Jacobian of the curve and $\Theta$ the $\Theta$ divisor. See Vinnikov, "Complete description of determinantal representations of smooth irreducible curves", Linear Algebra Appl. 125 (1989), 103--140 –  David Speyer Sep 14 '10 at 21:21
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Thanks! I just noticed your comment. The question had continued to nag me, and I've been waiting for a comprehensible answer to fall out of the sky. Thanks, sky! I'm wondering also whether the determinant picture will help clarify the geometry of ratonal functions. –  Bill Thurston Sep 17 '10 at 6:10
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Consider a diagonal matrix A that has eigenvaues 1 and -1 with eigenvectors $e_1$ and $e_2$. Then A and -A have the same eigenvalues and eigenvectors as A and A. But A + -A = 0 and A + A $\ne$ 0

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