[proofplan] We fix a basis $\{e_1, \ldots, e_n\}$ for $V$ and exhibit explicit bases for $\operatorname{Sym}^r(V)$ and $\Lambda^r(V)$. For the symmetric power, the basis consists of symmetrised monomials $e_{i_1} \cdots e_{i_r}$ indexed by weakly increasing multi-indices $1 \le i_1 \le \cdots \le i_r \le n$, and we count these via a stars-and-bars argument. For the exterior power, the basis consists of wedge products $e_{i_1} \wedge \cdots \wedge e_{i_r}$ indexed by strictly increasing multi-indices $1 \le i_1 < \cdots < i_r \le n$, and the count is the binomial coefficient $\binom{n}{r}$. In both cases, spanning follows from multilinearity and (anti)symmetry, while linear independence is established via coordinate extraction using multilinear functionals. [/proofplan] [step:Identify spanning sets for $\operatorname{Sym}^r(V)$ and $\Lambda^r(V)$] Let $\{e_1, \ldots, e_n\}$ be a basis for $V$. Denote by $\pi: V^{\otimes r} \to \operatorname{Sym}^r(V)$ the symmetrisation quotient map, so $\pi(v_1 \otimes \cdots \otimes v_r) = v_1 \cdots v_r$ is the symmetric product. Similarly, let $\alpha: V^{\otimes r} \to \Lambda^r(V)$ denote the alternating quotient map, so $\alpha(v_1 \otimes \cdots \otimes v_r) = v_1 \wedge \cdots \wedge v_r$. Every element of $\operatorname{Sym}^r(V)$ is a sum of symmetric products $v_1 \cdots v_r$, and expanding each $v_k = \sum_{i=1}^n a_{k,i}\, e_i$ in the basis and using multilinearity gives a linear combination of products $e_{i_1} \cdots e_{i_r}$ with $1 \le i_1, \ldots, i_r \le n$. Since the symmetric product is commutative ($e_i e_j = e_j e_i$), we may reorder each multi-index to be weakly increasing. Therefore $\operatorname{Sym}^r(V)$ is spanned by the set \begin{align*} \mathcal{B}_S := \{e_{i_1} \cdots e_{i_r} : 1 \le i_1 \le i_2 \le \cdots \le i_r \le n\}. \end{align*} By the same argument, every element of $\Lambda^r(V)$ is a linear combination of wedge products $e_{i_1} \wedge \cdots \wedge e_{i_r}$. Since the wedge product is alternating, swapping two factors introduces a sign, and any repeated factor gives zero. Therefore we may reorder to a strictly increasing multi-index (absorbing signs), and $\Lambda^r(V)$ is spanned by \begin{align*} \mathcal{B}_\Lambda := \{e_{i_1} \wedge \cdots \wedge e_{i_r} : 1 \le i_1 < i_2 < \cdots < i_r \le n\}. \end{align*} When $r > n$, no strictly increasing sequence of length $r$ in $\{1, \ldots, n\}$ exists, so $\mathcal{B}_\Lambda = \varnothing$ and $\Lambda^r(V) = \{0\}$. [guided] Let $\{e_1, \ldots, e_n\}$ be a basis for $V$. We construct spanning sets for both $\operatorname{Sym}^r(V)$ and $\Lambda^r(V)$. **Symmetric power.** Every element of $\operatorname{Sym}^r(V)$ is a sum of symmetric products $v_1 \cdots v_r$ with $v_k \in V$. Expanding each factor in the basis: $v_k = \sum_{i=1}^n a_{k,i}\, e_i$. By multilinearity of the symmetric product (it is a quotient of the tensor product, which is multilinear), we get a linear combination of terms $e_{i_1} \cdots e_{i_r}$ where each $i_k$ ranges over $\{1, \ldots, n\}$. Now, the symmetric product satisfies $e_i e_j = e_j e_i$ for all $i, j$ (this is the defining relation of $\operatorname{Sym}^r(V)$ — it quotients out by the ideal generated by $v \otimes w - w \otimes v$). So any product $e_{i_1} \cdots e_{i_r}$ can be rewritten with indices in weakly increasing order. This means $\operatorname{Sym}^r(V)$ is spanned by the set $\mathcal{B}_S = \{e_{i_1} \cdots e_{i_r} : 1 \le i_1 \le \cdots \le i_r \le n\}$. **Exterior power.** By the same expansion argument, $\Lambda^r(V)$ is spanned by wedge products $e_{i_1} \wedge \cdots \wedge e_{i_r}$. The wedge product is alternating: $e_i \wedge e_j = -e_j \wedge e_i$, and in particular $e_i \wedge e_i = 0$. So if any index repeats, the product vanishes. For products with distinct indices, we can reorder to make the indices strictly increasing, at the cost of a sign factor $\operatorname{sgn}(\sigma)$ where $\sigma$ is the permutation that sorts the indices. Therefore $\Lambda^r(V)$ is spanned by $\mathcal{B}_\Lambda = \{e_{i_1} \wedge \cdots \wedge e_{i_r} : 1 \le i_1 < \cdots < i_r \le n\}$. When $r > n$, there is no way to choose $r$ strictly increasing indices from $\{1, \ldots, n\}$, so $\mathcal{B}_\Lambda = \varnothing$ and $\Lambda^r(V) = \{0\}$. [/guided] [/step] [step:Prove linear independence of $\mathcal{B}_\Lambda$ via alternating multilinear forms] We show $\mathcal{B}_\Lambda$ is linearly independent. For $r > n$ the set is empty and there is nothing to prove, so assume $r \le n$. For each strictly increasing multi-index $I_0 = (i_1^0 < \cdots < i_r^0)$, define the $r$-linear alternating form \begin{align*} \omega_{I_0}: V \times \cdots \times V \to k, \qquad \omega_{I_0}(v_1, \ldots, v_r) = \det\!\left[\varepsilon_{i_p^0}(v_q)\right]_{1 \le p, q \le r}, \end{align*} where $\varepsilon_i: V \to k$ is the $i$-th coordinate functional defined by $\varepsilon_i(e_j) = \delta_{ij}$. The map $\omega_{I_0}$ is alternating and multilinear (the determinant is alternating and multilinear in its columns, and each column $(\varepsilon_{i_1^0}(v_q), \ldots, \varepsilon_{i_r^0}(v_q))$ depends linearly on $v_q$). By the [universal property of the exterior power](/theorems/???), $\omega_{I_0}$ factors through a unique linear functional \begin{align*} \tilde{\omega}_{I_0}: \Lambda^r(V) \to k \end{align*} with $\tilde{\omega}_{I_0}(v_1 \wedge \cdots \wedge v_r) = \omega_{I_0}(v_1, \ldots, v_r)$. Evaluating on a basis element $e_{j_1} \wedge \cdots \wedge e_{j_r}$ with $J = (j_1 < \cdots < j_r)$: \begin{align*} \tilde{\omega}_{I_0}(e_{j_1} \wedge \cdots \wedge e_{j_r}) = \det\!\left[\varepsilon_{i_p^0}(e_{j_q})\right]_{p,q} = \det\!\left[\delta_{i_p^0, j_q}\right]_{p,q}. \end{align*} If $I_0 = J$, this is $\det(I_r) = 1$. If $I_0 \neq J$, then some row of the matrix $[\delta_{i_p^0, j_q}]$ is entirely zero (since $i_p^0 \notin \{j_1, \ldots, j_r\}$ for at least one $p$), giving determinant $0$. Therefore $\tilde{\omega}_{I_0}(e_{j_1} \wedge \cdots \wedge e_{j_r}) = \delta_{I_0, J}$. Suppose $\sum_I c_I\, (e_{i_1} \wedge \cdots \wedge e_{i_r}) = 0$ where the sum ranges over all strictly increasing $r$-element subsets $I$ of $\{1, \ldots, n\}$. Applying $\tilde{\omega}_{I_0}$: \begin{align*} 0 = \tilde{\omega}_{I_0}\!\left(\sum_I c_I\, (e_{i_1} \wedge \cdots \wedge e_{i_r})\right) = \sum_I c_I\, \delta_{I_0, I} = c_{I_0}. \end{align*} Since $I_0$ was arbitrary, all coefficients vanish. Therefore $\mathcal{B}_\Lambda$ is linearly independent. [guided] To establish linear independence, we construct a family of "coordinate extraction" functionals on $\Lambda^r(V)$ — one for each basis element — such that each functional picks out a single coefficient and kills all others. Fix a strictly increasing multi-index $I_0 = (i_1^0 < \cdots < i_r^0)$. We define an alternating $r$-linear form $\omega_{I_0}: V^r \to k$ by \begin{align*} \omega_{I_0}(v_1, \ldots, v_r) = \det\!\left[\varepsilon_{i_p^0}(v_q)\right]_{1 \le p, q \le r}, \end{align*} where $\varepsilon_i \in V^*$ is the $i$-th coordinate functional satisfying $\varepsilon_i(e_j) = \delta_{ij}$. Why is this alternating and multilinear? The matrix $A$ with entries $A_{pq} = \varepsilon_{i_p^0}(v_q)$ has its $q$-th column equal to $(\varepsilon_{i_1^0}(v_q), \ldots, \varepsilon_{i_r^0}(v_q))^\top$, which depends linearly on $v_q$ (because each $\varepsilon_{i_p^0}$ is linear). The determinant is multilinear and alternating in the columns of $A$, and therefore $\omega_{I_0}$ is multilinear and alternating in $(v_1, \ldots, v_r)$. By the [universal property of the exterior power](/theorems/???), there is a unique linear map $\tilde{\omega}_{I_0}: \Lambda^r(V) \to k$ with $\tilde{\omega}_{I_0}(v_1 \wedge \cdots \wedge v_r) = \omega_{I_0}(v_1, \ldots, v_r)$. Now we evaluate $\tilde{\omega}_{I_0}$ on a basis element $e_{j_1} \wedge \cdots \wedge e_{j_r}$ with $J = (j_1 < \cdots < j_r)$. The $r \times r$ matrix has entries $\delta_{i_p^0, j_q}$. If $I_0 = J$, then $i_p^0 = j_p$ for all $p$, so the matrix is the identity $I_r$, which has determinant $1$. If $I_0 \neq J$, then there exists some $p$ with $i_p^0 \notin \{j_1, \ldots, j_r\}$. Row $p$ of the matrix is then entirely zero, forcing the determinant to be $0$. In summary: $\tilde{\omega}_{I_0}(e_{j_1} \wedge \cdots \wedge e_{j_r}) = \delta_{I_0, J}$. Now suppose $\sum_I c_I\, (e_{i_1} \wedge \cdots \wedge e_{i_r}) = 0$. Applying the linear functional $\tilde{\omega}_{I_0}$ to both sides and using $\delta_{I_0, I}$: \begin{align*} 0 = \sum_I c_I\, \delta_{I_0, I} = c_{I_0}. \end{align*} Since $I_0$ was an arbitrary strictly increasing multi-index, every coefficient is zero, establishing linear independence of $\mathcal{B}_\Lambda$. [/guided] [/step] [step:Prove linear independence of $\mathcal{B}_S$ via the polynomial ring isomorphism] We establish linear independence of $\mathcal{B}_S$ in arbitrary characteristic by constructing an explicit isomorphism between $\operatorname{Sym}^r(V)$ and the space of homogeneous polynomials of degree $r$. Let $\{e_1, \ldots, e_n\}$ be the chosen basis for $V$, and let $k[x_1, \ldots, x_n]_r$ denote the $k$-vector space of homogeneous polynomials of degree $r$ in $n$ variables over $k$. Define the $k$-linear map \begin{align*} \Phi: V^{\otimes r} &\to k[x_1, \ldots, x_n]_r \\ e_{i_1} \otimes \cdots \otimes e_{i_r} &\mapsto x_{i_1} \cdots x_{i_r} \end{align*} extended by linearity. Since polynomial multiplication is commutative, $\Phi(v_1 \otimes \cdots \otimes v_r)$ is unchanged under any permutation of the factors $v_1, \ldots, v_r$. Therefore $\Phi$ vanishes on all elements of the form $v_1 \otimes \cdots \otimes v_r - v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(r)}$ for $\sigma \in S_r$, which generate the kernel of the quotient map $\pi: V^{\otimes r} \to \operatorname{Sym}^r(V)$. By the universal property of quotients, $\Phi$ descends to a well-defined $k$-linear map \begin{align*} \bar{\Phi}: \operatorname{Sym}^r(V) &\to k[x_1, \ldots, x_n]_r \\ e_{i_1} \cdots e_{i_r} &\mapsto x_{i_1} \cdots x_{i_r}. \end{align*} **Surjectivity.** Every monomial $x_{i_1} \cdots x_{i_r} \in k[x_1, \ldots, x_n]_r$ is the image of $e_{i_1} \cdots e_{i_r} \in \operatorname{Sym}^r(V)$. Since the monomials span $k[x_1, \ldots, x_n]_r$, the map $\bar{\Phi}$ is surjective. **Dimension count on the target.** The monomials $x_1^{\mu_1} \cdots x_n^{\mu_n}$ with $\mu_k \ge 0$ and $\mu_1 + \cdots + \mu_n = r$ form a basis for $k[x_1, \ldots, x_n]_r$ over $k$. This is a standard fact: distinct monomials are linearly independent because they are distinguished by their exponent vectors (the coefficient of any given monomial in a polynomial is well-defined). The number of such monomials is the number of solutions to $\mu_1 + \cdots + \mu_n = r$ in non-negative integers, which by stars-and-bars equals $\binom{n+r-1}{r}$. Therefore $\dim_k k[x_1, \ldots, x_n]_r = \binom{n+r-1}{r}$. **Injectivity from dimension comparison.** Since $\mathcal{B}_S$ spans $\operatorname{Sym}^r(V)$ (established in Step 1) and $|\mathcal{B}_S| = \binom{n+r-1}{r}$ (the stars-and-bars count), we have $\dim \operatorname{Sym}^r(V) \le \binom{n+r-1}{r} = \dim k[x_1, \ldots, x_n]_r$. Since $\bar{\Phi}$ is a surjection between finite-dimensional spaces with $\dim \operatorname{Sym}^r(V) \le \dim k[x_1, \ldots, x_n]_r$, and surjectivity forces $\dim \operatorname{Sym}^r(V) \ge \dim k[x_1, \ldots, x_n]_r$, we conclude $\dim \operatorname{Sym}^r(V) = \binom{n+r-1}{r}$ and $\bar{\Phi}$ is an isomorphism. In particular, since $\bar{\Phi}$ maps the spanning set $\mathcal{B}_S$ onto the basis of monomials bijectively, $\mathcal{B}_S$ is linearly independent. [guided] The permanent-based coordinate extraction that worked for the exterior power does not generalise cleanly to the symmetric power in arbitrary characteristic: the permanent of the relevant matrix evaluates to $\mu_1! \cdots \mu_n!$ (a product of factorials of the index multiplicities), which may vanish in positive characteristic. Instead, we use a characteristic-free strategy: we construct a natural map from $\operatorname{Sym}^r(V)$ to a space whose dimension we already know, and use it to pin down $\dim \operatorname{Sym}^r(V)$ exactly. Let $k[x_1, \ldots, x_n]_r$ denote the $k$-vector space of homogeneous polynomials of degree $r$ in $n$ indeterminates over $k$. We define a map $\Phi: V^{\otimes r} \to k[x_1, \ldots, x_n]_r$ on pure tensors of basis elements by \begin{align*} \Phi(e_{i_1} \otimes \cdots \otimes e_{i_r}) = x_{i_1} \cdots x_{i_r}, \end{align*} extended by $k$-linearity to all of $V^{\otimes r}$. Why does this descend to the symmetric power? The key point is that polynomial multiplication is commutative: $x_{i_{\sigma(1)}} \cdots x_{i_{\sigma(r)}} = x_{i_1} \cdots x_{i_r}$ for any permutation $\sigma \in S_r$. Therefore $\Phi$ sends every generator of the symmetrising ideal — elements of the form $v_1 \otimes \cdots \otimes v_r - v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(r)}$ — to zero. By the universal property of quotients, $\Phi$ factors through the quotient $\operatorname{Sym}^r(V) = V^{\otimes r} / \langle \text{symmetrising relations} \rangle$, yielding a well-defined $k$-linear map \begin{align*} \bar{\Phi}: \operatorname{Sym}^r(V) &\to k[x_1, \ldots, x_n]_r, \\ e_{i_1} \cdots e_{i_r} &\mapsto x_{i_1} \cdots x_{i_r}. \end{align*} **Surjectivity of $\bar{\Phi}$.** Every monomial $x_{i_1} \cdots x_{i_r}$ in $k[x_1, \ldots, x_n]_r$ lies in the image: it equals $\bar{\Phi}(e_{i_1} \cdots e_{i_r})$. Since the monomials span the space of homogeneous polynomials, $\bar{\Phi}$ is surjective. **Why do we know $\dim k[x_1, \ldots, x_n]_r$?** The distinct monomials $x_1^{\mu_1} \cdots x_n^{\mu_n}$ with $\mu_1 + \cdots + \mu_n = r$ and each $\mu_k \ge 0$ form a basis for $k[x_1, \ldots, x_n]_r$ — they span by definition, and they are linearly independent because distinct monomials have distinct exponent vectors (if $\sum_\mu c_\mu x^\mu = 0$ as a polynomial, then every coefficient $c_\mu$ is zero, since the polynomial ring $k[x_1, \ldots, x_n]$ is defined as the free $k$-module on monomials). The number of solutions to $\mu_1 + \cdots + \mu_n = r$ in non-negative integers is $\binom{n+r-1}{r}$ by stars-and-bars. So $\dim k[x_1, \ldots, x_n]_r = \binom{n+r-1}{r}$. This holds over any field — no characteristic restriction is needed. **The dimension squeeze.** From Step 1, we know $\mathcal{B}_S$ spans $\operatorname{Sym}^r(V)$, so $\dim \operatorname{Sym}^r(V) \le |\mathcal{B}_S| = \binom{n+r-1}{r}$. From surjectivity of $\bar{\Phi}$, we get $\dim \operatorname{Sym}^r(V) \ge \dim k[x_1, \ldots, x_n]_r = \binom{n+r-1}{r}$. Therefore equality holds: $\dim \operatorname{Sym}^r(V) = \binom{n+r-1}{r}$, and $\bar{\Phi}$ is an isomorphism. Since the spanning set $\mathcal{B}_S$ has exactly $\binom{n+r-1}{r}$ elements and spans a space of dimension $\binom{n+r-1}{r}$, it must be linearly independent (a spanning set of size equal to the dimension is a basis). So $\mathcal{B}_S$ is a basis for $\operatorname{Sym}^r(V)$, as required. [/guided] [/step] [step:Count the spanning sets to obtain the dimension formulas] **Exterior power.** The set $\mathcal{B}_\Lambda$ consists of all strictly increasing $r$-element subsequences of $\{1, \ldots, n\}$. The number of such subsequences is the number of $r$-element subsets of an $n$-element set, which is $\binom{n}{r}$. Since $\mathcal{B}_\Lambda$ is a basis: \begin{align*} \dim \Lambda^r(V) = \binom{n}{r}. \end{align*} When $r > n$, $\binom{n}{r} = 0$, confirming $\Lambda^r(V) = \{0\}$. When $r = n$, $\binom{n}{n} = 1$, so $\Lambda^n(V)$ is one-dimensional, spanned by the single element $e_1 \wedge \cdots \wedge e_n$. **Symmetric power.** The set $\mathcal{B}_S$ consists of all weakly increasing sequences $(i_1 \le \cdots \le i_r)$ with entries in $\{1, \ldots, n\}$. Such a sequence is determined by the multiplicity vector $(\mu_1, \ldots, \mu_n)$ with $\mu_k \ge 0$ and $\mu_1 + \cdots + \mu_n = r$. The number of such vectors is the number of ways to distribute $r$ identical objects into $n$ distinct bins, which by the stars-and-bars combinatorial identity equals $\binom{n + r - 1}{r}$. Since $\mathcal{B}_S$ is a basis: \begin{align*} \dim \operatorname{Sym}^r(V) = \binom{n + r - 1}{r}. \end{align*} [guided] We now count the elements in each basis. **Exterior power.** Each element of $\mathcal{B}_\Lambda$ corresponds to a choice of $r$ distinct indices from $\{1, \ldots, n\}$ arranged in increasing order. The number of ways to choose $r$ elements from an $n$-element set (without regard to order, since the increasing ordering is determined) is $\binom{n}{r}$. Therefore $|\mathcal{B}_\Lambda| = \binom{n}{r}$ and $\dim \Lambda^r(V) = \binom{n}{r}$. The special cases: when $r > n$, we cannot choose $r$ distinct elements from $\{1, \ldots, n\}$, so $\binom{n}{r} = 0$ and $\Lambda^r(V) = \{0\}$. When $r = n$, there is exactly one way to choose all $n$ indices, so $\dim \Lambda^n(V) = 1$ and $\Lambda^n(V) = \operatorname{span}\{e_1 \wedge \cdots \wedge e_n\}$. This one-dimensional space is the determinant line of $V$. **Symmetric power.** Each element of $\mathcal{B}_S$ is a weakly increasing sequence $(i_1 \le \cdots \le i_r)$ with $i_k \in \{1, \ldots, n\}$. Such a sequence is equivalent to a multiplicity vector $(\mu_1, \ldots, \mu_n)$ where $\mu_k$ counts the occurrences of index $k$, satisfying $\mu_k \ge 0$ and $\sum_{k=1}^n \mu_k = r$. This is a standard combinatorial problem: distributing $r$ identical objects (the tensor factors) into $n$ distinct bins (the basis indices). By the stars-and-bars counting principle, the number of solutions is \begin{align*} \binom{n + r - 1}{r}. \end{align*} To see why: arrange $r$ stars (objects) and $n - 1$ bars (dividers between bins) in a row. The total number of symbols is $r + (n-1) = n + r - 1$, and we choose which $r$ positions are stars. Therefore $|\mathcal{B}_S| = \binom{n+r-1}{r}$ and $\dim \operatorname{Sym}^r(V) = \binom{n+r-1}{r}$. [/guided] [/step]