> [!NOTE] Lemma > Let $V,W$ be [[Vector spaces|vector spaces]] over a [[Field (Algebra)|field]] $\mathbb{F}.$ Let $\varphi:V\to W$ be a [[Linear Isomorphism|linear isomorphism]]. Let $B\subset V$ be a [[Basis of Vector Space|basis]] of $V.$ Then the [[Image of a set under a function|image]] of $B$ under $\varphi$ is a basis of $W.$ **Proof**: We first show that $\varphi(B)$ spans $W$. Let $w\in W$. Consider $v = \varphi^{-1}(w)\in V$. Since $B$ is a basis of $V$, there are elements $v_{1},\dots,v_{s}\in B$ and scalars $\lambda_{1},\dots\lambda_{s}\in\mathbb{F}$ such that $v=\lambda_{1}v_{1}+\dots\lambda_{s}v_{s}.$Applying $\varphi$ to both sides gives $w=\lambda_{1} \varphi(v_{1})+\dots+\lambda_{s} \varphi(v_{s}) $by linearity which does indeed express $w$ as a linear combination of elements of $\varphi(B)$. To show that $\varphi(B)$ is linearly independent, suppose $0_{W}= \lambda_{1} \varphi(v_{1})+\dots+\lambda_{s} \varphi(v_{s})$for elements $\varphi(v_{i})\in \varphi(B)$. Since [[Linear Map is Injective iff Kernel Only Contains Zero]], $\ker\varphi=\{ 0_{V} \}$ so applying $\varphi^{-1}$ to both sides and using its linearity gives: $\begin{align} 0_{V} = \varphi^{-1} (0_{W}) = \lambda_{1} v_{1} +\dots + \lambda_{s} v_{s} \end{align}.$However $v_{1},\dots,v_{s}\in B$ and $B$ is linearly independent, so we conclude that $\lambda_{1}=\dots=\lambda_{s}=0$. $\square$