Let $H=\mathbb{M}_{n}(k)$ be a matrix algebra over a field $k$ with standard multiplication and assume that $n>1$ . Assume that $H$ can be turned into a Hopf algebra. In particular, there is $\varepsilon:H\to k$ such that $\varepsilon$ is a morphism of algebras. It can be shown that matrix algebra is simple, i.e. if $I\subseteq H$ is a two-sided ideal, then $I=0$ or $I=H$ . Thus we have that $\mathrm{ker}\varepsilon=0$ (because $\varepsilon(1)=1$ ). Contradiction, because $\mathrm{dim}_{k}H>1=\mathrm{dim}_{k}k$ .

Now consider $H=\mathbb{M}^{c}(n,k)$ a vector space of all $n\times n$ matrices over $k$ . We introduce coalgebra structure on $H$ . Let $E_{ij}$ be a matrix in $H$ with $1$ in $(i,j)$ place and $0$ everywhere else. Of course $\{E_{ij}\}$ forms a basis of $H$ and it is sufficient to define comultiplication and counit on it. Define $$\Delta(E_{ij})=\sum_{p=1}^{n} E_{ip}\otimes E_{pj};$$ $$\varepsilon(E_{ij})=\delta_{ij},$$ where $\delta_{ij}$ denotes Kronecker delta. It can be easily checked, that $\big(\mathbb{M}^c(n,k),\Delta,\varepsilon\big)$ is a coalgebra known as the matrix coalgebra. Also, is well known that the dual algebra $\mathbb{M}^c(n,k)^*$ is isomorphic to the standard matrix algebra.

Now assume that matrix coalgebra $H=\mathbb{M}^c(n,k)$ (where $n>1$ ) can be turned into a Hopf algebra. Since $H$ is finite dimensional, then we can take dual Hopf algebra $H^*$ . But the underlaying algebra structure of $H^*$ is isomorphic to a matrix algebra (as we remarked earlier), which we've already shown to be impossible. Thus matrix coalgebra cannot be turned into a Hopf algebra.