[proofplan] We prove both directions. For the forward direction, we show that any isomorphism maps a basis to a basis, preserving dimension. For the reverse direction, given bases of equal cardinality, we construct an explicit isomorphism by mapping basis vectors bijectively and extending linearly, then verify it is well-defined, linear, and bijective. [/proofplan]

[step:Show that an isomorphism preserves dimension]Assume $T: V \to W$ is an isomorphism, i.e., a bijective linear map. Let $\mathcal{B} = \{b_1, \ldots, b_n\}$ be a basis of $V$, so that $\dim V = n$. [claim:The image of a basis under an isomorphism is a basis] Let $T: V \to W$ be an isomorphism and let $\mathcal{B} = \{b_1, \ldots, b_n\}$ be a basis of $V$. Then $T(\mathcal{B}) = \{T(b_1), \ldots, T(b_n)\}$ is a basis of $W$. [/claim] [proof] **Linear independence.** Suppose $\sum_{i=1}^n \alpha_i T(b_i) = 0_W$ for some $\alpha_1, \ldots, \alpha_n \in F$. By linearity of $T$: \begin{align*} T\!\left(\sum_{i=1}^n \alpha_i b_i\right) = \sum_{i=1}^n \alpha_i T(b_i) = 0_W. \end{align*} Since $T$ is injective, $\ker T = \{0_V\}$, so $\sum_{i=1}^n \alpha_i b_i = 0_V$. Since $\mathcal{B}$ is linearly independent, $\alpha_i = 0$ for all $i \in \{1, \ldots, n\}$. Hence $\{T(b_1), \ldots, T(b_n)\}$ is linearly independent. **Spanning.** Let $w \in W$. Since $T$ is surjective, there exists $v \in V$ with $T(v) = w$. Since $\mathcal{B}$ spans $V$, write $v = \sum_{i=1}^n \alpha_i b_i$ for some $\alpha_i \in F$. Then \begin{align*} w = T(v) = T\!\left(\sum_{i=1}^n \alpha_i b_i\right) = \sum_{i=1}^n \alpha_i T(b_i). \end{align*} So $w \in \operatorname{span}\{T(b_1), \ldots, T(b_n)\}$. Since $w$ was arbitrary, $T(\mathcal{B})$ spans $W$. [/proof] Since $T(\mathcal{B})$ is a basis of $W$ with $n$ elements, $\dim W = n = \dim V$.[/step]

[guided]Assume $T: V \to W$ is an isomorphism. We want to show $\dim V = \dim W$. The strategy is to take a basis $\mathcal{B} = \{b_1, \ldots, b_n\}$ of $V$ and show that its image $T(\mathcal{B}) = \{T(b_1), \ldots, T(b_n)\}$ is a basis of $W$. Why should an isomorphism map bases to bases? Because an isomorphism is bijective and linear, it preserves exactly the two properties that define a basis: linear independence (preserved by injectivity) and spanning (preserved by surjectivity). **Linear independence of $T(\mathcal{B})$.** Suppose $\sum_{i=1}^n \alpha_i T(b_i) = 0_W$. By linearity, $T\!\left(\sum_{i=1}^n \alpha_i b_i\right) = 0_W$. Since $T$ is injective, $\ker T = \{0_V\}$, so $\sum_{i=1}^n \alpha_i b_i = 0_V$. Linear independence of $\mathcal{B}$ in $V$ forces $\alpha_i = 0$ for all $i$. The key point: injectivity of $T$ "transfers" the linear independence from $V$ to $W$. **Spanning.** Let $w \in W$ be arbitrary. Surjectivity of $T$ provides $v \in V$ with $T(v) = w$. Since $\mathcal{B}$ spans $V$, we write $v = \sum_{i=1}^n \alpha_i b_i$, and linearity gives $w = \sum_{i=1}^n \alpha_i T(b_i)$. So every element of $W$ is a linear combination of $T(\mathcal{B})$. Therefore $T(\mathcal{B})$ is a basis of $W$ with exactly $n$ elements, giving $\dim W = n = \dim V$.[/guided]

[step:Construct an isomorphism from two bases of equal cardinality]Assume $\dim V = \dim W = n$. Let $\mathcal{B}_V = \{b_1, \ldots, b_n\}$ be a basis of $V$ and $\mathcal{B}_W = \{c_1, \ldots, c_n\}$ be a basis of $W$. Define the map \begin{align*} T: V &\to W \\ v = \sum_{i=1}^n \alpha_i b_i &\mapsto \sum_{i=1}^n \alpha_i c_i. \end{align*} This map is well-defined because, by the [Basis Criterion](/theorems/3308), the representation $v = \sum_{i=1}^n \alpha_i b_i$ is unique, so the coefficients $\alpha_1, \ldots, \alpha_n$ are uniquely determined by $v$.[/step]

[guided]Assume $\dim V = \dim W = n$. Choose a basis $\mathcal{B}_V = \{b_1, \ldots, b_n\}$ of $V$ and a basis $\mathcal{B}_W = \{c_1, \ldots, c_n\}$ of $W$. The idea is natural: since both bases have $n$ elements, we "match them up" and extend linearly. Define \begin{align*} T: V &\to W \\ v = \sum_{i=1}^n \alpha_i b_i &\mapsto \sum_{i=1}^n \alpha_i c_i. \end{align*} Is this well-defined? The concern is that $v$ might have multiple representations as a linear combination of $\mathcal{B}_V$, yielding different images under $T$. But by the [Basis Criterion](/theorems/3308), every $v \in V$ has a unique representation in the basis $\mathcal{B}_V$, so the coefficients $\alpha_1, \ldots, \alpha_n$ are uniquely determined by $v$. Thus $T$ is well-defined.[/guided]

[step:Verify that $T$ is linear] Let $v = \sum_{i=1}^n \alpha_i b_i$ and $u = \sum_{i=1}^n \beta_i b_i$ be vectors in $V$, and let $\lambda \in F$. Then $v + u = \sum_{i=1}^n (\alpha_i + \beta_i) b_i$ and $\lambda v = \sum_{i=1}^n (\lambda \alpha_i) b_i$. Applying the definition of $T$: \begin{align*} T(v + u) &= \sum_{i=1}^n (\alpha_i + \beta_i)\, c_i = \sum_{i=1}^n \alpha_i c_i + \sum_{i=1}^n \beta_i c_i = T(v) + T(u), \\ T(\lambda v) &= \sum_{i=1}^n (\lambda \alpha_i)\, c_i = \lambda \sum_{i=1}^n \alpha_i c_i = \lambda\, T(v). \end{align*} Therefore $T$ is linear. [/step]

[step:Verify that $T$ is bijective]**Injectivity.** Suppose $T(v) = 0_W$ for some $v = \sum_{i=1}^n \alpha_i b_i \in V$. Then $\sum_{i=1}^n \alpha_i c_i = 0_W$. Since $\mathcal{B}_W = \{c_1, \ldots, c_n\}$ is linearly independent, $\alpha_i = 0$ for all $i \in \{1, \ldots, n\}$. Hence $v = \sum_{i=1}^n 0 \cdot b_i = 0_V$, so $\ker T = \{0_V\}$ and $T$ is injective. **Surjectivity.** Let $w \in W$. Since $\mathcal{B}_W$ spans $W$, write $w = \sum_{i=1}^n \gamma_i c_i$ for some $\gamma_i \in F$. Define $v = \sum_{i=1}^n \gamma_i b_i \in V$. Then $T(v) = \sum_{i=1}^n \gamma_i c_i = w$, so $T$ is surjective. Since $T$ is a bijective linear map, $T$ is an isomorphism. Therefore $V \cong W$, completing the proof.[/step]

[guided]**Injectivity.** Suppose $T(v) = 0_W$ for some $v = \sum_{i=1}^n \alpha_i b_i$. By definition of $T$, this means $\sum_{i=1}^n \alpha_i c_i = 0_W$. Since $\{c_1, \ldots, c_n\}$ is a basis of $W$, it is linearly independent, so the only way a linear combination of its elements equals $0_W$ is if all coefficients vanish: $\alpha_i = 0$ for all $i$. This means $v = 0_V$, so $\ker T = \{0_V\}$ and $T$ is injective. Why does this work? The map $T$ "translates" the coordinate representation of $v$ in $\mathcal{B}_V$ into the same coordinates in $\mathcal{B}_W$. If those coordinates produce zero in $W$, they must all be zero (by independence of $\mathcal{B}_W$), and hence $v$ itself was zero. **Surjectivity.** Let $w \in W$ be arbitrary. Since $\mathcal{B}_W$ spans $W$, there exist $\gamma_1, \ldots, \gamma_n \in F$ with $w = \sum_{i=1}^n \gamma_i c_i$. Define $v = \sum_{i=1}^n \gamma_i b_i \in V$. Then by definition of $T$: \begin{align*} T(v) = T\!\left(\sum_{i=1}^n \gamma_i b_i\right) = \sum_{i=1}^n \gamma_i c_i = w. \end{align*} So every $w \in W$ is in the image of $T$, confirming surjectivity. Since $T: V \to W$ is linear, injective, and surjective, it is a bijective linear map -- an isomorphism. Therefore $V \cong W$, completing the reverse direction and the proof.[/guided]