How would I take the jacobian of a map [math] f:{\mathbb{R}^{nxn}}

google

>>7920709
i did

>>7920715
try turning off your computer and turning it on again, then tell me if your problem is fixed

is the domain the set of n by n matrices?

>>7920830
ye

>>7920707
Just take the partial derivative with respect to A_{mn} for each mn. You get 2 A, no?

>>7920707
the jacobian is 2A.

>>7920835
>>7920837
Ok good, thats what I thought.

Now what about [math] d{\left( {f{|_{O\left( n \right)}}} \right)_I} [/math] when [math]{T_I}O\left( n \right) = \left\{ {A \in {\mathbb{R}^{n \times n}}|A = - {A^T}} \right\} [/math]

>>7920842
what the fuck is that notation
i like linear algebra problems but that's just disgusting

>>7920847
The differential of that same map f with its domain restricted to orthogonal matrices. Evaluated at the the identity matrix.

Where the tangent space of the orthogonal group at the identity is the space of skew-symmetric matrices.

>>7920852
ah that makes sense
this problem is a little more involved than the previous one, this might take a while

>>7920842
what the fuck is this
I know it's linear algebra judging by [math]\mathbb{R}^{n \times n}[/math] but D A M N

File: 5172396.jpg (255KB, 961x759px) Image search: [Google]

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>>7920842
wait a minute...

>>7920842
Isn't it still 2A?

File: Screen Shot 2016-03-09 at 11.06.44 PM.png (155KB, 2116x236px) Image search: [Google]

155KB, 2116x236px

>>7920867
Prep for midterm

>>7920867
It's smooth manifold theory. But yes, [math]T_I O(n)[/math] is a vector space, called the tangent space. It's sort of like the space of "derivatives" at a point, and that business with [math]d(f_{O(n)})_I[/math] is sort of like a derivative (hence the notation).

>>7920943
So the answer isn't 2A then. For the unrestricted problem, the tangent space at I is essentially R^{nxn} again. Say A is a tangent vector (in this case an nxn matrix) at I, then df_I A should be an element of the tangent space at f(I)=n, i.e. essentially a real number. We've seen that the gradient of f at I is just 2 I. So df_I (A) = 2 trace(A), the inner product of 2 I and A thought of as vectors.

>>7921896
As OP knows, for the O(n) group, the tangent space at I consists of skew-symmetric matrices. These have zeros along the diagonal, thus zero trace.

So d(f_{O(n)})_I = 0 identically it seems.

Intuitively, if you start with I, and add small epsilon bits to the off-diagonal elements, the sum of squares will only change by order epsilon^2.

>>7921896
>>7921902
Thanks. I figured it was still 2I b/c [math] d{\left( {f{|_M}} \right)_p} = d{f_p}{|_{{T_p}M}} [/math]. But was sure how to describe the way it acted on elements of the tangent space.

>>7921927
>was sure
wasn't sure*

On a scale of 1 to 10, how advanced is the linear algebra seen in this thread?
It looks so frightening.

>>7922068
Will I be able to understand these concepts with rudimentary handling of linear algebra? Linear algebra I - II?

>>7922068
>>7922483
No. While the tangent space to O(n) is a vector space, O(n) is not. It is a smooth manifold (and also a group).

So you need a course/book in Smooth Manifolds.

>>7921896
>>7921902

You could also see that f is constrant on On, since if you had the squares of the coefficients of one colon of an On Matrix you get one.
Thus, f = n identically over On, thus the derivative is 0 identically.

>>7924301
Interesting

>>7924301
ha! obvious once you see it. nice job anon.