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Lesson 0: Technology that is Acceptable in a PLEM Degree

Show f(t) = t has laplace F(s) = 1/s^2
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Show f(t) = t has laplace F(s) = 1/s^2

Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof) Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof) Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof)
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Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof)

Title: Construct a Matrix with Given Column Space and Nullspace (Step-by-Step Proof)

In this walkthrough, we show how to construct a matrix whose column space contains (1,1,1) and whose nullspace is the line of multiples of (1,1,1,1).

Step 1: Understand the Problem

We want:

  • Column space to contain the vector (1,1,1)

  • Nullspace to be the set of all scalar multiples of (1,1,1,1)

This means we are looking for a matrix A such that:

  • A * x = 0 has solution space spanned by (1,1,1,1)

  • The matrix A has 3 rows (since the column vector has 3 entries)

  • The matrix A has 4 columns (since the nullspace vector has 4 components)

Step 2: Definitions

Column Space: All linear combinations of the columns
Nullspace: All solutions to A * x = 0
So x = t(1,1,1,1) is the general solution of the homogeneous system.

Step 3: Deconstruct and Solve

We assume:

x = (1,1,1,1)t
So, x₁ = x₂ = x₃ = x₄

This gives:

  • x₁ - x₂ = 0

  • x₂ - x₃ = 0

  • x₃ - x₄ = 0

Which gives us a system of equations:

x₁ - x₂ = 0  
x₂ - x₃ = 0  
x₃ - x₄ = 0

Written in matrix form:

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

Thus, a correct matrix A is:

A = [ 1  0  0  -1  
      0  1  0  -1  
      0  0  1  -1 ]

This has:

  • Column space containing (1,1,1)

  • Nullspace spanned by (1,1,1,1)

Step 4: Validation

Any vector in the nullspace must satisfy A * x = 0.
Let x = (1,1,1,1), then:

A * x = [1 -1 0 0]·x = 1 - 1 = 0  
        [0 1 -1 0]·x = 1 - 1 = 0  
        [0 0 1 -1]·x = 1 - 1 = 0  

✅ Valid

And since (1,1,1) can be expressed as a linear combination of the columns, it's in the column space.

Conclusion

One valid matrix is:

A = [ 1  0  0  -1  
      0  1  0  -1  
      0  0  1  -1 ]

This matches the conditions exactly. There are infinitely many matrices that can satisfy these conditions, but this form is simplest and easily validated.

This lesson aligns with Chapter 3.2 in Strang’s Linear Algebra (4th ed.). Always cite sources and avoid shortcut tools like AI or uncredited platforms during exams unless explicitly allowed.

Visit:
https://author-jonathan-david-shop.fourthwall.com for more STEM guides.
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