Can You Find a Matrix Where Column Space Equals Null Space?

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A 4 x 4 matrix can have its column space equal to its null space if both dimensions are 2, as indicated by the relationship dim(col A) + dim(null A) = 4. For this to hold, the matrix must satisfy the condition A² = 0, meaning it is nilpotent. This implies that every vector x in R4 will result in Ax being in both col A and null A. Therefore, a matrix that meets these criteria will have a column space that is a subset of its null space. Finding such a matrix involves ensuring it is nilpotent with the appropriate dimensions.
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Hey fellas could use a hand on a linear question that has been bugging me.

Give an example of a 4 x 4 matrix such that col A = nul A
Appreciate any help!
Thanks,
ib
 
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To start with, remember that dim(col A) + dim(null A) = 4, so we better have dim(col A) = dim(null A) = 2.
 
If col A = null A, then given any vector x in R4, Ax ∈ col A = null A, so A2x = 0. Since x is arbitrary, A2 = 0. Conversely, if A2 = 0, you can show that col A ⊆ null A, and you need dim(col A) = 2. This might help a bit in finding such a matrix.

(Matrices A such that Ak = 0 for some integer k are called nilpotent.)
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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