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Topic: Failing Linear Algebra:
Replies: 54   Last Post: Jan 10, 2007 12:47 PM

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 Guest
Re: Failing Linear Algebra:
Posted: Apr 28, 2004 6:40 PM

Grey Knight:

>If you calculate det(<>-l*<>) for an n-by-n matrix <>, you
>will get a degree-n polynomial in l; generally you'll be expected to
>deal with 2-by-2 matrices, so this polynomial is just a quadratic. The
>solutions of this polynomial are your eigenvalues.

Thanks, Grey Knight. I've got that part down.

>The eigenvectors
>can then be found by solving the problem <> <x_i> = l_i <x_i>; that
>is (<> - l_i<>)<x_i> = <0>

This is the part I don't get. I'm sorry but your notation loses me. I see the
"<>", and I think "dot product".

We find the eigenvalues by solving matrix A....det (A-[lambda]*I ), where I is
the identity matrix.

>Two matrices are similar iff
> <> = <<X>> <> <<X'>>

OK. So I = X*I*X'. So, X' is any matrix that when multiplied by X gives the
identity element as well?

What's the difference between A transpose and A inverse? A specific example(s)
would help. I'd love to test my skills.

Date Subject Author
4/22/04 Guest
4/22/04 Michael N. Christoff
1/10/07 Gerry Myerson
1/10/07 Jonathan Miller
1/10/07 Guest
1/10/07 David C. Ullrich
1/10/07 Acid Pooh
1/10/07 Guest
4/23/04 Brian Borchers
4/27/04 Guest
1/10/07 maky m.
4/26/04 David Ames
1/10/07 Guest
1/10/07 Michael Stemper
1/10/07 maky m.
4/23/04 Porker899
4/27/04 Guest
1/10/07 Abraham Buckingham
1/10/07 Mitch Harris
1/10/07 Guest
1/10/07 Grey Knight
1/10/07 Guest
1/10/07 Toni Lassila
1/10/07 Thomas Nordhaus
1/10/07 George Cox
4/28/04 Dave Rusin
4/28/04 George Cox
4/28/04 George Cox
4/29/04 Marc Olschok
4/29/04 Mitch Harris
4/29/04 Robert Israel
4/28/04 Guest
4/29/04 Guest
1/10/07 Dave Rusin