Exercise 1.6.1

Question

Prove that if $A: V \rightarrow{} W$ is an isomorphism (i.e. an invertible linear transformation) and $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ is a basis in $V$, then $A\mathbf{v}_1, A\mathbf{v}_2, ..., A\mathbf{v}_n$ is a basis in $W$.

Jing Huang · Accepted Answer

\begin{proof}
For any $w\in W$, $A^{-1}w = v \in V$, 
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\begin{equation*}
\begin{split}
    w = Av &= A[\mathbf{v}_1 \; \mathbf{v}_2 \; ...\; \mathbf{v}_n][v_1 \;  v_2 \; ... \; v_n]^{\mathsf{T}} \
    &= [A\mathbf{v}_1 \; A\mathbf{v}_2 \; ...\; A\mathbf{v}_n][v_1 \;  v_2 \; ... \; v_n]^{\mathsf{T}}.
\end{split}
\end{equation*}
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So we can see $[A\mathbf{v}_1 \; A\mathbf{v}_2 \; ... \; A\mathbf{v}_n]$ is in the form of a basis in $W$. Next we show that $A\mathbf{v}_1 \; A\mathbf{v}_2 \; ...\; A\mathbf{v}_n$ is linearly independent. If not, suppose $A\mathbf{v}_1$ can be expressed as a linear combination of $ A\mathbf{v}_2 \; A\mathbf{v}_3 \; ... \; A\mathbf{v}_n$ without loss of generality. Multiplying them with $A$ in the left side, it results in that $\mathbf{v}_1$ can be expressed by $ \mathbf{v}_2 \; ... \; \mathbf{v}_n$, which contradicts the fact that $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ is a basis in $V$. So the proposition is proved.
\end{proof}

Exercise 1.6.1

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Exercise 1.6.1

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