Let $\|u\|=1$ and $Q=I-2uu^\top$.
Of course, the first condition above says that $u^\top u=1$. Secondly, the meaning of $v$ is perpendicular to $u$" is that $v^\top u= u^\top v=0$.
compute Qu and simplify as much as possible. Does this just mean move the equation around to get Qu?
No, $Qu$ is a matrix product. Starting with the above, you have $Qu=(I-2uu^\top)u=Iu-2uu^\top u$. What does this reduce to?
Suppose v is orthogonal to u. Compute Qv. Im not sure how this is done.
Again, it's just a matrix product. $Qv=(I-2uu^\top)v=Iv-2uu^\top v$. What does this reduce to?
Then I'm asked to explain in plain English which subspace $Q$ is reflecting across.
This is an interesting question which is not as mechanical as the rest of the problem. For one of the two parts above, you'll discover that $Qx=-x$. Geometrically, this means that $Q$ just reversed the direction of that vector.
One of the other computations is going to come out to $Qy=y$, meaning that $Q$ didn't alter the vector at all! But remember that the above had you assume that $x$ and $y$ are perpendicular to each other, so this means that one direction was reversed, and all perpendicular directions were left alone. If you fix a vector $x$, what does the collection of all perpendicular vectors look like?
Compute the reflection matrix $Q_1=I−2u_1u^\top_1$ where $u_1=(0,1)$. Compute $Q_1x_1$, where $x_1=(0,1)$ and sketch the vectors $u_1, x_1$ and $Q_1x_1$ in the plane.
This is one is easy to start because it begins with "do this computation." What part is holding you back? Putting the givens into the equations? The matrix addition and multiplication?
Any real eigenvalue is $\pm 1,$ and it is $-1$ as you are in dimension 3 with negative determinant. And there is a real eigenvalue because the dimension is odd, the degree pf the characteristic polynomial is odd.
So, there is a -1 eigenvector $v,$ if the matrix is called $M$ we have $Mv=-v.$
$M$ also preserves angles, easy enough to prove. So, the plane orthogonal to $v$ is setwise fixed. Note that $M$ may also cause a rotation within that plane.
For example, compare
$$ M \; = \;
\left( \begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}
\right)
$$
versus
$$ M \; = \;
\left( \begin{array}{rrr}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & -1
\end{array}
\right)
$$
versus
$$ M \; = \;
\left( \begin{array}{rrr}
\frac{1}{\sqrt 2} & \frac{-1}{\sqrt 2} & 0 \\
\frac{1}{\sqrt 2} & \frac{1}{\sqrt 2} & 0 \\
0 & 0 & -1
\end{array}
\right) .
$$
Thought question, or called a "Gesundheit" experiment in German, what if $M=-I?$ Possibly Gedanken experiment, or Weltanschauung. Could be Schadenfreude.
Best Answer
If $\bf n$ is a unit normal to the plane (as a column vector), the reflection across that plane is $I - 2 {\bf n} {\bf n}^T$.
EDIT: In your case $${\bf n} = \pmatrix{1/\sqrt{3}\cr 1/\sqrt{3}\cr 1/\sqrt{3}\cr}$$