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If you have a situation where every row-sum is equal in a matrix A, this sum equals one of the eigenvalues of the matrix. It might make a difference in cases where the matrix is symmetrical?

Using this fact, is there any easy procedure/shortcut for finding the rest of the eigenvalues?

Or do you have to solve the characteristic equation? You could use the eigenvalue we found at first and use this as a factor when doing long division on the charactersitic equation, however when the long division doesn't end up with a nice expression for the factorization you start to wonder if this really is the right way to go (in case of 3rd or more degreee polynomials). I got this characteristic equation for n = n for the matrix below: $\lambda{^3} - 3 \lambda{^3} +3(1-n^2) \lambda + 2(n^2 -n^3) - 1$

EDIT: Now going over to a specific example with a symmetric matrix where all rowsums are equal, but where everything not on the main diagonal is equal to n (any real number). For example we could say that n = 1, then this would be a duplicate question of another thread here, but n can be any real number, so would the same apply if n was equal to 5 instead? This thread for instance is similar, however the OPs non-main-diagonal elements are in the range 0 < r < 1, which are not any of the real numbers: Find the eigenvalues of a matrix with ones in the diagonal, and all the other elements equal

So the matrix would looks like this:

$ A = \left(\begin{array}{rrr} 1 & n & n\\ n & 1 & n\\ n & n & 1 \end{array}\right) $

Finding the characteristic equation for this matrix returned a pretty ugly result, which results in a nightmare if you try doing longdivision on this expression using the rowsum as the first eigenvalue = 1 + 2n

Therefore I'm convinced there must be some shorter way for finding the eigenvalues/eigenvectors, wondering what it might be? Any link to any exisiting thread that might adress or apply to this specific scenario? Perhaps here?: Determinant of a rank $1$ update of a scalar matrix, or characteristic polynomial of a rank $1$ matrix

So I tried with some different values for n and solved in a computer program.

We know that 1 of the eigenvalues equals the rowsum. tr(A) = sum of all eigenvalues = 3 for any n (real)

So for n=1 we get eigenvalues 0, 3 and 0.

So for n=2 we get eigenvalues -1, 5 and -1.

So for n=3 we get eigenvalues -2, 7 and -2.

But I can't seem to see how to present a general solution for any real n, for the reasons mentioned further above.

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jibo
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1 Answers1

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Nothing else can be said exactly, without running an eigenvalue routine (which usually is not solving the characteristic equation in practice).

Indeed, the spectral theory of left-stochastic matrices, meaning matrices with row sum $1$ and nonnegative entries, is a subject of significant interest and complexity in probability theory. Yet this is itself a special case of your situation.

Ian
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  • hmm I might have forgotten that the matrix might be symmetric as well. Does such a scenario change anything? – jibo May 12 '15 at 15:46
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    @Mollart In that case you are in a somewhat better situation in that you can sensibly subtract off the part of $A$ which contributes to the eigenspace that you've found. Specifically you can subtract off $xx^T$ where $x$ is the eigenvector you found; the resulting matrix will have the same eigenvectors but the eigenvalue $1$ that you found will now be zero. But it is not that much better. – Ian May 12 '15 at 16:42
  • You mention "...running an eigenvalue routine (usually is not solving the characteristic equation in practice)..." Do you mean that there is an alternative to solving the characteristic equation, in that case which are you thinking about? – jibo May 18 '15 at 15:27
  • @Mollart Analytically there is usually not except in very special situations. Numerically there are many possibilities which are used in practice depending on the matrix in question. – Ian May 18 '15 at 17:23
  • I see, I added a concrete matrix to the original post, do you know the method for that specific case? BTW how do you add the [@]username to the comments, when I do it always vanishes. – jibo May 18 '15 at 17:33
  • @Mollart It is removed when you are responding to a post which will always notify the user in question (for instance when you are commenting on their answer). As for your special case, there might be some special structure to the characteristic equation? – Ian May 18 '15 at 17:41
  • I got this characteristic equation for that specific case when n = n: $\lambda{^3} - 3 \lambda{^3} +3(1-n^2) \lambda + 2(n^2 -n^3) - 1$

    And the 1st eigenvalue $\lambda = 1 + 2n $:

     – jibo May 18 '15 at 17:49