MHB LU decomposition: Total pivoting

[mathmari]

Hey!

I want to determine the LU decomposition of
$A=\begin{pmatrix}0 & 2 & 1\\1 & 10 & 1 \\1 & 1 & 1\end{pmatrix}$ with total pivoting. I have done the following:

The biggest element of the whole matrix is $10$, so we exchange the first two rows and the first two columns and then we get $\begin{pmatrix}10 & 1 & 1 \\2 & 0 & 1\\1 & 1 & 1\end{pmatrix}$.
Applying now the Gauss algorithm we get $\begin{pmatrix}10 & 1 & 1 \\0 & -\frac{1}{5} & \frac{4}{5}\\0 & \frac{9}{10} & \frac{9}{10}\end{pmatrix}$.
The biggest element of the submatrix is $\frac{9}{10}$ and so we exchange the last two rows and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & -\frac{1}{5} & \frac{4}{5} \end{pmatrix}$. Now we apply the Gauss algorithm and get: $\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.

The matrix $U$ is the resulting matrix, $U=\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.

The matrix $L$ is $L=P\cdot P_0\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$, or not? (Wondering)

The matrices $G_i^{-1}$ are defined as:
$$G_1^{-1}=\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix} \ \text{ and } \ G_2^{-1}=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}$$ or not? (Wondering)


Are the matrices $P_i$ defined as follows?
$$P_0=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$ since this describes the step at which we exchanged the first two rows and the first two columns. (Wondering)
$$P_1=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}$$ since this describes the step at which we exchanged the two last rows. (Wondering) If these are correct, it doesn't hold that $LU=PA$, does it? (Wondering)

 

[I like Serena]

Are the matrices $P_i$ defined as follows?
$$P_0=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$ since this describes the step at which we exchanged the first two rows and the first two columns. (Wondering)
$$P_1=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}$$ since this describes the step at which we exchanged the two last rows.

Hey mathmari!

It should be clear that
$$P_0=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$$
cannot be correct. It is the identity matrix, isn't it? (Worried)

Let's define instead:
$$P_0=\begin{pmatrix}0 & 1 & 0 \\1 & 0 & 0\\0 & 0 & 1\end{pmatrix}$$
This is the matrix that swaps either 2 rows or 2 columns, isn't it?
And it's its own inverse. That is, $P_0^{-1}=P_0$.

Swapping 2 rows is a permutation on the left.
However, swapping 2 columns is a permutation on the right.
So after those 2 permutations we have $P_0 A P_0$. (Thinking)

If these are correct, it doesn't hold that $LU=PA$, does it?

Shouldn't it be $LU=PAQ$, where $P$ reorders the rows of $A$ and $Q$ reorders the columns of $A$? (Wondering)

The matrix $L$ is $L=P\cdot P_0\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$, or not?

I don't think so. We have:
$$R = G_2 \cdot P_1 \cdot G_1 \cdot P_0 \cdot A \cdot P_0$$
From this we have to deduce $L$ such that:
$$LR = PAQ$$
Don't we? (Wondering)

 

[mathmari]

We have $R=\begin{pmatrix}10 & 1 & 1 \\ 0 & \frac{9}{10} & \frac{9}{10} \\ 0 & 0 & 1 \end{pmatrix}$.

Since $R = G_2 \cdot P_1 \cdot G_1 \cdot P_0 \cdot A \cdot P_0$ we get that $$L\cdot G_2 \cdot P_1 \cdot G_1 \cdot P_0 \cdot A \cdot P_0=P\cdot A\cdot P_0 \Rightarrow L =P\cdot P_0^{-1}\cdot G_1^{-1}\cdot P_1^{-1}\cdot G_2^{-1} \Rightarrow L=P_1\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$$

We have the matrices $$G_1^{-1}=\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix} \ \text{ and } \ G_2^{-1}=\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}$$ We also have the matrices $$P_0=\begin{pmatrix}0 & 1 & 0 \\1 & 0 & 0\\0 & 0 & 1\end{pmatrix} \ \text{ and } \ P_1=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}$$

So we get $$L=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ \frac{1}{10} & 1 & 0 \\ \frac{1}{5} & -\frac{2}{9} & 1\end{pmatrix}$$

Is this correct? (Wondering)

 

[I like Serena]

Since $R = G_2 \cdot P_1 \cdot G_1 \cdot P_0 \cdot A \cdot P_0$ we get that $$L\cdot G_2 \cdot P_1 \cdot G_1 \cdot P_0 \cdot A \cdot P_0=P\cdot A\cdot P_0 \Rightarrow L =P\cdot P_0^{-1}\cdot G_1^{-1}\cdot P_1^{-1}\cdot G_2^{-1} \Rightarrow L=P_1\cdot G_1^{-1}\cdot P_1\cdot G_2^{-1}$$

How did you get $P\cdot P_0^{-1}=P_1$ ?
I think that the $P$ we will get, will be incorrect. (Worried)

So we get $$L=\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\\frac{2}{10} & 1 & 0\\\frac{1}{10} & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & -\frac{2}{9} & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ \frac{1}{10} & 1 & 0 \\ \frac{1}{5} & -\frac{2}{9} & 1\end{pmatrix}$$

Is this correct?

Yes. (Nod)

 

[mathmari]

How did you get $P\cdot P_0^{-1}=P_1$ ?
I think that the $P$ we will get, will be incorrect. (Worried)

Do we not have $P=P_1P_0 \Rightarrow PP_0^{-1}=P_1$ ? (Wondering)

 

[I like Serena]

Do we not have $P=P_1P_0 \Rightarrow PP_0^{-1}=P_1$ ?

Ah yes, $P$ has to be whatever row permutation it takes to ensure that $L$ is a lower triangular matrix.

We have:
$$L =P\cdot P_0^{-1}\cdot G_1^{-1}\cdot P_1^{-1}\cdot G_2^{-1}$$
And $G_1^{-1}$ and $G_2^{-1}$ are lower triangular.
Since $P_1\cdot G_1^{-1}\cdot P_1^{-1}$ is a conjugation, it is lower triangular as well.
That is, we swap 2 rows and we also swap the corresponding 2 columns, so that the triangular form is retained. (Nerd)

So we can indeed pick $\smash{P\cdot P_0^{-1}=P_1}$.
I must have made a calculation mistake earlier. (Blush)