1. **Problem Statement:** We have a linear map $T : V \to W$ between $F$-vector spaces with $\dim V = \dim W = n$. Given a basis $\{\vec{b_1}, \ldots, \vec{b_n}\}$ of $V$, the student tries to prove that $\{T(\vec{b_1}), \ldots, T(\vec{b_n})\}$ is a basis of $W$ using four lines of reasoning. 2. **Recall:** A set of vectors is a basis if it is linearly independent and spans the space. Since $\dim W = n$, any linearly independent set of $n$ vectors in $W$ is automatically a basis. 3. **Line 1 Analysis:** - The equation $\sum_{i=1}^n \alpha_i T(\vec{b_i}) = \vec{0}$ means the zero vector in $W$ because $T(\vec{b_i}) \in W$. - By linearity, $T(\sum_{i=1}^n \alpha_i \vec{b_i}) = \vec{0}$ in $W$. - So, (A) is **true**. 4. **Line 2 Analysis:** - It states if $\sum_{i=1}^n \alpha_i \vec{b_i} = \vec{0}$ in $V$, then all $\alpha_i = 0$. - This is the definition of linear independence of $\{\vec{b_i}\}$, the basis of $V$. - It does **not** use linear independence of $T(\vec{b_i})$. - So, (B) is **false**. 5. **Line 3 Analysis:** - It claims any linearly independent sequence in $W$ of length $n$ is a basis. - This uses the fact that $\dim W = n$. - Since $\dim V = \dim W = n$, this dimension fact is crucial. - So, (C) is **true**. 6. **Line 4 Analysis:** - Lines 1 and 2 together imply that if $\sum_{i=1}^n \alpha_i T(\vec{b_i}) = \vec{0}$, then $\alpha_i = 0$ for all $i$. - This shows $\{T(\vec{b_i})\}$ is linearly independent. - So, (D) is **true**. **Final answers:** - (A) True - (B) False - (C) True - (D) True