Problem Breakdown and Solution:

Let $V$ be a finite-dimensional vector space, and let $U$ be a proper subspace of $V$, i.e., $U \neq V$ and $U \neq \{0\}$. A basis for $V$ is given as $\{u, v_1, \ldots, v_k\}$, where $u \notin U$ and $\{v_1, \ldots, v_k\}$ is a basis for $U$.

We define a linear mapping $T: V \to V$ by $T(a u + b_1 v_1 + \cdots + b_k v_k) = b_1 v_1 + \cdots + b_k v_k,$ where $a, b_1, \ldots, b_k \in \mathbb{F}$, the field over which $V$ is defined.

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Step 1: Understand $T$'s Action

• $T$ maps $u$ to $0$, since the $u$-coefficient does not appear in $T$'s definition.
• $T$ acts as the identity map on the subspace spanned by $\{v_1, \ldots, v_k\}$, which is $U$.

Thus, $T(u) = 0$ and $T(v_i) = v_i$ for all $i = 1, \ldots, k$.

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Step 2: Check if $U$ is $T$-invariant

A subspace $U$ is $T$-invariant if $T(x) \in U$ for all $x \in U$.

• Let $x \in U$. Since $U = \text{span}\{v_1, \ldots, v_k\}$, we can write $x = c_1 v_1 + \cdots + c_k v_k$, where $c_i \in \mathbb{F}$.
• Applying $T$ to $x$, we have: $T(x) = T(c_1 v_1 + \cdots + c_k v_k) = c_1 v_1 + \cdots + c_k v_k.$ This result lies in $U$, since $T(x) \in \text{span}\{v_1, \ldots, v_k\} = U$.

Hence, $U$ is $T$-invariant.

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Conclusion:

The subspace $U$ is invariant under the mapping $T$.

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Would you like a deeper explanation or examples for invariance, linear maps, or related concepts? Let me know!

5 Related Questions:

1. What does it mean for a subspace to be invariant under a linear transformation?
2. How can we find the kernel and image of $T$ in this scenario?
3. What are some properties of projections onto subspaces, and how does $T$ relate to such projections?
4. How would the result change if $T(u) \neq 0$?
5. Can we generalize this idea to infinite-dimensional vector spaces?

Tip:

When checking invariance under a linear map, always confirm that applying the map to any vector in the subspace results in a vector still within that subspace.