Linear Algebra Quiz Help: Find Canonical Basis & Solution Space Dimension

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SUMMARY

The discussion focuses on finding the canonical basis and dimension of the solution space for a homogeneous system of equations represented by the augmented matrix: [3 1 1 1 0] and [5 -1 1 -1 0]. The matrix is reduced to its reduced row echelon form, yielding [1 0 2 0 0] and [0 1 -1 0 0]. The solution space is expressed in terms of two free variables, x3 and x4, leading to the canonical basis B = {(-2, 1, 0, 0), (0, 0, 1, 0)}. The dimension of the solution space is determined to be 2, corresponding to the number of free variables.

PREREQUISITES
  • Understanding of homogeneous systems of linear equations
  • Familiarity with augmented matrices and row operations
  • Knowledge of reduced row echelon form (RREF)
  • Concept of free variables in linear algebra
NEXT STEPS
  • Study the process of converting matrices to reduced row echelon form (RREF)
  • Learn about canonical bases in linear algebra
  • Explore the concept of dimensions in vector spaces
  • Practice solving homogeneous systems of equations with multiple free variables
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Students studying linear algebra, particularly those preparing for quizzes or exams involving systems of equations and vector spaces.

JasonJo
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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?
 
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Yep. Sweep the matrix.
 


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!
 

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