To show that the set

$\text{ker}(T) = \{\vec{v} \in V : T(\vec{v}) = 0\}$

is a subspace of $V$, we need to verify that it satisfies the conditions required to be a subspace. These conditions are:

1. The zero vector of $V$ is in $\text{ker}(T)$.
2. The set is closed under vector addition.
3. The set is closed under scalar multiplication.

Step 1: Check if the zero vector is in $\text{ker}(T)$

By the definition of a linear transformation, we have $T(\vec{0}) = 0$, where $\vec{0}$ is the zero vector in $V$ and $0$ is the zero vector in $W$. Therefore, $\vec{0} \in \text{ker}(T)$.

Step 2: Check if $\text{ker}(T)$ is closed under addition

Let $\vec{v_1}, \vec{v_2} \in \text{ker}(T)$. By definition, $T(\vec{v_1}) = 0$ and $T(\vec{v_2}) = 0$. Now, consider the sum $\vec{v_1} + \vec{v_2}$:

$T(\vec{v_1} + \vec{v_2}) = T(\vec{v_1}) + T(\vec{v_2}) = 0 + 0 = 0.$

Since $T(\vec{v_1} + \vec{v_2}) = 0$, it follows that $\vec{v_1} + \vec{v_2} \in \text{ker}(T)$. Hence, $\text{ker}(T)$ is closed under addition.

Step 3: Check if $\text{ker}(T)$ is closed under scalar multiplication

Let $\vec{v} \in \text{ker}(T)$ and let $c$ be any scalar. By definition, $T(\vec{v}) = 0$. Now, consider the scalar multiple $c\vec{v}$:

$T(c\vec{v}) = cT(\vec{v}) = c \cdot 0 = 0.$

Since $T(c\vec{v}) = 0$, it follows that $c\vec{v} \in \text{ker}(T)$. Hence, $\text{ker}(T)$ is closed under scalar multiplication.

Conclusion

Since $\text{ker}(T)$ contains the zero vector, is closed under addition, and is closed under scalar multiplication, it satisfies the conditions to be a subspace of $V$.

Would you like further details or clarifications?

Here are some related questions to expand on this topic:

1. What are some examples of kernels in different vector spaces?
2. How does the dimension of $\text{ker}(T)$ relate to the rank-nullity theorem?
3. Can the kernel of a non-linear transformation form a subspace?
4. How does the kernel of a matrix transformation relate to the null space of the matrix?
5. What is the significance of the kernel in determining whether a transformation is injective?

Tip: The kernel of a linear transformation is crucial in understanding the structure of the transformation, particularly its injectivity.