Recall that an algebra $A$ is said to be quadratic if it is unital and the set $\left\{{1, x, x^2}\right\}$ is linearly dependent for every $x \in A$. {{explain|This needs to be reviewed, as Definition:Quadratic Algebra as currently posted up does not match the above.}} Thus, for every $x \in A$ there exist $t \left({x}\right), n \left({x}\right) \in \R$ such that: :$x^2 - t \left({x}\right) x + n \left({x}\right) = 0$ Obviously, $t \left({x}\right)$ and $n \left({x}\right)$ are uniquely determined if $x \notin \R$. {{handwaving|Obviously?}} {{Proofread}} Suppose $x \notin \R$. Then $x$ can be expressed as $x = a + b i$, with $a, b \in \R$ and $b \ne 0$. Then: :$x^2 = a^2 - b^2 + 2 a b i$ and: :$x^2 - t \left({x}\right) x + n \left({x}\right) =  a^2 - b^2 - a t \left({x}\right) + n \left({x}\right) + \left({2 a b - b t \left({x}\right)}\right) i = 0$ It follows that both: : $(1): \quad 2 a b - b t \left({x}\right) = 0$ and: : $(2): \quad a^2 - b^2 - a t \left({x}\right) + n \left({x}\right) = 0$ $(1)$ leads to: : $t \left({x}\right) = 2 a$ and $(2)$ leads to; : $n \left({x}\right) = a^2 + b^2$ Setting $t \left({\lambda}\right) = 2 \lambda$ and $n \left({\lambda}\right) = \lambda^2$ for $\lambda \in \R$, we can then consider $t$ and $n$ as maps from $A$ into $\R$. (In this way $t$ becomes a linear functional). We call $t \left({x}\right)$ and $n \left({x}\right)$ the trace and the norm of $x$ respectively. {{explain|Trace most certainly has not been disambiguated, and I'm not sure that "norm" is directly the same as the complex modulus}} {{Disambiguate|Definition:Norm}} {{Disambiguate|Definition:Trace}} From $x^2 - \left({x + x^*}\right) x + x^* x = 0$ we see that all algebras $\Bbb A_n$ are quadratic. Further, every real division algebra $A$ that is algebraic and power-associative (this means that every subalgebra generated by one element is associative) is automatically quadratic. Indeed, if $x \in A$ then there exists a nonzero polynomial $f \left({X}\right) \in \R \left[{X}\right]$ such that $f \left({x}\right) = 0$. Writing $f \left({X}\right)$ as the product of linear and quadratic polynomials in $\R \left[{X}\right]$ it follows that $p \left({x}\right) = 0$ for some $p \left({X}\right) \in \R \left[{X}\right]$ of degree $1$ or $2$. In particular, algebraic alternative (and hence associative) real division algebras are quadratic. Finally, if $A$ is a real unital algebra, that is, an algebra over $\R$ with unity $1$, then we shall follow a standard convention and identify $\R$ with $\R 1$. Thus we shall write $\lambda$ for $\lambda 1$, where $\lambda \in \R$. === Lemma 1 === {{:Frobenius's Theorem/Lemma 1}} === Lemma 2 === {{:Frobenius's Theorem/Lemma 2}} We have from above that $A$ is quadratic. We may assume that $n = \dim A \ge 2$. By Lemma 1: Assertion 4 we can fix $i \in A$ such that $i^2 = -1$. Thus, $A \cong \C$ if $n = 2$. Let $n > 2$. By Lemma 2: :$\exists j \in A: j^2 = -1, i j = -j i$ Set $k = ij$. It can immediately be checked that: : $k^2 = -1$ : $k i = j = -i k$ : $j k = i = -k j$ : $\left\{{i, j, k}\right\}$ is a linearly independent set. Therefore $A$ contains a subalgebra isomorphic to $\Bbb H$. Finally, suppose $n > 4$. By Lemma 2 there would exist $e \in A, e \ne 0$ such that: :$(a): \quad e i = -i e$ :$(b): \quad e j = -j e$ :$(c): \quad e k = -k e$ However, from the $(a)$ and $(b)$ it follows that $e i j = -i e j = i j e$. Since $i j = k$, this contradicts $(c)$. It follows that $n \le 4$, and so $\Bbb H$ is the highest order  {{qed}} {{MissingLinks}} {{namedfor|Ferdinand Georg Frobenius|cat = Frobenius}}