Finding the Directional Derivative in Multivariable Calculus

[Carl140]

Homework Statement

Let f: R^n -> R be defined as follows:

f(x) = x*L(x) where * denotes the standard inner product and L: R^n -> R^n is a linear
function.
I'm trying to find the directional derivative f'(x;u).

Homework Equations

I know that f'(x;u) (the directional derivative of f(x) in the direction of the unit vector u)
is equal to gradient(f(x)) * u where * denotes inner product.

The Attempt at a Solution

In this case gradient(f(x)) = gradient(x*L(x)) = gradient(x)*L(x) + L(x)*gradient(x)
(Not sure if this step is correct).
Then because L: R^n -> R^n is a linear map the differential of L is L again, no?
So we get: gradient(f(x)) = gradient(x)*L(x) + L(x)*gradient(x) = 2L(x)*gradient(x).

Conclusion: the directional derivative in the direction of u is then (2L(x)*gradient(x))*u where u is a unit vector.

Is this correct? I'm a little bit confused about the part of L being a linear map, is it correct
to state that the gradient is again L itself?

 

[Dick]

You are writing expressions that don't really make sense. Think about what f(x) looks like in components. It's f(x)=x_i*L_ij*x_j (where x_i are the components of x, L_ij is the matrix representing L and i and j are summed from 1...n). The kth component of the gradient is df/x_k. Can you think how to write that concisely using matrices?

 

[Carl140]

Ok Dick, thanks for your reply. So
I think f(x) = x^t * A * x where x represents the column vector and A is the matrix
which represents L. Is this correct so far?
Then how can I take the gradient? I'm really confused about this matrices thing,
all I'm used to is functions.

 

[Dick]

Yes, now write it in indices. x_i*A_ij*x_j (summed over i and j). It's a quadratic form. You need to use the product rule. Take d/dx_k of it to get the kth component of the gradient. You get something nonzero if either i=k or j=k. So d/dx_k is A_kj*x_j+x_i*A_ik where the first is summed over i and the second is summed over j. Express that in terms of matrix products. I know I'm repeating myself, but I don't know what else to say.