How to easily find coefficents of a set of matricies to find if the set spans a linear space?

38 Views Asked by At

I apologise for my very badly worded question. But what im trying to say is that how would you easily find that the coefficents are (-b12 + b21) and (-b11+ b22) e.c.t as show in the image that ive attached? I understand what goign on but is there a quick method to determine the coefficents without trial and error?

enter image description here

1

There are 1 best solutions below

1
On BEST ANSWER

Two things to see when you are dealing with problems like whether a given set of vectors is a basis or not:

  1. The vectors should be linearly independent, i.e. let $A_{1},A_{2},A_{3},A_{4}$ represent the vectors given, $a_1 A_1 + a_2 A_2 +a_3 A_3 +a_4 A_4 = 0 \implies a_1 = a_2 = a_3 = a_4 = 0$
  2. Given any $A \in \mathbb{R}^{2x2}$ (the given space), $A$ can be written uniquely in terms of the given vectors.

Part 1 can be easily shown and has been shown in your image. The only set of values of $c_1, c_2, c_3, c_4$ that will satisfy $c_1 A_1 + c_2 A_2 + c_3 A_3 + c_4 A_4 = 0_{2x2}$ is $c_1=c_2=c_3=c_4 = 0$.

For part 2 let $A = [b_{11}, b_{12} ; b_{21}, b_{22}]$ can be written as a linear combination of the given vectors $A_{1},A_{2},A_{3},A_{4}$.

$$A = c_1 A_1 + c_2 A_2 +c_3 A_3 +c_4 A_4$$ Now you have 4 equations and 4 unknowns $c_1, c_2, c_3 ,c_4$. Solve!

Cheers!