Linearly independence of vector function

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
HAMJOOP
Messages
31
Reaction score
0
Given two vectors
x(t) = (e^t te^t)^T

y(t) = (1 t)^T


a) Show that x and y are linearly dependent at each point in the interval [0, 1]

b) Show that x and y are linearly independent on [0, 1]


I compute det([x y]) = 0, so they are linearly dependent
how about part b. Isn't a) and b) are contradictory


The above problem comes from Elementary Differential Equations and Boundary Value Problems 9th ed.




Another question
given two vectors depends on t, v and w each has two components

det([v w]) = 0 at some points only
Can I say v and w are linearly dependent at those points ??
 
Physics news on Phys.org
HAMJOOP said:
Given two vectors
x(t) = (e^t te^t)^T

y(t) = (1 t)^T


a) Show that x and y are linearly dependent at each point in the interval [0, 1]

b) Show that x and y are linearly independent on [0, 1]


I compute det([x y]) = 0, so they are linearly dependent
how about part b. Isn't a) and b) are contradictory
No, there is no contradiction.

In part (a), you are fixing a value of ##t##, call it ##t = t_0##, so the elements of the vectors are simply numbers. The linear dependence means that there exist coefficients ##a## and ##b## such that ##a x(t_0) + b y(t_0) = 0##. But the coefficients will vary with ##t_0##.

Part (b) is asking you to show that there are no coefficients ##a## and ##b## for which ##ax(t) + by(t) = 0## is true simultaneously for all ##t \in [0,1]##.