Let $\mathbb{O}$ be the octonion algebra (say over $\mathbb{R}$) and let $J\_{3}(\mathbb{O})$ be the set of $3 \times 3$ hermitian matrices with octonion coefficients, that is:


$$ J\_3(\mathbb{O}) = \left\{ \begin{pmatrix} \lambda\_1 & a & b \\ \overline{a} & \lambda\_2 & c \\ \overline{b} & \overline{c} & \lambda\_3 \end{pmatrix}, \ \ \lambda\_i \in \mathbb{R}, \ a,b,c \in \mathbb{O} \right\}$$


The group $\mathrm{E}\_6$ is the group of linear automorphisms of $J\_{3}(\mathbb{O})$ which preserve the cubic form:
$$\lambda\_1 \lambda\_2 \lambda\_3 + 2 \mathrm{Re}(a\overline{b}c) - \lambda\_2 N(b)^2 - \lambda\_3 N(a)^2 - \lambda\_1 N(c)^2,$$ where $N$ is the norm over $\mathbb{O}$.


There are many interesting subgroups of $\mathrm{E}\_6$ related to this description. $\mathrm{SL}\_3(\mathbb{R})$ is one of them. The action of $\mathrm{SL}\_3(\mathbb{R})$ on $J\_3(\mathbb{O})$ is given by :


$$ \forall g \in \mathrm{SL}\_3(\mathbb{R}), \forall A \in J\_{3}(\mathbb{O}), \ g\cdot A = g A\,^{t}\! g,$$
where $^{t}\!g$ is the transpose of $g$.


The group $\mathrm{Spin\_8}$ can also be seen as a subgroup of $\mathrm{E}\_6$ with the action:
\begin{align\*} &\forall (g\_1,g\_2,g\_3) \in \mathrm{Spin}\_8,\quad \forall A = \begin{pmatrix} \lambda\_1 & a & b \\ \overline{a} & \lambda\_2 & c \\ \overline{b} & \overline{c} & \lambda\_3 \end{pmatrix} \in J\_{3}(\mathbb{O}), \\ &g\cdot A = \begin{pmatrix} \lambda\_1 & g\_1(a) & g\_2(b) \\ \overline{g\_1(a)} & \lambda\_2 & g\_3(c) \\ \overline{g\_2(b)} & \overline{g\_3(c)} & \lambda\_3 \end{pmatrix},
\end{align\*}
where we identify $\mathrm{Spin\_8}$ with $\{(g\_1,g\_2,g\_3) \in \mathrm{SO}\_8^3, \ \forall (x,y) \in \mathbb{O}, \ g\_3(xy) = g\_1(x)g\_2(y) \}$.


It is well-known (see Harvey's *Spinor and calibrations* for instance) that the subgroup of $E\_6$ generated by $\mathrm{SO\_3}$ and $\mathrm{Spin}\_8$ is $\mathrm{F}\_4$. I think it is equally well-known (I don't have a reference at hand, but it seems to be an easy corollary of the previous statement) that $\mathrm{E}\_6$ itself is generated by $\mathrm{SL}\_3$ and $\mathrm{Spin\_8}$.


**Question :** Is there an explicit description the subgroup of $\mathrm{E}\_6$ (resp. $\mathrm{F}\_4$) generated by $\mathrm{SL}\_{3}$ and $\mathrm{Spin}\_7$ (resp. $\mathrm{SO}\_3$ and $\mathrm{Spin}\_7$), where $\mathrm{Spin\_7}$ is seen in $\mathrm{Spin}\_8$ as $\{(g\_1,g\_2,g\_3) \in \mathrm{Spin}\_8, \ g\_1(1) = 1\}$?



Subgroup $\mathrm{E}_6$ generated by $\mathrm{Spin_7}$ and $\mathrm{SL}_3$
**N.B.:** I am revising my response for clarity. (The actual answer to the question asked by the OP is still the same, but I think that this re-organization, particularly at the end, makes the structure of the argument for the answer more clear. I was inspired to do this because some people had some difficulty following the original.) I should also say that the main idea is essentially the one that Theo Johnson-Freyd proposed in his first comment on the question.


I'll use the more usual notation


$$ 
J\_3(\mathbb{O}) = \left\{\ \left.\begin{pmatrix} \lambda\_1 & a\_3 & {\overline{a\_2}} \\ \overline{a\_3} & \lambda\_2 & a\_1 \\ a\_2 & \overline{a\_1} & \lambda\_3 \end{pmatrix}\ \right| \ \ \lambda\_i \in \mathbb{R}, \ a\_i \in \mathbb{O} \right\}
\tag 1
$$
and the cubic form given by
$$
C = \lambda\_1\lambda\_2\lambda\_3 + 2\,\mathrm{Re}(a\_1a\_2a\_3) 
 - \lambda\_1\,a\_1\overline{a\_1} - \lambda\_2\,a\_2\overline{a\_2}
 - \lambda\_3\,a\_3\overline{a\_3}\,.
$$
Then $\mathrm{E}\_6\subset\mathrm{GL}\bigl(J\_3(\mathbb{O})\bigr)\simeq \mathrm{GL}(27,\mathbb{R})$ is the group of linear transformations of $J\_3(\mathbb{O})$ that preserve the cubic form $C$ and $\mathrm{F}\_4\subset\mathrm{E}\_6$ is the subgroup that also fixes $I\_3\in J\_3(\mathbb{O})$. (Explicitly, $\mathrm{F}\_4$ is a maximal compact in this noncompact real form $\mathrm{E}\_6^{(-26)}$ 
of $\mathrm{E}\_6$.)


The subgroup $\mathrm{Spin}(8)\subset{\mathrm{SO}(8)}^3$ is defined as the set of triples $g = (g\_1,g\_2,g\_3)$ that satisfy
$$
\mathrm{Re}\bigl(g\_1(a\_1)g\_2(a\_2)g\_3(a\_3)\bigr) = \mathrm{Re}(a\_1a\_2a\_3)
$$
for all $a\_i\in\mathbb{O}$. Let $K\_i\subset\mathrm{Spin}(8)$ for $1\le i\le 3$ be the subgroup that satisfies $g\_i(\mathbf{1}) = \mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative identity). Each of the $K\_i$ is isomorphic to $\mathrm{Spin}(7)$, any two of them generate $\mathrm{Spin}(8)$, and the intersection of any two of them is equal to the intersection of all three of them, which is a group isomorphic to $\mathrm{G}\_2$, diagonally embedded in ${\mathrm{SO}(8)}^3$ as the automorphism group of the octonions.


As has already been observed, $\mathrm{SL}(3,\mathbb{R})$ acts on $J\_3(\mathbb{O})$ preserving $C$ via $a\cdot A = a\,A\,^{t}a$ (usual matrix multiplication), where $a\in\mathrm{SL}(3,\mathbb{R})$ and $A\in J\_3(\mathbb{O})$ are arbitrary. This is a faithful action, so, in this way, $\mathrm{SL}(3,\mathbb{R})$ will be regarded as a subgroup of $\mathrm{E}\_6$. 


Meanwhile, by its very definition, $g = (g\_1,g\_2,g\_3)\in\mathrm{Spin}(8)$ acts on $A\in J\_3(\mathbb{O})$ via


$$
g\cdot \begin{pmatrix} \lambda\_1 & a\_3 & \overline{a\_2} \\ \overline{a\_3} & \lambda\_2 & a\_1 \\ a\_2 & \overline{a\_1} & \lambda\_3 \end{pmatrix}
= \begin{pmatrix} \lambda\_1 & g\_3(a\_3) & {\overline{g\_2(a\_2)}} \\ \overline{g\_3(a\_3)} & \lambda\_2 & g\_1(a\_1) \\ g\_2(a\_2) & \overline{g\_1(a\_1)} & \lambda\_3 \end{pmatrix}\tag 2
$$
and this faithful action preserves $C$ as well, so $\mathrm{Spin}(8)$ will also be regarded as a subgroup of $\mathrm{E}\_6$.


Now, as mentioned, $\mathrm{SO}(3)$ and $\mathrm{Spin}(8)$ together generate $\mathrm{F}\_4\subset \mathrm{E}\_6$. Consequently (since there is no connected Lie group that lies properly between $\mathrm{F}\_4$ and $\mathrm{E}\_6$), it follows easily that $\mathrm{SL}(3,\mathbb{R})$ and $\mathrm{Spin(8)}$ together generate $\mathrm{E}\_6$.


We want to show that $\mathrm{SO}(3)$ and $K\_1\simeq\mathrm{Spin}(7)$ also suffice to generate $\mathrm{F}\_4$ while $\mathrm{SL}(3,\mathbb{R})$ and $K\_1\simeq\mathrm{Spin}(7)$ suffice to generate $\mathrm{E}\_6$.


To do this, let $h\in\mathrm{SO}(3)\subset\mathrm{SL}(3,\mathbb{R})$ be 
$$
h = \begin{pmatrix}0&-1&0\\-1&0&0\\0&0&-1\end{pmatrix} = h^{-1} = {}^th.
$$
Then we have 
$$
h\cdot\begin{pmatrix} \lambda\_1 & a\_3 & \overline{a\_2} \\ \overline{a\_3} & \lambda\_2 & a\_1 \\ a\_2 & \overline{a\_1} & \lambda\_3 \end{pmatrix}
= \begin{pmatrix} \lambda\_2 & \overline{a\_3} & a\_1 \\ a\_3 & \lambda\_1 & \overline{a\_2} \\ \overline{a\_1} & a\_2 & \lambda\_3 \end{pmatrix}.
$$
Consequently, for $g = (g\_1,g\_2,g\_3)\in \mathrm{Spin}(8)$, computation yields
$$
h(g\_1,g\_2,g\_3)h = \bigl(\ cg\_2c,\ cg\_1c,\ cg\_3c\ \bigr)\in\mathrm{Spin}(8),
\tag 3
$$
where $c:\mathbb{O}\to\mathbb{O}$ is conjugation, i.e., $c(a) = \overline{a}$.
(Thus, conjugation by $h$ gives an involution of $\mathrm{Spin}(8)$ that, together with the order $3$ homomorphism $k(g\_1,g\_2,g\_3) = (g\_2,g\_3,g\_1)$, generates a group of automorphisms of $\mathrm{Spin}(8)$ isomorphic to $S\_3$ that maps isomorphically onto $\mathrm{Out}\bigl(\mathrm{Spin}(8)\bigr)$. I imagine that this is what Theo Johnson-Freyd had in mind with his initial comment on this question.)


Note that $g\_i(\mathbf{1}) =\mathbf{1}$ implies that $cg\_ic = g\_i$. Consequently, from the above formula $(3)$, it follows that if $g\in K\_1$, then $hgh\in K\_2$.


Thus the subgroup of $\mathrm{E}\_6$ generated by $\mathrm{SO}(3)$ and $K\_1 \simeq\mathrm{Spin}(7)$ contains $K\_2$ and, hence, $\mathrm{Spin}(8)$ (since $K\_1$ and $K\_2$ generate $\mathrm{Spin}(8)$). Thus, this group is $\mathrm{F}\_4$. Similarly, the subgroup of $\mathrm{E}\_6$ generated by $\mathrm{SL}(3,\mathbb{R})$ and $K\_1 \simeq\mathrm{Spin}(7)$ contains $K\_2$
and, hence, $\mathrm{Spin}(8)$. Thus, this group is $\mathrm{E}\_6$, as desired.


Similar arguments (using similar choices of $h$) suffice to show that, for any of $i= 1$, $2$, or $3$, the subgroup of $\mathrm{E}\_6$ generated by $\mathrm{SO}(3)$ and $K\_i$ is $\mathrm{F}\_4$, while $\mathrm{SL}(3,\mathbb{R})$ and $K\_i$ generate $\mathrm{E}\_6$.


