MHB Solving Linear Transformation: Find F Given 3 Equations

Petrus
Messages
702
Reaction score
0
Hello MHB,
given a linear transformation F so that this is known
$$\left\{
\begin{aligned}
\phantom{1}F(1,0,0)=(1,2,3) \\
F(1,1,0)=(0,0,1)\\
F(1,1,1)=(12,3,4)\\
\end{aligned}
\right.$$
Decide F

progress:
$$F(e_1)=(1,2,3)$$
$$F(e_2)=F(e_1)+F(e_2)-F(e_1)=(0,0,1)-(1,2,3)=(-1,-2,-2)$$
$$F(e_3)=F(e_1)+F(e_2)+F(e_3)-(F(e_1)+F(e_2))= (12,3,4)-(-1,-2,-2)=(13,5,6)$$
My $$f(e_3)$$ is wrong acording to facit and I don't understand

Regards,
$$|\pi\rangle$$
 
Physics news on Phys.org
Hello MHB,
I have seen what my misstake was... I forgot that $$F(e_2)=(-1,-2,-2)$$ that means $$F(e_1)+F(e_2)=(0,0,1)$$ so $$F(e_3)=(12,3,3)$$ I was not thinking clear, I did confuse this in my brain but now I see!

Regards,
$$|\pi\rangle$$
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K