By analogy with Lie algebra theory (see [17]), a proper
subalgebra $\mathcal{\mathcal{A}}$ of the matrix algebra
${\mathcal{\mathcal{M}}}_{n}(K)$ is called *parabolic* [16] if it
is similar to an algebra which contains ${\mathcal{\mathcal{U}}}_{n}(K)$, i.e.
there exists an invertible matrix *U* such that
${\mathcal{\mathcal{U}}}_{n}(K)\subseteq U\mathcal{\mathcal{A}}{U}^{-1}$. As we will
see, this concept will play an important role in determining the
maximal dimension of a subalgebra in a matrix algebra. It was
proved in [16] that $\mathcal{\mathcal{A}}$ is a parabolic subalgebra
of ${\mathcal{\mathcal{M}}}_{n}(K)$ if and only if there exists a set of
positive integers ${n}_{1}$, ${n}_{2},\mathrm{\dots},{n}_{s}$ with
${\sum}_{i=1}^{s}{n}_{i}=n$ such that $\mathcal{\mathcal{A}}$ is similar to
the algebra of all matrices having non-overlapping blocks of
${n}_{i}\times {n}_{i}$ matrices on the diagonal with non-zero
entries only in these blocks or above them. Moreover, it is
straightforward to see that $dimA=\frac{{n}^{2}}{2}+{\sum}_{i=1}^{s}\frac{{n}_{i}^{2}}{2}$.

A parabolic subalgebra $\mathcal{\mathcal{A}}$ of ${\mathcal{\mathcal{M}}}_{n}(K)$
determined by the set of positive integers ${n}_{1}$, ${n}_{2},\mathrm{\dots},{n}_{s}$ will be called a parabolic subalgebra of type $({n}_{1},{n}_{2},\mathrm{\dots},{n}_{s})$. If $s=2$ then $\mathcal{\mathcal{A}}$ will be called a
maximal parabolic subalgebra.

Our next result shows that a maximal parabolic subalgebra is in
fact a maximal proper subalgebra of ${\mathcal{\mathcal{M}}}_{n}(K)$.

*The maximal parabolic subalgebras are maximal proper subalgebras
of ${\mathcal{M}}_{n}\mathit{}\mathrm{(}K\mathrm{)}$.*

#### Proof.

Consider $\mathcal{\mathcal{A}}$ to be a parabolic subalgebra having on the
diagonal two blocks *U* and *V* of dimensions $l\times l$ and
respectively $(n-l)\times (n-l)$ with $l\ge 1$ and assume there
exists a subalgebra $\mathcal{\mathcal{T}}$ of ${\mathcal{\mathcal{M}}}_{n}(K)$ such
that $\mathcal{\mathcal{A}}\subset \mathcal{\mathcal{T}}$, $\mathcal{\mathcal{A}}\ne \mathcal{\mathcal{T}}$. The elements of $\mathcal{\mathcal{T}}$ are linear
combinations of the matrix units ${e}_{i,j}$ and there must be an element $x\in \mathcal{\mathcal{T}}-\mathcal{\mathcal{A}}$ that has a non-zero entry α in a position $(i,j)$ with $i>l$ and $j\u2a7dl$. Then ${e}_{i,j}\in \mathcal{\mathcal{T}}$ since ${e}_{i,j}={\alpha}^{-1}{e}_{i,i}x{e}_{j,j}$. Since $i>l$, ${e}_{m,i}\in \mathcal{\mathcal{A}}$ for all *m* and hence ${e}_{m,j}={e}_{m,i}{e}_{i,j}$ belongs to $\mathcal{\mathcal{T}}$ for all *m*. Similarly ${e}_{j,k}\in \mathcal{\mathcal{A}}$ for all *k* and so ${e}_{i,k}={e}_{i,j}{e}_{j,k}\in \mathcal{\mathcal{T}}$ for all *k*. Thus if ${e}_{i,j}$ belongs to $\mathcal{\mathcal{T}}$, then so do all the matrix units from row *i* and column *j*. Using the same reasoning for all these elements we can conclude that $\mathcal{\mathcal{T}}={\mathcal{\mathcal{M}}}_{n}(K)$ which finishes the proof.
∎

The parabolic subalgebras of ${\mathcal{\mathcal{M}}}_{3}(K)$ are similar to
the following subalgebras:

$\left(\begin{array}{ccc}\hfill K\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill 0\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill K\hfill \end{array}\right),\left(\begin{array}{ccc}\hfill K\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill K\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill K\hfill \end{array}\right),\left(\begin{array}{ccc}\hfill K\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill 0\hfill & \hfill K\hfill & \hfill K\hfill \\ \hfill 0\hfill & \hfill K\hfill & \hfill K\hfill \end{array}\right).$

The last two algebras are maximal proper subalgebras of
${\mathcal{\mathcal{M}}}_{3}(K)$.

If *K* is an algebraically closed field of characteristic zero and
$\mathcal{\mathcal{A}}$ is a finite-dimensional *K*-algebra then by Wedderburn’s theorem [7], $\mathcal{\mathcal{A}}$ has a semisimple part
which is a direct sum of matrix algebras over *K* of dimensions
${n}_{1}^{2}$, ${n}_{2}^{2},\mathrm{\dots},{n}_{s}^{2}$. In order for
$\mathcal{\mathcal{A}}$ to be a subalgebra of ${\mathcal{\mathcal{M}}}_{n}(K)$ we need
to have ${\sum}_{i=1}^{s}{n}_{i}\le n$. However, since we are
trying to maximize the dimension of $\mathcal{\mathcal{A}}$ we will assume
that ${\sum}_{i=1}^{s}{n}_{i}=n$.

*Let **K* be an algebraically closed field of characteristic zero
and $\mathcal{A}$ a subalgebra of ${\mathcal{M}}_{n}\mathit{}\mathrm{(}K\mathrm{)}$ whose
semisimple part *S* is a direct sum of matrix algebras over *K* of
dimensions ${n}_{\mathrm{1}}^{\mathrm{2}}$, ${n}_{\mathrm{2}}^{\mathrm{2}}\mathrm{,}\mathrm{\dots}\mathrm{,}{n}_{s}^{\mathrm{2}}$ with
${\mathrm{\sum}}_{i\mathrm{=}\mathrm{1}}^{s}{n}_{i}\mathrm{=}n$. Then $\mathrm{dim}\mathit{}\mathcal{A}\mathrm{\le}\frac{{n}^{\mathrm{2}}}{\mathrm{2}}\mathrm{+}{\mathrm{\sum}}_{i\mathrm{=}\mathrm{1}}^{s}\frac{{n}_{i}^{\mathrm{2}}}{\mathrm{2}}$. Moreover, if
equality holds then *A* is a parabolic subalgebra of
${\mathcal{M}}_{n}\mathit{}\mathrm{(}K\mathrm{)}$.

#### Proof.

By Wedderburn’s main theorem [7, Theorem 2.17] we have
$\mathcal{\mathcal{A}}=S\oplus rad\mathcal{\mathcal{A}}$. Let $dim\mathcal{\mathcal{A}}=d$ and let $dimrad\mathcal{\mathcal{A}}=r$. It
follows that ${\sum}_{i=1}^{s}{n}_{i}^{2}+r=d$. Remark that each
${n}_{i}\times {n}_{i}$ matrix subalgebra of *S* contains its
nilpotent subalgebra of strictly upper triangular matrices. This
implies that $\mathcal{\mathcal{A}}$ contains, all together, a subalgebra of
nilpotent matrices of dimension ${\sum}_{i=1}^{s}\frac{{n}_{i}({n}_{i}-1)}{2}+r$. Using Gerstenhaber’s result [5] on subspaces
of nilpotent matrices we must have

$\sum _{i=1}^{s}\frac{{n}_{i}({n}_{i}-1)}{2}+r\le \frac{n(n-1)}{2}.$

Since ${\sum}_{i=1}^{s}{n}_{i}^{2}+r=d$ we get

$d\le \frac{n(n-1)}{2}+\sum _{i=1}^{s}\frac{{n}_{i}({n}_{i}+1)}{2}.$

The conclusion now follows by using ${\sum}_{i=1}^{s}{n}_{i}=n$.

Suppose now that equality holds. It follows that $r=\frac{{n}^{2}}{2}-{\sum}_{i=1}^{s}\frac{{n}_{i}^{2}}{2}$ and so
$\mathcal{\mathcal{A}}$ contains a subalgebra of nilpotent matrices of
dimension $\frac{n(n-1)}{2}$. Gerstenhaber’s result implies that
any such subspace of ${\mathcal{\mathcal{M}}}_{n}(K)$ is conjugate to
$\overline{{\mathcal{\mathcal{U}}}_{n}(K)}$. Since for any invertible matrix
$C\in {\mathcal{\mathcal{M}}}_{n}(K)$ the map $u:\mathcal{\mathcal{A}}\to C\mathcal{\mathcal{A}}{C}^{-1}$ which takes any $A\in \mathcal{\mathcal{A}}$ to
$CA{C}^{-1}$ is an algebra isomorphism we may assume without loss of
generality that $\mathcal{\mathcal{A}}$ contains
$\overline{{\mathcal{\mathcal{U}}}_{n}(K)}$. However, by looking at dimensions
one sees that $\overline{{\mathcal{\mathcal{U}}}_{n}(K)}$ is not all of
$\mathcal{\mathcal{A}}$. Now since the semisimple part of $\mathcal{\mathcal{A}}$ is a
direct sum of matrix algebras of dimensions ${n}_{1}$, ${n}_{2},\mathrm{\dots},{n}_{s}$ with ${\sum}_{i=1}^{s}{n}_{i}=n$ it follows that each
${e}_{ii}\in \mathcal{\mathcal{A}}$ for all $i=1,2,\mathrm{\dots},n$. Therefore,
$\mathcal{\mathcal{A}}$ contains ${\mathcal{\mathcal{U}}}_{n}(K)$ and the proof is now finished.
∎

*Let **K* be an algebraically closed field of characteristic zero.
Then the proper subalgebras of maximum dimension in
${\mathcal{M}}_{n}\mathit{}\mathrm{(}K\mathrm{)}$ are the parabolic subalgebras of type $\mathrm{(}\mathrm{1}\mathrm{,}n\mathrm{-}\mathrm{1}\mathrm{)}$ and respectively $\mathrm{(}n\mathrm{-}\mathrm{1}\mathrm{,}\mathrm{1}\mathrm{)}$.

#### Proof.

By Theorem 2.4 we know that the upper bound for the dimension of a
subalgebra in ${\mathcal{\mathcal{M}}}_{n}(K)$ is attained when the subalgebra
is parabolic. In what follows we will prove that this bound is
maximal precisely when the subalgebra is parabolic of type $(1,n-1)$ and respectively $(n-1,1)$. Since the dimension of a
parabolic subalgebra of ${\mathcal{\mathcal{M}}}_{n}(K)$ is $\frac{{n}^{2}}{2}+{\sum}_{i=1}^{s}\frac{{n}_{i}^{2}}{2}$ in order to maximize its
dimension we need to have ${\sum}_{i=1}^{s}{n}_{i}^{2}$ as large as
possible. We denote by $S:={n}_{1}^{2}+{n}_{2}^{2}+\mathrm{\cdots}+{n}_{i}^{2}+{(n-{n}_{1}-{n}_{2}-\mathrm{\cdots}-{n}_{i})}^{2}$ and by ${S}^{\prime}:={n}_{1}^{2}+{n}_{2}^{2}+\mathrm{\cdots}+{n}_{i-1}^{2}+{(n-{n}_{1}-{n}_{2}-\mathrm{\cdots}-{n}_{i-1})}^{2}$, where ${n}_{t}$ are positive integers for all
$t=1,2,\mathrm{\dots},i$, such that ${n}_{1}+{n}_{2}+\mathrm{\cdots}+{n}_{i}<n$.
The proof will be finished once we show that $S<{S}^{\prime}$. Indeed,
this follows by noticing that $S={S}^{\prime}-2{n}_{i}(n-{n}_{1}-{n}_{2}-\mathrm{\cdots}-{n}_{i})<{S}^{\prime}$. Therefore we need to consider parabolic
subalgebras of type $(l,n-l)$, with $l\ge 1$. Then
${\sum}_{i=1}^{s}{n}_{i}^{2}={l}^{2}+{(n-l)}^{2}$. Now it can easily
be seen that for $1<l<n-1$ we have ${l}^{2}+{(n-l)}^{2}<1+{(n-1)}^{2}$ and the conclusion follows.
∎

*Let **K* be a field of characteristic zero. Then the maximal dimension of a proper subalgebra of the matrix algebra
${\mathcal{M}}_{n}\mathit{}\mathrm{(}K\mathrm{)}$ is ${n}^{\mathrm{2}}\mathrm{-}n\mathrm{+}\mathrm{1}$.

#### Proof.

If *K* is an algebraically closed field of characteristic zero then the assertion follows from Proposition 2.5. We will prove that the algebraically closed assumption on the field *K* can be dropped by simply extending the coefficients to the algebraic closure $\overline{K}$ of *K*. Indeed, let *A* be a *K*-subalgebra of ${\mathcal{\mathcal{M}}}_{n}(K)$. We have

$\overline{K}\simeq \overline{K}{\otimes}_{K}K\subset \overline{K}{\otimes}_{K}A\subseteq \overline{K}{\otimes}_{K}{\mathcal{\mathcal{M}}}_{n}(K)\simeq {\mathcal{\mathcal{M}}}_{n}(\overline{K}),$

where the last isomorphism follows from [12, Lemma 7.130]. Therefore, $\overline{K}{\otimes}_{K}A$ is a $\overline{K}$-subalgebra of ${\mathcal{\mathcal{M}}}_{n}(\overline{K})$ and since ${dim}_{K}(A)={dim}_{\overline{K}}(\overline{K}{\otimes}_{K}A)$ the conclusion follows.
∎

We end the paper by looking at the matrix coalgebra case. It is
well known that $X\subseteq {\mathcal{\mathcal{M}}}^{n}(K)$ is a coideal if
and only if ${X}^{\perp}$ is a subalgebra of ${\mathcal{\mathcal{M}}}_{n}(K)$
(see [15, Proposition 1.4.6] for a more general statement).
In light of this bijection we introduce the following:

A coideal $X\subseteq {\mathcal{\mathcal{M}}}^{n}(K)$ will be called
parabolic if ${X}^{\perp}$ is a parabolic subalgebra of
${\mathcal{\mathcal{M}}}_{n}(K)$. If ${X}^{\perp}$ is a parabolic subalgebra of
type $({n}_{1},{n}_{2},\mathrm{\dots},{n}_{s})$ then *X* will be called a
parabolic coideal of type $({n}_{1},{n}_{2},\mathrm{\dots},{n}_{s})$ as well.
If $s=2$ then *X* will be called a minimal parabolic coideal.

The parabolic coideals of ${\mathcal{\mathcal{M}}}^{n}(K)$ can be
characterized as follows:

*Let **X* be a parabolic coideal of type $\mathrm{(}{n}_{\mathrm{1}}\mathrm{,}{n}_{\mathrm{2}}\mathrm{,}\mathrm{\dots}\mathrm{,}{n}_{s}\mathrm{)}$ of the matrix coalgebra ${\mathcal{M}}^{n}\mathit{}\mathrm{(}K\mathrm{)}$. Then *X* is
the coideal of all matrices having non-overlapping blocks of
${n}_{i}\mathrm{\times}{n}_{i}$ matrices on the diagonal with non-zero
entries only below these blocks.

#### Proof.

If $\mathcal{\mathcal{A}}$ is a subalgebra of ${\mathcal{\mathcal{M}}}^{n}{(K)}^{*}\simeq {\mathcal{\mathcal{M}}}_{n}(K)$ (isomorphism of algebras) then
${\mathcal{\mathcal{A}}}^{\perp}$ is a coideal in ${\mathcal{\mathcal{M}}}^{n}(K)$. The
conclusion follows by a straightforward computation.
∎

*The minimal parabolic coideals are minimal proper coideals of
${\mathcal{M}}^{n}\mathit{}\mathrm{(}K\mathrm{)}$.*

#### Proof.

Let *X* be a minimal parabolic coideal of type $(l,n-l)$ and
assume there exists a coideal *Y* of ${\mathcal{\mathcal{M}}}^{n}(K)$ such
that $Y\subset X$, $Y\ne X$. Then ${X}^{\perp}\subset {Y}^{\perp}$
and ${X}^{\perp}$ is a maximal parabolic subalgebra of
${\mathcal{\mathcal{M}}}_{n}(K)$. Using Proposition 2.2 we obtain ${Y}^{\perp}={\mathcal{\mathcal{M}}}_{n}(K)$. This implies $Y=\{0\}$ and the proof is
finished.
∎

*Let **K* be an algebraically closed field of characteristic zero.
Then the non-zero coideals of minimal dimension in
${\mathcal{M}}^{n}\mathit{}\mathrm{(}K\mathrm{)}$ are those parabolic coideals *X* for which
${X}^{\mathrm{\perp}}$ is a parabolic subalgebra of type $\mathrm{(}\mathrm{1}\mathrm{,}n\mathrm{-}\mathrm{1}\mathrm{)}$ and
respectively $\mathrm{(}n\mathrm{-}\mathrm{1}\mathrm{,}\mathrm{1}\mathrm{)}$.

#### Proof.

For any finite-dimensional vector space *V*, and any subspace *X*
of ${V}^{*}$ we have ${dim}_{K}{X}^{\perp}={dim}_{K}{V}^{*}/X$. Therefore, for any coideal *X* in ${\mathcal{\mathcal{M}}}^{n}(K)$
we have ${dim}_{K}{X}^{\perp}={dim}_{K}{\mathcal{\mathcal{M}}}^{n}(K)/X$. Collaborating this result with the
bijection between the coideals of ${\mathcal{\mathcal{M}}}^{n}(K)$ and the
subalgebras of ${\mathcal{\mathcal{M}}}_{n}(K)$ (see [15]), it follows that a
coideal *X* in ${\mathcal{\mathcal{M}}}^{n}(K)$ has maximal codimension
precisely when the subalgebra ${X}^{\perp}$ of ${\mathcal{\mathcal{M}}}_{n}(K)$
has maximal dimension. The conclusion now follows by Proposition 2.5.
∎

*Let **K* be a field of characteristic zero. Then the minimal dimension of a
non-zero coideal in ${\mathcal{M}}^{n}\mathit{}\mathrm{(}K\mathrm{)}$ is $n\mathrm{-}\mathrm{1}$.

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