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Cake day: June 12th, 2023

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  • Here you go: https://www.gnu.org/fun/jokes/unix.errors.html

    (% represents the csh, $ represents the bourne shell)

    % "How poorly would you rate the Unix (so-called) user interface? Unmatched ".

    % rm congressional-ethics rm: congressional-ethics nonexistent

    % ar m God ar: God does not exist

    % [Where is Jimmy Hoffa? Missing ].

    % ^How did the sex change^ operation go? Modifier failed.

    % If I had a ( for every $ Congress spent, what would I have? Too many ('s.

    %make love Make: Don’t know how to make love. Stop.

    % sleep with me bad character

    % got a light? No match.

    % man: why did you get a divorce? man:: Too many arguments.

    % ^What is saccharine? Bad substitute.

    % (- (-: Command not found.

    % sh

    $ PATH=pretending! /usr/ucb/which sense no sense in pretending

    $ drink <bottle; opener bottle: cannot open opener: not found

    $ mkdir matter; cat >matter matter: cannot create

    Or, in a System V (att) universe:

    $ cat “can of food” cat: cannot open can of food











  • jxk@sh.itjust.workstoScience Memes@mander.xyzeigenspaces
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    2 months ago

    You want an answer?

    So you’ve probably learned that if u is an eigenvector, then multiplying u by any scalar gives you another eigenvector with the same eigenvalue. That means that the set of all a*u where a is any scalar forms a 1-dimensional space (a line if this is a real vector space). This is an eigenspace of dimension one. The full definition of an eigenspace is as the set of all eigenvectors of a given eigenvalue. Now, if an eigenvalue has multiple independent eigenvectors, then the set of all eigenvectors for that eigenvalue is is still a linear space, but of dimension more than one. So for a real vector space, if an eigenvalue has two sets of independent eigenvectors, its eigenspace will be a 2-dimensional plane.

    That’s pretty much it.