The key here is that it shuffles partition matricies.
The hw by example is done like so.
compute:
\(\begin{pmatrix}0 & 0 & 1\\ 1 & 0 & 0\\0 & 1 & 0\end{pmatrix} \begin{pmatrix}-2 & 3 & -1\\ 1 & 2 & 0\\2 & 1 & 3\end{pmatrix}\)
Rather than use R to calculate the result, we can look at each row of the left side. It says the result of the top row is row 2 of second matrix. Likewise, middle row is row 3 of second and bottom row is row 1 of second.
Put another way.
\[ \left( \begin{array}{c|c|c} 0 & 1 & 0 \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{c|c|c} -2 & 1 & 2 \\ \hline 3 & 2 & 1 \\ \hline -1 & 0 & -2 \\ \end{array} \right) \] = \[ \left( \begin{array}{c} 0 x (-2\>1\>2) + 1x(3\>2\>1) + 0x(-1\>0\>-3) \\ \hline 0 x (-2\>1\>2) + 0x(3\>2\>1) + 1x(-1\>0\>-3) \\ \hline 1 x (-2\>1\>2) + 0x(3\>2\>1) + 0x(-1\>0\>-3) \\ \end{array} \right) \] = \[ \left( \begin{array}{c} 3\>2\>1 \\ \hline -1\>0\>-3 \\ \hline -2\>1\>2 \\ \end{array} \right) \] = \[ \left( \begin{array}{ccc} 3 & 2 & 1 \\ -1 & 0 & -3 \\ -2 & 1 & 2 \\ \end{array} \right) \]
Check by simply doing the matrix multiplication in R
a=matrix(c(0,0,1,1,0,0,0,1,0),ncol=3)
b=matrix(c(-2,3,-1,1,2,0,2,1,-3),ncol=3)
print(a%*%b)## [,1] [,2] [,3]
## [1,] 3 2 1
## [2,] -1 0 -3
## [3,] -2 1 2Its odd, that he says compute the results. Perhaps there is something in matlab which does this easier than in R.
HOMEWORK 7.2.3.2 (1/3 points)
Example: If
p = \[ \left( \begin{array}{c} 0 \\ 1 \\ 2 \\ 3 \\ \end{array} \right) \]
then P(p) =
\[
\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array} \right)
\]
Throughout, P(p) is the permutation matrix that orders the components of the vector to which it is applied according to the permutation vector p.
if p = \[ \left( \begin{array}{c} 3 \\ 2 \\ 1 \\ 0 \\ \end{array} \right) \]
then P(p) =
\[
\left( \begin{array}{cccc}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
\end{array} \right)
\]
if p = \[ \left( \begin{array}{c} 1 \\ 0 \\ 2 \\ 3 \\ \end{array} \right) \]
then P(p) =
\[
\left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array} \right)
\]
if p = \[ \left( \begin{array}{c} 1 \\ 2 \\ 3 \\ 0 \\ \end{array} \right) \]
then P(p) =
\[
\left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
\end{array} \right)
\]
(He uses the word compute here as well?)
HOMEWORK 7.2.3.4 (1 point possible)
Example: Let
p = \[ \left( \begin{array}{c} 2 \\ 0 \\ 1 \\ \end{array} \right) \]
be a permutation vector and P=P(p) be a permutation matrix. Compute
\[
\left( \begin{array}{cccc}
-3 & 1 & 2 \\
3 & 2 & 1 \\
-1 & 0 & -3 \\
\end{array} \right)P^T =
\]
solution
# Ok, I am assuming since he is saying permutation matrix then
# I have to have a permutation matrix in play.
permutation_matrix=matrix(c(1,0,0,0,1,0,0,0,1),ncol=3)
permutation_matrix## [,1] [,2] [,3]
## [1,] 1 0 0
## [2,] 0 1 0
## [3,] 0 0 1# this is given for p...
p=matrix(c(2,0,1),ncol=1)
p## [,1]
## [1,] 2
## [2,] 0
## [3,] 1# But, I need a 3x3 matrix and I know that given p above
# it really translates to:
p=matrix(c(0,0,1, 1,0,0, 0,1,0),ncol=3)
p## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0#P=P(p)
P= permutation_matrix %*% p
P## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 1 0 0Ptranspose=t(P)
Ptranspose## [,1] [,2] [,3]
## [1,] 0 0 1
## [2,] 1 0 0
## [3,] 0 1 0a=matrix(c(-2,3,-1,1,2,0,2,1,-3),ncol=3)
a## [,1] [,2] [,3]
## [1,] -2 1 2
## [2,] 3 2 1
## [3,] -1 0 -3solution=a %*% Ptranspose
solution## [,1] [,2] [,3]
## [1,] 1 2 -2
## [2,] 2 1 3
## [3,] 0 -3 -1