We need to prove that if $T \in L(V, W)$ is a linear map between vector spaces $V$ and $W$, where $\dim V > \dim W$, then there exists a non-zero vector $v \in V$ such that $T(v) = 0$, i.e., $v$ lies in the kernel (null space) of $T$.

Key Concepts to Use

1. Linear Maps and the Kernel:

   • A linear map $T: V \to W$ sends vectors from $V$ (domain) to $W$ (codomain).
   • The kernel of $T$, denoted $\ker(T)$, consists of all vectors $v \in V$ such that $T(v) = 0$. Mathematically: $\ker(T) = \{ v \in V \mid T(v) = 0 \}.$
   • $\ker(T)$ is a subspace of $V$.

2. Rank-Nullity Theorem:

   • The rank-nullity theorem states that for a linear map $T: V \to W$: $\dim V = \dim(\ker(T)) + \dim(\operatorname{im}(T)),$ where:
     • $\dim(\ker(T))$ is the nullity (dimension of the kernel).
     • $\dim(\operatorname{im}(T))$ is the rank (dimension of the image of $T$).

3. Dimension Argument:

   • If $\dim V > \dim W$, then the map $T$ cannot be injective (one-to-one) because there are more vectors in $V$ than the dimensions of the space $W$ where they are mapped.
   • This implies that $\ker(T) \neq \{0\}$, meaning there must be a non-zero vector $v$ in $V$ such that $T(v) = 0$.

Proof

1. Setup the Linear Map:

   • Let $T: V \to W$ be a linear map.
   • Assume that $\dim V = n$ and $\dim W = m$, where $n > m$.

2. Using the Rank-Nullity Theorem:

   • By the rank-nullity theorem: $\dim V = \dim(\ker(T)) + \dim(\operatorname{im}(T)).$

   • Here:

     • $\dim(\ker(T))$ is the nullity of $T$,
     • $\dim(\operatorname{im}(T)) \leq \dim W$, since the image of $T$ is a subspace of $W$.

   • Therefore: $\dim(\ker(T)) + \dim(\operatorname{im}(T)) = n.$ Since $\dim(\operatorname{im}(T)) \leq m$, we have: $\dim(\ker(T)) \geq n - m.$

3. Analyzing the Dimensions:

   • Since $n > m$, we know: $n - m > 0.$
   • Thus: $\dim(\ker(T)) \geq n - m > 0.$
   • This implies that $\ker(T)$ is not trivial, meaning it contains vectors other than the zero vector.

4. Conclusion:

   • Since $\ker(T)$ is a subspace of $V$ with dimension at least 1, there exists a non-zero vector $v \in V$ such that: $T(v) = 0.$
   • This completes the proof.

Summary

We showed that if $\dim V > \dim W$, the kernel of the linear map $T: V \to W$ must be non-trivial (contain non-zero vectors). This follows from the rank-nullity theorem and the fact that $n > m$ forces $\ker(T)$ to have dimension at least 1.

Do you need further clarification on any step? Here are related questions for deeper understanding:

1. What is the formal definition of a linear map and its properties?
2. Can you prove the rank-nullity theorem in detail?
3. How does injectivity relate to the kernel of a linear map?
4. What happens when $\dim V = \dim W$ for a linear map $T$?
5. Can you give an example of a linear map where $\ker(T) \neq \{0\}$?

Tip:

When dealing with linear maps, always think of the rank-nullity theorem as a bridge between the dimensions of the domain, kernel, and image.