Differential equations. linear system.

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SUMMARY

The discussion focuses on the properties of solutions to the differential equation x' = G(t)x, where G(t) is an n x n matrix dependent on t. It establishes that the solutions form an n-dimensional subspace of C1(R+, Rn). The proof involves demonstrating closure under addition and scalar multiplication, as well as showing that the solutions x1(t), x2(t), ..., xn(t) are linearly independent. The only solution to the differential equation with initial condition x(0) = (0, 0, 0, ..., 0)T is the zero function, confirming the dimensionality of the solution space.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly n x n matrices.
  • Familiarity with differential equations and their solutions.
  • Knowledge of vector spaces and subspaces in Rn.
  • Proficiency in calculus, specifically in the context of continuous functions C1(R+, Rn).
NEXT STEPS
  • Study the properties of linear transformations represented by matrices.
  • Learn about the theory of linear independence in vector spaces.
  • Explore the existence and uniqueness theorems for solutions of differential equations.
  • Investigate the implications of the dimension of solution spaces in differential equations.
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, linear algebra, and vector spaces. This discussion is beneficial for anyone looking to deepen their understanding of the structure of solutions in linear systems.

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Homework Statement



G(t) is nxn matrix depends on t.
Show that solutions of x'=G(t)x form an n-dim subspace of C1(R+,Rn).


The Attempt at a Solution



So I can show closure, addition of solutions returns some combo inside R^n, and same with scalar multiplication. I need to show dimension..
 
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Since Gn is an n by n matrix, x must be a column matrix with n rows. Let x1(t) be the solution with x(0)1= (1, 0, 0, ..., 0)T. Let x2(t) be the solution with x2(0)= (0, 1, 0, ..., 0)T. Let x3(t) be the solution with x3(0)= (0, 0, 1, ..., 0)T. Continue until you have xn(t) defined as the solution with xn(t)= (0, 0, 0, ..., 1)T. Show that they are independent, by showing that the only solution to the differential equation with x(0)= (0, 0, 0, ..., 0)T is the 0 function, and that the solution with x(t)= (a1, a2, a3, ..., an) is equal to a1x1(t)+ a2x2(t)+ a3x3(t)+ ...+ anxn(t) by showing that they both satisfy the differential equation and the same initial condition.
 

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