What does this mean? (1 Viewer)

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Two things:

1) If we say that the space of all 2 by 2 matrices is identified with R^4, what does that mean?

2) Suppose f is a function from GL(n, R) to GL(n, R) (the space of all real n by n invertible matrices) identified with [tex] \mathbb{R}^{n^2} [/tex] I am asked to prove that [tex] df_{A_0} (X) = -X [/tex] where [tex] A_0 [/tex] is the identity matrix. My question is, [tex] df_{A_0} [/tex] would usually denote that derivative of f at the point [tex] A_0 [/tex], so where does that (X) part come into play?

I know that I should be asking my prof this, but I wanna do these homework questions before my next class (Wednesday), so it would be great if you guys could help me out.
 
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mathman

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For the first question: a (real) 2 by 2 marix is specified by four numbers, which defines a point in R^4.

Second question: I am not familiar with the notation.
 

Office_Shredder

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For the second question: The derivative of a function is a linear function. In this case the question is asking you to prove that the linear function that the derivative is is the function df(X)=-X
 
726
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OfficerShredder, I was thinking that. But usually my prof would use the notation [tex] df_{A_0} [/tex] to denote the derivative of f at [tex] A_0 [/tex]. Why add in the extra [tex] (X) [/tex] ?

And mathman, do I read off the entries of the matrix row by row or column by column
 
726
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In case it helps, OfficerShredder, f is defined by [tex] f(A) = A^{-1} [/tex] if [tex] A \in GL(n,r) [/tex]
 

Landau

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And mathman, do I read off the entries of the matrix row by row or column by column
That's up to you. You can identify the space of nxm-matrices with R^{mn} in a lot (namely (nm)!) of ways, there's not really a preferred way.
 

HallsofIvy

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OfficerShredder, I was thinking that. But usually my prof would use the notation [tex] df_{A_0} [/tex] to denote the derivative of f at [tex] A_0 [/tex]. Why add in the extra [tex] (X) [/tex] ?

And mathman, do I read off the entries of the matrix row by row or column by column
For the same reason that to talk about the "squaring function" we say [itex]f(x)= x^2[/itex] rather than just "[itex]f= ( )^2[/itex]". A function is defined by what it does to values of x.
 
726
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I'm still not getting it. In my prof's usual notation [tex] df_{A_0} [/tex] would mean the derivative of f at [tex] A_0 [/tex]. If you write it using the prime notation, [tex] df_{A_0} = f'(A_0) [/tex]. I still don't see why you would need the matrix [tex] X [/tex] when we're evaluating the derivative function at the point [tex] A_0 [/tex]
 
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726
1
lol nevermind guys. I totally forgot that my proof uses that notation to mean the directional derivative of f at the point X with respect to A_0
 

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