I agree with you, Jotaf, that disproving it for 2-by-2 was enough for me to say that a general statement about diagonalizing with lower triangular matrices is false. But I also liked seeing Steve L.’s calculation.

I have a feeling that there is a general linear algebra way to see that the statement must be false in arbitrary dimensions. But instead of thinking hard to figure that out, I just did this little symbolic calculation, and it was good enough for me to say that the economics paper had a misstatement.

Alan

]]>The Symbolic Toolbox is already an invaluable tool, but I can’t help but feel that it would be immensely more useful if the basic construct was a matrix (arbitrarily sized, or with properties such as square/symmetric/etc) instead of a scalar. As it stands, it is limited to small (2×2 or 3×3) examples, otherwise the expressions blow up. But I realize that this would be a major project on its own, rivaling in complexity with the current symbolic engine!

]]>A = sym('A', [3 3]); % Make A symmetric A = tril(A)+tril(A, -1).'; % Use only the lower-triangular piece of G G = tril(sym('G', [3 3])); D = G*A*inv(G); D(1, 3) ans = (A3_1*G1_1)/G3_3

By the reasoning above, A3_1 must be 0 since G1_1 can’t be. It also means G3_3 can’t be 0 either, which makes sense because if it were the inverse wouldn’t exist. Let’s update A with that information and see how that changes D.

A = subs(A, A(3, 1), 0); D = G*A*inv(G); D(1, 3) % This should be 0 ans = 0 D(1, 2) ans = (A2_1*G1_1)/G2_2

This fixes A2_1 at 0, since again G1_1 can’t be. This also means G2_2 can’t be 0, since if it were we would be dividing by 0.

A = subs(A, A(2, 1), 0); D = G*A*inv(G); D(1, 2) % This should be 0 ans = 0 D(2, 3) ans = (A3_2*G2_2)/G3_3

If A3_2 were 0, A would be diagonal already.

subs(A, A(3, 2), 0) ans = [ A1_1, 0, 0] [ 0, A2_2, 0] [ 0, 0, A3_3]

On the other hand, if G2_2 were 0 (which we’ve already ruled out above, but let’s double-check), G would be:

subs(G, G(2, 2), 0) ans = [ G1_1, 0, 0] [ G2_1, 0, 0] [ G3_1, G3_2, G3_3]

which is no longer invertible. This isn’t a proof that no such thing exists for an arbitrarily sized matrix, but it does support that statement. Often 2 is an unusual case — as two examples, it’s the only even prime number and it’s the only positive integer for which no magic square of that order can exist (the MAGIC function specifically calls this out.)

]]>What comes to my mind is the charts included in this recent blog post http://blogs.mathworks.com/community/2014/09/02/expo-conversations-part-1/ this system processed live feed from RFID sensors and continually updated the status of conference attendees using MATLAB Production Server with D3.js for the web front end.

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Thanks also Goose, I didn’t think about Uber’s surge pricing, but it is driven by demand as you say. However, because it is a dynamic pricing, it is hard to tell when it kicks in. I would imagine people taking Uber rides early morning hours would pay more, because supply would be limited. But apparently that didn’t stop them from using the service! ]]>