Eigen Decomposition Check

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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?

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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.

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Your answer is correct except that you forgot to switch the elements on the diagonal of $T$ when finding its inverse.