Verify that $\| T(x) \| = \| x \|$ Let $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ be multiplication by the matrix 
$$A=
\begin{bmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\[0.3em]
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\[0.3em]
-\frac{2}{3} & -\frac{1}{3} & \frac{2}{3}
\end{bmatrix}
$$
Find $T(x)$ for the vector $x=(1,-3,4)$. Using the Euclidean inner product on $\mathbb{R}^3$ verify that $\| T(x) \| = \| x \|$.
For the first part I have 
$$
\begin{bmatrix}
w_1 \\
w_2 \\
w_3
\end{bmatrix}=
\begin{bmatrix}
\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\[0.3em]
\frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\[0.3em]
-\frac{2}{3} & -\frac{1}{3} & \frac{2}{3}
\end{bmatrix}
\begin{bmatrix}
1 \\
-3 \\
4
\end{bmatrix}=
\begin{bmatrix}
1 \\
4 \\
3
\end{bmatrix}$$
But can anyone help with the second part?

Let $x$ be the vector $$\begin{bmatrix} a \\ b \\ c \end{bmatrix}$$, and let $u_1, u_2, u_3$ be the three columns of the matrix $A$. Then 
$$
T(x) = au_1 + b u_2 + c u_3
$$
and
$$
\|T(x)\|^2 = T(x) \cdot T(x) = a^2 u_1 \cdot u_1 + 2ab u_1 \cdot u_2 + \ldots + c^2 u_3 \cdot u_3
$$ 
Computing all pairwise dot products of columns, we see that $u_i \cdot u_j = 0$ if $i \ne j$, but $u_i \cdot u_1 = 1$. Hence the right hadn side above simplifies to just
$$
\|T(x)\|^2 = T(x) \cdot T(x) = a^2 u_1 \cdot u_1 + b^2 u_2 \cdot u_2 + c^2 u_3 \cdot u_3 = a^2 + b^2 + c^2 = \| x \|^2
$$ 
and you're done. 
A slightly more sophisticated way of saying this is to note that $$a \cdot b = a^t b,$$
so that 
\begin{align}
\|T(x) \|^2 
&= (Ax) \cdot (Ax)\\
&= (Ax)^t (Ax)\\
&= x^tA^t Ax\\
&= x^t( A^t A) x.
\end{align}
You now observe that $A^t A = I$, so that this last expression simplifies to $x^t x = \|x \|^2$. 
