The fact that A is NOT diagonalizable means that it does NOT have a complete set of eigenvectors. You need to form S from eigenvectors as much as possible and use "generalized" eigenvectors to fill out S.
In your example, the eigenvalue equation is
\left|\begin{array}{cc} 5-\lambda & 1 \\ -1 & 3-\lambda\end{array}\right|
= \lambda^2- 8\lambda+ 16= (\lambda- 4)^2= 0[/itex] <br />
so 4 is a double eigenvalue (which you knew). <br />
<br />
An eigenvector corresponding to eigenvalue 4 must satisfy <br />
\left[\begin{array}{cc}5 &amp; 1 \\ -1 &amp; 3\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{c}4x \\ 4y\end{array}\right]<br />
and so must satisfy 5x+ y= 4x and -x+ 3y= 4y, both of which reduce to y= -x. Any vector of the form <x, -x>= x<1, -1> is an eigenvector. But that's only one eigenvector which is why we cannot diagonalize this matrix.<br />
<br />
Now, every matrix satisfies it own characteristic equation: (A- 4I)^2= 0 which means (A- 4I)^2v= 0 for every vector v. Obviously, if v is an eigenvector with eigenvalue 4, (A- 4I)v= 0 so (A- 4I)(A- 4I)v= (A- 4I)0= 0. But it might also be possible that (A-4I)v is not 0 but (A- 4I)((A-4I)v)= 0. Such a vector is a "generalized" eigenvector In that case, (A-4I)v must be an eigenvector! To find another vector such that (A- 4I)^2v= 0, we must find a vector such that (A-4I)v= <1, -1>. That gives the equation<br />
\left[\begin{array}{cc}5- 4 &amp; 1 \\ -1 &amp; 3- 5\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{cc}1 &amp; 1 \\ -1 &amp; -2\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right]= \left[\begin{array}{c}1 \\ -1\end{array}\right]<br />
which gives the two equations x+ y= 1 and -x- 2y= -1. Adding the two equations, -y= 0 so y= 0. Then x= 1. A "generalized" eigenvector is <1, 0><br />
Take<br />
S= \left[\begin{array}{cc} 1 &amp; 1 \\ -1 &amp; 0\end{array}\right]<br />
taking the eigenvector and "generalized" eigenvector as columns. Then<br />
S^{-1}= \left[\begin{array}{cc} 0 &amp; -1 \\ 1 &amp; 1\end{array}\right]<br />
and<br />
S^{-1}AS= \left[\begin{array}{cc} 0 &amp; -1 \\ 1 &amp; 1\end{array}\right]\left[\begin{array}{cc}5 &amp; 1 \\ -1 &amp; 3\end{array}\right]\left[\begin{array}{cc} 1 &amp; 1 \\ -1 &amp; 0\end{array}\right]<br />
= \left[\begin{array}{cc}1 &amp; -3 \\ 4 &amp; 4\end{array}\right]\left[\begin{array}{cc} 1 &amp; 1 \\ -1 &amp; 0\end{array}\right]= \left[\begin{array}{cc} 4 &amp; 1 \\ 0 &amp; 4\end{array}\right]<br />
<br />
If your matrix, A, had \lambda as a <b>triple</b> eigenvalue but only one eigenvector corresponding to that eigenvalue, say, v_1, then you would look for a vector v_2 such that (A-\lambda I)v_2= v_1 and a vector v_3 such that (A- \lambda I)v_3= v_2.<br />
<br />
By the way, your title "Symmetric Matrices to Jordan Blocks" puzzled me. All symmetric (real) matrices are <b>diagonalizable</b> so the question of Jordan Blocks doesn't arize with them. And, of course, your example is not a symmetric matrix.
