# Reflection Matrix linear algebra

I am practicing some linear algebra question to prepare for my test. I have come across one question that has given me much trouble. It states:

If $\lVert u\rVert = 1$, then $Q = I - 2uu^T$ is a reflection matrix.

I have established this:

If you reflect twice of across the same axis twice $Q^2 = I$.

First, it says to compute $Qu$ and simplify as much as possible. Does this just mean move the equation around to get $Qu$?

Suppose $v$ is orthogonal to $u$. Compute $Qv$. Im not sure how this is done.

Then im asked to explain in plain English which subspace $Q$ is reflecting across.

The last one states:

Compute the reflection matrix $Q_1 = I - 2u_1 u_1^T$ where $u_1 = (0,1)$. Compute $Q_1 x_1$, where $x_1 = (0,1)$ and sketch the vectors $u1$, $x1$ and $Q_1x_1$ in the plane.

I'm not sure how to even start that one.

If anyone could clear this up for me that would be much help !

• Please review my edits to see if I interpreted your text correctly, and also take note of correct Mathjax usage. – dfeuer Oct 31 '13 at 16:38
• Yes that all makes sense. – user104536 Oct 31 '13 at 16:39

## 1 Answer

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?

• Thanks for the post! I actually did some some work after asking this question and got the same thing you got for the first two. I just did not reduce it anymore. I am having trouble getting it any smaller. Would u⊤u go to I or is that only for the dot product? – user104536 Oct 31 '13 at 17:39
• That question is answered within my solution. It looks like you have trouble with matrix multiplication, possibly. You should notice that $x^\top y=x\cdot y$. – rschwieb Oct 31 '13 at 17:41
• Thats what I was thinking. Thats why I am having trouble reducing this even more – user104536 Oct 31 '13 at 17:44