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.

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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.

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