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


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.