Find a isometry such that the matrix in respect to the canonical basis is: I need to find a isometry such that the matrix in respect to the canonical basis is:
$$\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}& 0\\0 & 0 & 1\\x & y & z\end{bmatrix}$$
However, my definition of isometry is a linear transformation that preserves length. So, the linear transformation corresponding to this matrix is:
$$a = a_1e_1+ a_2e_2 + a_3e_3\\T(a) = T(a_1e_1+ a_2e_2 + a_3e_3) = a_1(\frac{1}{\sqrt{2}},0,x) + a_2(\frac{1}{\sqrt{2}},0,y)+a_3(0,1,z)$$
Then:
$$||a||^2 = ||a_1(\frac{1}{\sqrt{2}},0,x) + a_2(\frac{1}{\sqrt{2}},0,y)+a_3(0,1,z)||^2$$
But this requires too much computation. I remember that my professor made something like this:
$$||(1,0,0)|| = ||(\frac{1}{\sqrt{2}},0,x)||\\||(0,1,0)|| = ||(\frac{1}{\sqrt{2}},0,y)||\\||(0,0,1)|| = ||(0,1,z)||$$
But I can't see why a linear transformation that preserves the basis lengths also preserves the entire vector length. How to deal with it?
 A: It is not sufficient that preserving basis lengths is enough.  For example the transformation with matrix
$$\begin{bmatrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
preserves basis lengths but does not preserve the length of $(1, 1, 0)$.
But it is necessary that the transformation preserve basis lengths.  The wording of the question implies that there is some choice of $x$, $y$, and $z$ such that the resulting matrix defines an isometry.  Since an isometry must preserve basis lengths and the resulting equations can be almost solved for $x$, $y$, and $z$ this is a good first step in finding the answer.
I say "almost" solved because while you can solve for $x$, you can't solve for $y$ and $z$.  You can only solve for $y^2$ and $z^2$ which means you know $y$ and $z$ up to a choice of sign.  To choose that sign you can use the following additional condition: An isometry cannot have a kernel.  So it must be the case that the columns of the matrix are linearly independent.  In particular the first two columns cannot be equal.  You'll find that this means you can choose any sign for $x$ but then you must choose the opposite sign for $y$.  And yes, this means there are two correct solutions, one with a positive $x$ and one with a negative $x$.
