Finding canonical basis and dimension for a homogeneous system

[JasonJo]

I just got my quiz back and i bombed it, and my teacher neglected going through it, hoping you guys could help me out with it

1) Find the canonical basis for the solution space of the homogeneous system, and state the dimension of the space.

3x1 + x2 + x3 + x4 = 0
5x1 - x2 + x3 - x4 = 0

how do i even begin this problem? I believe I set up the augmented matrix and reduce it to reduced row echelon form or no?

 

[Galileo]

Yep. Sweep the matrix.

 

[thepatientmental]

First of all, don't worry if you didn't do well on the quiz. It's important to learn from mistakes and seek help when needed. Let's go through the problem step by step.

1) To find the canonical basis for the solution space, we need to solve the system of equations. To do this, we can set up the augmented matrix:

[3 1 1 1 0]
[5 -1 1 -1 0]

Note that the last column is all zeros since the system is homogeneous. Then, we can reduce the matrix to its reduced row echelon form using row operations. This will give us:

[1 0 2 0 0]
[0 1 -1 0 0]

2) Now, we can express the solution space in terms of the free variables. In this case, we have two free variables, x3 and x4. So, our solution space can be represented as:

x1 = -2x3
x2 = x3
x4 = x4

3) To find the canonical basis, we need to express the solution space in terms of vectors. We can choose two vectors, one for each free variable, and use them to represent the solution space. So, our canonical basis is:

B = {(-2, 1, 0, 0), (0, 0, 1, 0)}

4) Finally, the dimension of the solution space is equal to the number of free variables, which in this case is 2. So, the dimension of the solution space is 2.

I hope this helps you understand the problem better. Remember to practice more problems and seek help when needed. Good luck on your future quizzes and exams!