Let $k$ be a field. Let $A$ be the $k$-subspace of $M_3(k)$ spanned by

$1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \alpha =\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix},\beta =\begin{pmatrix}0&0&0\\0&0&0\\1&1&0\end{pmatrix}$,

which can be shown to actually be a $k$-algebra.

Determine the Jacobson radical of $A$, which is the intersection of all maximal left ideals.

I know at least $\beta$ (and hence the left ideal it generates) is in the Jacobson radical, because it is nilpotent (it is known that the Jacobson radical contains all nilpotent elements). I don't know where to proceed from here, though...

Originally Posted by topspin1617
Let $k$ be a field. Let $A$ be the $k$-subspace of $M_3(k)$ spanned by

$1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}, \alpha =\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix},\beta =\begin{pmatrix}0&0&0\\0&0&0\\1&1&0\end{pmatrix}$,

which can be shown to actually be a $k$-algebra.

Determine the Jacobson radical of $A$, which is the intersection of all maximal left ideals.

I know at least $\beta$ (and hence the left ideal it generates) is in the Jacobson radical, because it is nilpotent (it is known that the Jacobson radical contains all nilpotent elements). I don't know where to proceed from here, though...
first of all, this claim that "the Jacobson radical contains all nilpotent elements" is true in commutative rings not in all rings. for example $J(M_2(\mathbb{C}))=(0)$ but obviously $M_2(\mathbb{C})$ has non-zero nilpotent elements.

what is true in general is that the Jacobson radical contains all nilpotent ideals.

to answer your question, just use this fact that $a \in J(A)$ if and only if $1-xa$ is invertible for all $x \in A.$ you'll eventually need to consider two cases: $\text{char}(k) \neq 2$ and $\text{char}(k)=2.$