# What will H(^T)*H be?

1. Oct 12, 2012

### hivesaeed4

Problem:

If P is a n x 1 matrix such that P( ^T)*P = 1 and H = I - 2PP(^T), then H(^T)*H is ?

Note :

A(^T) means the transpose of matrix A.

Confusion:

What I don't get is are we supposed to do I (identity matrix-I'm assuming of order 1 x 1) - 2 ? If that is the way we're supposed to do the question then are we supposed to treat 2 as a 1 x 1 matrix. If that's the case then is the following correct:

Since P( ^t)*P = PP(^T) = 1 = [1]

So,

H = I - 2PP(^T) = I - 2PP(^T) = I - 2[1] = [1] - [2] =[-1]

thus,

H(^T)*H = [-1] * [-1](^T) = [-1] * [-1] = [1]

Help?

2. Oct 12, 2012

### Simon Bridge

$$\renewcommand{\vb}[1]{\mathbf{#1}} \renewcommand{\hb}[1]{\hat{\mathbf{#1}}}$$

If P is a n x 1 matrix such that P( ^T)*P = 1 and H = I - 2PP(^T), then H(^T)*H is ?

If P is nx1, then it is a vector p.
so, switching to LaTeX, we have $\vb{p}$ so that $\vb{p}^t\vb{p}=1$ so it is a unit vector?

$\vb{H}=\vb{I}-2\vb{p}\vb{p}^t$

Then $\vb{H}^t\vb{H}=?$

So does this translate to 1x1 matrices ... then
$p_{(1\times 1)}=1$, and $p^tp=1\times 1=1$, and
$H_{(1\times 1)}=I-2p^tp = 1-2=-1$ so
$H^tH = (-1)(-1)=1=I_{(1\times 1)}$
... seems reasonable to me so far.
The trick is to articulate your confusion.
The above is consistent with the matrix formulation if $\vb{H}^t\vb{H}=\vb{I}$ for any $\vb{p}$.

So can you find a $\vb{p}$ so that:
$\vb{H}^t\vb{H}\neq\vb{I}$

... but I suspect the central question is whether it is (or "when is it...") sensible to treat a scalar as a 1x1 matrix.

For instance - you can multiply a nxn matrix by a scalar - but can yout multiply an nxn matrix by a 1x1 matrix?

3. Oct 12, 2012

### Vaedoris

$PP^T$ is an n-by-n matrix. (prove: $P$ is n-by-1 and $P^T$ is 1-by-n).

So $PP^T\neq P^TP$ unless if n = 1.

$H$ is symmetric.

$H^TH=I_{n}$ which has dimension n-by-n.

4. Oct 12, 2012

### Muphrid

The linear operator you have here is equivalent to

$$\underline H(a)=a-2(p \cdot a)p$$

This is a reflection operator, and since it is symmetric, all you have to do is think about what you get when you reflect twice.

5. Oct 12, 2012

### Vaedoris

Reflecting twice off the same "surface" gets you the original vector back.

6. Oct 12, 2012

### AlephZero

@Simon Bridge, I think you misread the question or made typo part way through your answer. if should be $H = I - 2pp^t$ not $H = I - 2 p^t p$.

@the OP, as post #3 said, $pp^T$ is symmetric, so $H$ is symmetric. So $H^T = H$.

To answer the question, you just have to multiply out $H^T H$ = $H.H$ and simplify it using $P^T P = 1$.

2P means a scalar (2) times a matrix (P). The matrix I has the same dimension as $P P^T$, which is n by n.

7. Oct 12, 2012

### hivesaeed4

Thanks.

8. Oct 13, 2012

### Simon Bridge

<sigh> typo - thanks Aleph.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}