Linear maps between finite dimensional spaces

  • Thread starter aurorasky
  • Start date
  • #1
7
0
Is it true that any linear map between two arbitrary finite-dimensional vector spaces is continuous? Is it differentiable?
 

Answers and Replies

  • #2
22,129
3,297
Hi aurorasky! :smile:

Is it true that any linear map between two arbitrary finite-dimensional vector spaces is continuous? Is it differentiable?

This is indeed true! Take an arbitrary linear map [itex]f:\mathbb{R}^n\rightarrow \rightarrow{R}^m[/itex]. Then it suffices that the components f1,...,fm are continuous. And this is true since

[tex]\begin{eqnarray*}
|f_1(x_1,...,x_n)|
& = & |x_1f_i(1,0,...,0)+...+x_nf_i(0,...,0,1))|\\
& \leq & |x_1||f(1,0,...,0)|+...+|x_n||f(0,...,0,1)|\\
& \leq & \|(x_1,...,x_n)\|_\infty(|f(1,0,...,0)|+...+|f(0,...,0,1)|)\\
\end{eqnarray*}[/tex]

which implies continuity. For differentiability, we have to find a linear map u such that

[tex]\lim_{h\rightarrow 0}{\frac{f(x_0+h)-f(x_0)-u(h)}{\|h\|}}=0[/tex]

take u=f and it's easy to see that this is true.
 
Last edited:
  • #3
7
0
Thanks! Also, if a map is differentiable, it is also continuous and so the proof can be made really simple, right?

BTW, could you explain to me how to type latex code?
 
  • #4
22,129
3,297
Thanks! Also, if a map is differentiable, it is also continuous and so the proof can be made really simple, right?

Yes, but the proof that a differentiable map is continuous uses that linear maps are continuous... (at least the proof I've seen)

BTW, could you explain to me how to type latex code?

Click on the LaTeX symbols to see what I did.
 
  • #5
mathwonk
Science Advisor
Homework Helper
2020 Award
11,184
1,382
of course you have to give the topology to decide continuity. presumably you meant the norm topology, i.e. the one induced from R^n by the isomorphism given by any basis. In that case a linear map is given by a polynomial hence continuous.

then you have to give the differentiable structure, which again is rpesumably the one defined by a linear isomorphism with R^n. Now of coiurse for it to be meaningful, you need to know that every linear isomorphism defines the same differentiable structure, which is equivalent to your original question for R^n, i,.e is every linear map on R^n differentible,

well by definition a map if differentiable if locally it is approximated very well by a linear map, so duhh,... yes a linear map is differentiable.

more precisely a map f is tangent to zero if |f(x)|/|x|-->0 as x-->0. and a map g is differentiable at a if there is a linear map L such that the map g(a+x)-g(a)-L(x) is tangent to zero. Obviously if g is linear and we take L = g, then

g(a+x)-g(a)-L(x) = 0, hence is certainly tangent to zero.

in infinite dimensions some linear maps are not (norm) continuous, and differentiability may be defined as "approximable by a continuous linear map". is this your confusion? i.e., is some nincompoop teaching you in finite dimensional differential calculus before finite dimensional calculus? (that happened to me.) (I was the nincompoop.)
 

Related Threads on Linear maps between finite dimensional spaces

Replies
5
Views
2K
Replies
8
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
8
Views
1K
Replies
8
Views
10K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
3K
Replies
1
Views
3K
Top