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In this paper I found following characterization of the Jacobson radical of a matrix algebra $A$ of dimension $n$: $$\operatorname{Rad}(A) =\{x \in A \mid x^n = 0 \text{ and } \forall_{y\in A}(xy)^n=0\} $$ I tried to start a proof based on the principle that an $A$-module $M$ is simple $\iff M \cong A/I$ for some maximal left ideal $I$. Let $I$ be such an ideal and $\pi:A \rightarrow A/I$ the natural projection. The properties of $x$ and $y$ project onto the quotient giving $\pi(x)^n=0$ and $\pi(xy)^n = 0$. Hoping that this contradicts the simplicity of $M$ unless $\pi(x) = 0$ this shows that $x \in I$ for every maximal left ideal $I$ so $x \in \operatorname{Rad}(A)$. Being an absolute novice in non commutative rings I was unable to find such a contradiction.

On the other hand if $x \in \operatorname{Rad}(A)$ then $x^n = 0$ since all the elements of the radical are nilpotent and also $(xy)^n=0$ because $xy$ also belongs to the radical.

Marc Bogaerts
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On a finite dimensional $F$-algebra, every element of the Jacobson radical is nilpotent, because the Jacobson radical of an Artinian ring is nilpotent.

If $x$ is nilpotent, then $1-x$ is invertible. So, if $xy$ is nilpotent for every $y\in A$, then $1-xy$ is right invertible for every $y\in A$ and therefore $x$ belongs to the Jacobson radical.

An element $x\in M_n(F)$ is nilpotent if and only if $x^n=0$.

Let's prove the lemma: If $R$ is a ring and $x\in R$, then $x\in J(R)$ if and only if, for every $y\in R$, $1-xy$ is right invertible.

Suppose $x\in J(R)$ and consider $1-xy$. If $1-xy$ is not right invertible, then $1-xy$ belongs to a maximal right ideal $I$; since $xy\in I$ too, we get $1\in I$: impossible.

Conversely, if $1-xy$ is right invertible for every $y\in R$ and $I$ is a maximal right ideal, then $x\notin I$ implies $1=xy+z$, with $z\in I$: but then $z=1-xy$ is right invertible: contradiction. Therefore $x\in I$ and, since $I$ is arbitrary, $x\in J(R)$ (the intersection of all maximal right ideals).

egreg
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