6. Given that $A$ is the upper vertex of the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, and $B$ and $C$ are two other points on the ellipse, and $\triangle A B C$ is an isosceles right triangle with $A$ as the right-angle vertex. If there is only one solution for $\triangle A B C$, then the range of the eccentricity of the ellipse is $\qquad$
6. $\left(0, \frac{\sqrt{6}}{3}\right]$.

Let the equation of the line on which one side of an isosceles right triangle lies be $y=k x+1(k>0)$, and its intersection point $B$ with the ellipse has a horizontal coordinate of $-\frac{2 k a^{2}}{1+a^{2} k^{2}}$.
Thus, the horizontal coordinate of point $C$ is $\frac{2 k a^{2}}{a^{2}+k^{2}}$.
From $A B=A C$, we get
$$
\begin{array}{l}
\left(1+k^{2}\right) \frac{4 k^{2} a^{4}}{\left(1+a^{2} k^{2}\right)^{2}}=\left(1+\frac{1}{k^{2}}\right) \frac{4 k^{2} a^{4}}{\left(a^{2}+k^{2}\right)^{2}} \\
\Rightarrow k^{3}-a^{2} k^{2}+k a^{2}-1=0 .
\end{array}
$$

According to the problem, the solution to this equation is only $k=1$.
And $k^{3}-a^{2} k^{2}+k a^{2}-1$
$$
=(k-1)\left[k^{2}-\left(a^{2}-1\right) k+1\right]=0 \text {, }
$$

To make the above equation have a unique positive solution $k=1$, then
$$
\left(a^{2}-1\right)^{2}-4 \leqslant 0 \Rightarrow 1<a \leqslant \sqrt{3} \text {, }
$$

When $a=\sqrt{3}$, the solution to the equation is unique.
Therefore, the range of the eccentricity is $\left(0, \frac{\sqrt{6}}{3}\right)$.