# Directional Derivative

Hi:

Can someone point how to approach this problem- we had 5 problems on directional derivatives and I solved 4. I understand the concept but in this question I don't know where to begin

Problem Statement
Assume that f:R$$^{n}$$ -> R$$^{m}$$ is a linear map, with matrix A with respect to the canonical bases. Show that Df(xo) = f for every xo $$\in$$ R$$^{n}$$

Plz advise - I will probably post follow-up questions to any answers

Thanks

Asif

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Hurkyl
Staff Emeritus
Gold Member
I understand the concept but in this question I don't know where to begin
You should never be at a loss of how to begin a problem -- definitions are almost always a reasonable starting point.

HallsofIvy
Homework Helper
Specifically, what is the definition of "Df" and what happens when you apply that definition to a linear map?

I will give what I have done so far...

Definition of a directional derivative is its partial derivatives wrt to all the variables in the given function.

So in this case the question is for f:R$$^{n}$$->R$$^{m}$$ there is an mxn matrix which would look like the following

Df(x0) = (assume this is equation 1)

df1/dx1 df1/dx2... df1/dxn
df2/dx1 df2/dx2... df2/dxn
. ........ . ........ .
. ........ . ........ .
dfm/dx1 dfm/dx2...dfm/dxn

Where in above matrix dfm/dx1 is the partial derivative of function wrt x1,x2... I couldn't find symbol for partial derivative

Now, if I use the definition of a linear map then I know that
D($$\alpha1$$+$$\alpha2$$ ) f(x0) = $$\alpha1$$Df(x0) + $$\alpha2$$ Df(x0)

I can also prove by continuity and as t->0 and $$\varsigma$$->0 that
Df(x0 + $$\varsigma$$p1 + t$$\alpha2$$p2) -> Df(x0) ----- equation 2

Now since this is a canonical map the above matrix of Df(x0) in equation 1 reduces to the following
1 0... 0
0 1... 0
. ...
. ...
0 0...1

So in equation2 since D is essentially the above matrix, I can say the following:
Df(x0 + $$\varsigma$$p1 + t$$\alpha2$$p2) -> f(x0) which is what I think the question wants.

Is this correct?

Thanks

Asif

HallsofIvy
(in LaTex, [ tex ]\partial[ /tex ] gives $$\partial$$.)