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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 \(V\).

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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.

Let T: V → W be a linear transformation from a vector space V to a vector space W. Show that the set ker(T) = {v ∈ V: T(v) = 0} is a subspace of V.

This problem involves proving that the kernel of a linear transformation, ker(T) = {v ∈ V: T(v) = 0}, is a subspace of V. The solution checks the conditions for subspaces, including the zero vector, closure under addition, and scalar multiplication. Suitable for linear algebra students.

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