<s>[INST] Eigen Decomposition Check I am following the wiki entry on eigen dicomposition with the following matrix:
$$A = \begin{pmatrix}
  0 & 1  \\
  1 & 0  \\
 \end{pmatrix}$$
I wish to find a diagonalizing matrix T.S.
$$T^{-1}AT=\Lambda$$
$$AT=T\Lambda$$
where,
$$\Lambda = \begin{pmatrix}
  x & 0  \\
  0 & y  \\
 \end{pmatrix}$$
$$T = \begin{pmatrix}
  a & b  \\
  c & d  \\
 \end{pmatrix}$$
I am following with the wiki:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=x\begin{pmatrix}a \\c \\\end{pmatrix}$$
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=y\begin{pmatrix}b \\d \\\end{pmatrix}$$
and the eigenvalues are:
$$\left|A-I\lambda\right| = \left|\begin{pmatrix}-\lambda & 1  \\1 & -\lambda  \\\end{pmatrix}\right|=(\lambda^2-1)=(\lambda-1)(\lambda+1)$$
Hence, $\lambda_{1,2}=\pm 1$
Now I am finding the eigenvectors with:
$$A\begin{pmatrix}a \\c \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}a\\c \\\end{pmatrix}=\begin{pmatrix}c \\a \\\end{pmatrix}=1\begin{pmatrix}a \\c \\\end{pmatrix}$$
Hence, $a=c$. And,
$$A\begin{pmatrix}b \\d \\\end{pmatrix}=\begin{pmatrix}0 & 1\\1 & 0 \\\end{pmatrix}\begin{pmatrix}b\\d \\\end{pmatrix}=\begin{pmatrix}d \\b \\\end{pmatrix}=-1\begin{pmatrix}b \\d \\\end{pmatrix}$$
Hence, $b=-d$.
So I take $T$ to be:
$$T = \begin{pmatrix}
  a & b  \\
  a & -b  \\
 \end{pmatrix}=\begin{pmatrix}
  1 & 1  \\
  1 & -1  \\
 \end{pmatrix}$$
with:
$$T^{-1} = \frac{1}{det(T)}\begin{pmatrix}
  1 & -1  \\
  -1 & -1  \\
 \end{pmatrix}=\frac{1}{-1-1}\begin{pmatrix}
  1 & -1  \\
  -1 & -1  \\
 \end{pmatrix}=\frac{1}{-2}\begin{pmatrix}
  1 & -1  \\
  -1 & -1  \\
 \end{pmatrix}=\frac{1}{2}\begin{pmatrix}
  -1 & 1  \\
  1 & 1  \\
 \end{pmatrix}$$
Now, when I try to calculate $T^{-1}AT$ I get:
$$\Lambda =\frac{1}{2}\begin{pmatrix}
  -1 & 1  \\
  1 & 1  \\
 \end{pmatrix}\begin{pmatrix}
  0 & 1  \\
  1 & 0  \\
 \end{pmatrix}\begin{pmatrix}
  1 & 1  \\
  1 & -1  \\
 \end{pmatrix}=\frac{1}{2}\begin{pmatrix}
  1 & -1  \\
  1 & 1  \\
 \end{pmatrix}\begin{pmatrix}
  1 & 1  \\
  1 & -1  \\
 \end{pmatrix}=\frac{1}{2}\begin{pmatrix}
  0 & -2  \\
  2 & 0  \\
 \end{pmatrix}=\begin{pmatrix}
  0 & -1  \\
  1 & 0  \\
 \end{pmatrix}$$
Where have I gone wrong here?
 [/INST] As a good check you can compute $TT^{-1}$, and you'll find you don't get the identity matrix. So the mistake must be in your computation of $T^{-1}$.
It looks to me like you've set it up wrong. Seems you're trying to use the formula $$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$$ but you've mixed up some of the terms. 
</s>