<s>[INST] Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal. Find the eigenvalues and eigenvectors of the linear transformation $T(x,y,z)=(x+y,x-y,x+z)$. Verify that the eigenvectors are orthogonal.
Part A:
$$T(x,y,z)=\begin{pmatrix} 1 & 1 & 0 \\ 1 & -1 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ \end{pmatrix}$$
\begin{equation*}
\begin{split}
det(T(x,y,z)-I_n \lambda ) & =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =(1-\lambda)\begin{vmatrix} -1-\lambda & 0 \\ 0 & 1-\lambda \\ \end{vmatrix}-\begin{vmatrix} 1 & 0 \\ 1 & 1-\lambda \\ \end{vmatrix}=(1-\lambda)^2(-1-\lambda)-(1-\lambda) \\
& =\begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & -1-\lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix} \\
& =-(1-\lambda)[(1-\lambda)(1+\lambda)+1] \\
& =-(1-\lambda)[1-\lambda^2+1] \\
& =(\lambda-1)[2-\lambda^2]=0 \\
\end{split}
\end{equation*}
Hence $\lambda=1, \pm \sqrt{2}$.
When $\lambda=1$:
$\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\  0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1,0)^T, (0,0,1)^T \}$. 
When $\lambda=\sqrt{2}$:
$\begin{pmatrix} 1-\sqrt{2} & 1 & 0 \\ 1 & -1-\sqrt{2} & 0 \\ 1 & 0 & 1-\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1-\sqrt{2} \\  0 & 1-\sqrt{2} & 1-\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,\sqrt{2}-1,0)^T, (1-\sqrt{2},1+\sqrt{2},0)^T \}$. 
When $\lambda=-\sqrt{2}$:
$\begin{pmatrix} 1+\sqrt{2} & 1 & 0 \\ 1 & -1+\sqrt{2} & 0 \\ 1 & 0 & 1+\sqrt{2} \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 1+\sqrt{2} \\  0 & -1+\sqrt{2} & 1+\sqrt{2} \\ 0 & 0 & 0 \\ \end{pmatrix}$
Hence the eigenvectors for $\lambda =1$ are $\{ (1,0,0)^T, (0,1-\sqrt{2},0)^T, (1+\sqrt{2},1-\sqrt{2},0)^T \}$. 
Are my eigenvectors correct? Is there a better way to find see if the two vectors are orthogonal other than using the cross-product method?
 [/INST] Eigenvectors solve the equation $(A-\lambda I) v = 0$; eigenvectors are vectors, not matrices.
For $\lambda = 1$, you should have
$$\begin{align*}
(A-I )v &= 0 \\
\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{align*}$$
This means that the vector $v$ is in the null space of the matrix $\begin{pmatrix} 0 & 1 & 0 \\ 1 & -2 & 0 \\ 1 & 0 & 0 \end{pmatrix}.$ It should be apparent that any vector of the form $\begin{pmatrix} 0 \\ 0 \\ v_3\end{pmatrix}$ satisfies this relation. We shall choose $v_3 = 1$, giving us the eigevector $v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ corresponding to $\lambda = 1$.
Repeat this process for your other matrices, and be sure to perform the matrix subtraction properly. You should find that the eigenvectors are simply the unit vectors.
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