Great question, Chris! Let’s break this down rigorously.

Axioms for a Vector Subspace

Your textbook is correct: to verify that a subset $W$ of a vector space $V$ is a subspace, you only need to check four properties (though some sources combine or derive some from others):

1. Closure under addition: If $\mathbf{u}, \mathbf{v} \in W$, then $\mathbf{u} + \mathbf{v} \in W$.
2. Closure under scalar multiplication: If $\mathbf{u} \in W$ and $c$ is a scalar, then $c\mathbf{u} \in W$.
3. Existence of a zero vector: $\mathbf{0} \in W$.
4. Existence of additive inverses: For every $\mathbf{u} \in W$, there exists $-\mathbf{u} \in W$ such that $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$.

Can We Prove the Existence of Additive Inverses from the Zero Vector?

No, not directly. The existence of a zero vector alone does not guarantee the existence of additive inverses for every element in $W$.

However, if you already know that $W$ is closed under scalar multiplication and that $\mathbf{0} \in W$, you can prove the existence of additive inverses as follows:

• Let $\mathbf{u} \in W$.
• By closure under scalar multiplication, $(-1)\mathbf{u} \in W$.
• But $(-1)\mathbf{u} = -\mathbf{u}$, which is the additive inverse of $\mathbf{u}$.

So, if you have closure under scalar multiplication and the zero vector, you can derive the existence of additive inverses. This is why some sources only list three axioms (closure under addition, closure under scalar multiplication, and the zero vector), because the fourth (additive inverses) is implied by the other two.

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Why Do Some Sources List Four?

Some textbooks explicitly list all four for clarity, especially for beginners, to emphasize that a subspace must satisfy all the vector space axioms inherited from the parent space. However, in practice, you only need to check the three (closure under addition, closure under scalar multiplication, and the zero vector), because the fourth follows from the others.

In short:

• Three axioms are sufficient for a rigorous proof.
• Four axioms are sometimes listed for pedagogical reasons.

Would you like to see a formal proof of how additive inverses follow from the other three? Or is there another aspect you’d like to explore?

¡De nada, Chris!

Yes, the properties you mentioned for matrix transformations are deeply connected to the subspace axioms!

Connection Between Subspace Axioms and Matrix Transformations

The two properties of a matrix transformation $T: \mathbb{R}^n \to \mathbb{R}^m$:

1. Additivity (or "additive property"): $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. Homogeneity (or "scalar multiplication property"): $T(k\mathbf{u}) = kT(\mathbf{u})$

are exactly the conditions that define a linear transformation.

How This Relates to Subspaces

• The image (or range) of a linear transformation $T$, denoted $\text{Im}(T)$, is always a subspace of $\mathbb{R}^m$.
• The kernel (or null space) of $T$, denoted $\ker(T)$, is always a subspace of $\mathbb{R}^n$.

Why? Because these properties ensure that the image and kernel are closed under addition and scalar multiplication, and they always contain the zero vector (since $T(\mathbf{0}) = \mathbf{0}$ for any linear transformation).

Key Insight

• The additivity and homogeneity of $T$ directly imply that $\text{Im}(T)$ and $\ker(T)$ satisfy the subspace axioms:
  • Closure under addition: $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$, so the sum of any two vectors in the image/kernel is also in the image/kernel.
  • Closure under scalar multiplication: $T(k\mathbf{u}) = kT(\mathbf{u})$, so scalar multiples of vectors in the image/kernel are also in the image/kernel.
  • Zero vector: $T(\mathbf{0}) = \mathbf{0}$, so the zero vector is always in both the image and kernel.

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In Short

The properties of matrix transformations (additivity and homogeneity) are what make their images and kernels subspaces. This is why linear algebra often feels like a dance between transformations and subspaces!

Would you like to explore an example, or is there another connection you’re curious about?

Yes, matrix transformations are a fundamental example of a functional vector space—specifically, the space of linear transformations (or linear operators) between vector spaces.

How Matrix Transformations Fit into Functional Vector Spaces

1. Vector Space of Linear Transformations

• Let $\mathcal{L}(V, W)$ be the set of all linear transformations from a vector space $V$ to a vector space $W$.
• Addition: You can add two linear transformations $T$ and $S$ by defining $(T + S)(\mathbf{v}) = T(\mathbf{v}) + S(\mathbf{v})$.
• Scalar multiplication: You can multiply a linear transformation $T$ by a scalar $c$ by defining $(cT)(\mathbf{v}) = cT(\mathbf{v})$.
• Zero transformation: The transformation that sends every vector to $\mathbf{0}$ acts as the zero vector in this space.
• Additive inverses: For any linear transformation $T$, the transformation $-T$ defined by $(-T)(\mathbf{v}) = -T(\mathbf{v})$ is its additive inverse.

With these operations, $\mathcal{L}(V, W)$ satisfies all the axioms of a vector space.

2. Matrix Representation

• When $V = \mathbb{R}^n$ and $W = \mathbb{R}^m$, every linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ can be represented by an $m \times n$ matrix $A$, where $T(\mathbf{v}) = A\mathbf{v}$.
• The set of all $m \times n$ matrices, with standard matrix addition and scalar multiplication, forms a vector space isomorphic to $\mathcal{L}(\mathbb{R}^n, \mathbb{R}^m)$.

3. Why This Matters

• This means the space of all $m \times n$ matrices is a concrete example of a functional vector space, where the "vectors" are functions (in this case, linear transformations).
• The axioms of a vector space (closure under addition, scalar multiplication, etc.) are satisfied by these operations on matrices/transformations.

Summary Table

Key Takeaway

Matrix transformations are not just mappings—they are vectors in a functional vector space. This perspective is central to advanced linear algebra and functional analysis.

Would you like to see an explicit example, or does this clarify the connection?

You’re so welcome, Chris! That “brain lobes tingling” feeling is the best—it means the pieces are clicking into place, and there’s nothing quite like it. You’re absolutely nailing these connections, and your curiosity is what’s driving you forward.

If you ever want to dive deeper, explore examples, or tackle another “aha!” moment, I’m here for it. What’s next on your mathematical adventure? More linear algebra, or is there another topic you’re itching to explore? 🚀