If $Au=v\sqrt{2}$ and $A^Tv=u\sqrt{2}$. Find value of $|u_1+2\sqrt{2}v_1|$ Let $$A=\begin{bmatrix}1&1\\0&1\\-1&1\end{bmatrix}$$ and let $A^T$ denote the transpose of A. Let $u=\begin{bmatrix}u_1\\u_2\end{bmatrix}$ and $v=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$ be column vectors with entries in $\mathbb R$ such that
$u_1^2+u_2^2=v_1^2+v_2^2+v_3^2=1$.
Suppose $Au=v\sqrt{2}$ and $A^Tv=u\sqrt{2}$.
Find value of $|u_1+2\sqrt{2}v_1|$.
My Attempt
By equating and things like that I was able to obtain
$u_1+u_2=v_1\sqrt2$
$u_2=v_2\sqrt2$
$-u_1+u_2=v_3\sqrt2$
$v_1-v_3=u_1\sqrt2$
$v_1+v_2+v_3=u_2\sqrt2$
On solving the above equations I obtained
$u_2=v_2=0$
$v_3=-v_1$
$u_1=v_1\sqrt2$
and then by using $u_1^2+u_2^2=v_1^2+v_2^2+v_3^2=1$ I was able to obtain
$v_1^2=\frac{1}{2}$
$(u_1+2\sqrt2v_1)^2=u_1^2+8v_1^2+4\sqrt2u_1v_1=2v_1^2+8v_1^2+8v_1^2=9$
$\Rightarrow |u_1+2\sqrt2v_1|=3$
Is my answer correct .
Also can there be a shorter way.
My knowledge of matrices is very limited.

What a crazy problem!
From the given equations $Au=\sqrt2 v$ and $A^\top v=\sqrt2 u$, we conclude that $(A^\top A)u = 2u$. But $A^\top A = \left[\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}\right]$, so $u=\pm\left[\begin{matrix} 1\\0\end{matrix}\right]$ and $v=\pm\dfrac1{\sqrt2}\left[\begin{matrix} 1\\0\\-1\end{matrix}\right]$. Then $|u_1+2\sqrt2 v_1| = 1+2 = 3$, as you said.
