Jacobi Method and Frobenius Norm Question. I have this linear algebra question concerning the jacobi method and the frobenius norm that I am having a lot of trouble on, I have an exam soon and I would appreciate any help. NOTE: I have read the entire article of wikipedia on jacobi method and frobnenius norm but I have just started the linear algebra topic and there's only so much I can make out of it.
Consider the following system of equations...
$$
\left\{ 
\begin{array}{c}
2x + y =5\\
x - 3y =6
\end{array}
\right. 
$$
1. Compute the Frobenius norm for the Matrix C under the Jacobi iterative method(would you expect this problem to converge)?
2.Starting at (x, y) = 
(0, 0) perform 2 iterations of Jacobi Iterative method for this problem.
I dont understand what the Frobenius norm is exactly, so I started at question 2...
First I converted the system of equations to the respective augmented matrix...
$$
        \begin{bmatrix}
        2 & 1 & 5 \\
        1 & -3 & 6 \\
        \end{bmatrix}
$$
From here I can see that...
$$
\left\{ 
\begin{array}{c}
x^{new} = (-y^{old} + 5)/2\\
y^{new} = (6 - x^{old})/-3
\end{array}
\right. 
$$
So I set up the array...
$$
\begin{array}{c|lcr}
Iteration & \text{x^{old}} & \text{y^{old}} & \text{x^{new}} & \text{y^{new}}\\
\hline
1 & 0 & 0 & 2.5 & -2 \\
2 & 2.5 & -2 & 3.5 & -7/6 \\
\end{array}
$$
And that is my answer to number 2, any help with how to do number 1 would be greatly appreciated, thank you for your time.
 A: Hint: The Frobenius norm of an $m \times n$ matrix $A$ is defined as the square root of the sum of the absolute squares of its elements.
Example: Consider matrix 
$$A = \left(
  \begin{array}{ccc}
    ~~~~1 & -2 & ~~~~~~3 \\
    -4 & ~~5 & ~-6 \\
    ~~~7 & -8 & ~~~~~~9 \\
  \end{array}
\right)$$
$\lVert A \rVert _{F} = \sqrt (|1|^2+|-2|^2+|3|^2+|-4|^2+|5|^2+|-6|^2+|7|^2+|-9|^2 + |-8|^2) = \sqrt(285) = 16.8819$
For convergence of Jacobi iteration method you  need to find iteration matrix $P = -D^{-1}(L+U)$. Note that the Jacobi iterative scheme will converge  if $\lVert P\rVert_{F} $ is strictly less than $1$, where $F$ stands for the Frobenius norm. There may be some other matrix norm (such as the $1$-norm ) that is strictly less than $1$, in which case convergence is still guaranteed. In any case, however, the condition $\lVert P\rVert <1 $ is only a sufficient condition for convergence, not a
necessary one.
For the given example
$D = \left(
  \begin{array}{ccc}
    ~~1 & ~0 & ~0 \\
    ~~0 & ~5 & ~0 \\
    ~~0 & ~0 & ~9 \\
  \end{array}
\right)$ 
$L = \left(
  \begin{array}{ccc}
    ~~~0 & ~~0 & ~0 \\
    -4 & ~~0 & ~0 \\
    ~~~~7 & -8 & ~0 \\
  \end{array}
\right)$ 
$U = \left(
  \begin{array}{ccc}
    ~~0 & -2 &~~3 \\
    ~~0 & ~~~0 & -6 \\
    ~~0 & ~~~0 & ~~~0 \\
  \end{array}
\right)$ 
Added: Consider to solve  $3\times 3$ size system of linear equation $Ax = b$, where coefficient matrix 
$A = \left(
  \begin{array}{ccc}
      a_{11} & a_{12} & a_{13} \\
    a_{21}& a_{22} & a_{23} \\
    a_{31}& a_{32} & a_{32} \\
  \end{array}
\right)$
Assume coefficient matrix $A$ has no zeros on its main diagonal i.e. $a_{11}$, $a_{22}$, $a_{23}$ are non zeros, then 
$D = \left(
  \begin{array}{ccc}
    ~~\frac{1}{a_{11}} & 0 & 0 \\
    0 & \frac{1}{a_{22}} & ~0 \\
    ~~0 & 0 & \frac{1}{a_{33}}\\
  \end{array}
\right)$ 
$L = \left(
  \begin{array}{ccc}
    0 & 0 & 0 \\
    a_{21} & 0 & ~0 \\
   a_{31}& a_{32} & 0\\
  \end{array}
\right)$ 
$U = \left(
  \begin{array}{ccc}
    0 & a_{12} & a_{13} \\
    0 & 0 & a_{23} \\
   0 & 0 & 0\\
  \end{array}
\right)$ 
A: To solve $Ax=b$, you can split $A=D+R$, where $D$ is the diagonal, and $R=A-D$. If $D$ is sufficiently 'dominant', the Jacobi method rewrites the equation as a fixed point computation as in $(D+R)x = b$, $D x = (b-Rx)$, and then $x= D^{-1}(b-Rx)$. I am guessing that by $C$ you mean $C=D^{-1} R$, so the equation becomes $x = D^{-1} b -Cx$.
If we write $f(x) = D^{-1} b -Cx$, then a sufficient condition for the fixed point iteration to converge is that $f$ is Lipschitz with rank less than one. We have $\|f(x)-f(y) \| = \|C (x-y) \|$ for any norm, if we choose a norm that satisfies $\|Ax\|\le \|A\| \|x\|$ (a 'compatible' norm), then we have $\|f(x)-f(y) \| \le \|C\| \|x-y\|$, and if that norm satisfies $\|C\| < 1$, then we see that $f$ is a contraction map and the fixed point iteration will converge to the unique fixed point (which is a solution of $Ax=b$).
The Frobenius norm is compatible with the Euclidean norm, and is easy to compute. That is, we have $\|A x \|_2 \le \|A \|_F \|x\|_2$, and $\|A\|_F = \sqrt{\sum_i \sum_j [A]_{ij}^2}$.
In your case, $C = \begin{bmatrix} 2&0\\0&-3\end{bmatrix}^{-1} \begin{bmatrix} 0&1\\1&0\end{bmatrix} = \begin{bmatrix} 0& \frac{1}{2}\\ -\frac{1}{3} &0\end{bmatrix}$, and $\|C\|_F = \sqrt{\frac{1}{9}+\frac{1}{4}} < 1$. Hence the scheme will converge.
