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Okay, I'm given

$$P=\begin{bmatrix} 1 & 1 \end{bmatrix}^T$$

and given 4 different sections of problem. Before I continue, I want make sure my suspicion is correct:

$$P=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

Right? The rest of the question will work on that assumption until I'm corrected. The first section is to compute $PP^T$ and $P^TP$ which I've computed (and confirmed on Wolfram|Alpha) as

$$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

and

$$\begin{bmatrix} 2 \end{bmatrix}$$

The second section, which is where my main trouble is, tells me to compute

$$M_P=I-{2 \over P^TP}PP^T$$

where $I$ is the 2x2 identity matrix

$$I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Using the above, ${2 \over P^TP}={2 \over \begin{bmatrix} 2 \end{bmatrix}}$ and I don't understand where to go from there to work on the problem. I barged ahead, assuming it would simplify to $1$ and found that this made one of the next sections impossible (prove that $M_P$ is invertible).

Nakaan
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  • I'm really suspecting that $[2]$ is, well, $2$. When you take a dot product of two vectors, doesn't that amount to creating a $1 \times 1$ matrix, but expressing it as a scalar? – John Feb 06 '14 at 22:12

1 Answers1

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You were right; just simplify $\displaystyle \frac{2}{[2]} = 1$. This gives you $$ M_P = I - 1 \left[ \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right] = \left[ \begin{array}{cc} 0 & -1 \\ -1 & 0 \\ \end{array} \right] $$ which is indeed invertible, just as the next section requires. (Try squaring it.)

Caleb Stanford
  • 45,895
  • I snagged a friend of mine to help me while waiting for answers here (attack on all fronts, so to speak) and we managed to work through it, coming up with $A^{-1} = A$ as well. Thank you! Edit: finally figured out how to group characters to be super-scripted. – Nakaan Feb 06 '14 at 22:24