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?

Transformation T(x,y,z) is defined  as in the  question posted $$
 T(x,y,z) =
  \left[ {\begin{array}{cc}
   1 & 1 & 0\\
   1 &-1 & 0\\
   1 & 0 & 1\\ 
  \end{array} } \right]\left[ {\begin{array}{cc}
   v_1\\
   v_2\\
   v_3\\ 
  \end{array} } \right]
$$ 
which is of the form $T(x,y,z)=Ax.$ Such that $A=\left[ {\begin{array}{cc}
   1 & 1 & 0\\
   1 &-1 & 0\\
   1 & 0 & 1\\ 
  \end{array} } \right]$ and the vector $x =\left[ {\begin{array}{cc}
   v_1\\
   v_2\\
   v_3\\ 
  \end{array} } \right] $ respectively. For a matrix A of 3*3 order,there are three Eigen values that can be determined from it. For each Eigen value, a vector is determined which is named as Eigen vector, hence three vectors are determined from three Eigen values. Meaning of Eigen values and Eigen vectors can be well-described as below: As mentioned above, Eigen values of A can be determined by the following the equation $$ Ax = \lambda I_{n} x$$ where $I_n$ is the unit matrix with n as the order of square matrix involved. Here, $n =3.$ From this equation, It can be understood that the Matrix $A$ operates on a vector $x$ so as to result in a vector which is parallel to the original vector $x.$ And $\lambda$ is any scalar that operates on the vector $x.$ It can be rephrased as: what are vectors (here, say three values for vectors $x$) for which Matrix $A$ operates on each of them to give a vector parallel to it. Upon simplifying the above equation, we get the equation (i), $$(A-\lambda I_3)x=0$$ $$ det\left|A-\lambda x\right|=0$$ $$\left| {\begin{array}{cc}
   1-\lambda & 1 & 0\\
   1 &-1-\lambda & 0\\
   1 & 0 & 1-\lambda\\ 
  \end{array} } \right| =0 $$
solving this equation results in third order polynomial of the following form (calculations procedure shown in the question is sufficient),$$(\lambda -1)(2-\lambda^{2})=0.$$ obviously, $\lambda=1,\sqrt{2},-\sqrt{2}.$  Case 1: $\lambda=1$$$\left| {\begin{array}{cc}
   0 & 1 & 0\\
   1 &-2 & 0\\
   1 & 0 & 0\\ 
  \end{array} } \right|\left[ {\begin{array}{cc}
   v_1\\
   v_2\\
   v_3\\ 
  \end{array} } \right] =\left[ {\begin{array}{cc}
   0\\
   0\\
   0\\ 
  \end{array} } \right]$$ On solving this equation, we get vector $x$ is of the form $\left[ {\begin{array}{cc}
   0\\
   0\\
   v_3\\ 
  \end{array} } \right] $ such that $v_3$ is independent of both $v_1$ and $v_2$
.By choosing $v_3=1$, results in Eigen vector $(\epsilon_1)$  $\left[ {\begin{array}{cc}
   0\\
   0\\
   1\\ 
  \end{array} } \right]$ corresponding to Eigen value $\lambda_1=1.$ By substituting other two Eigen values, corresponding Eigen vectors are obtained. Case 2: $\lambda=\sqrt{2}$ Eigen vector, $\epsilon_2$ is $\left[ {\begin{array}{cc}
   1\\
   \sqrt{2}-1\\
   \sqrt{2}+1\\ 
  \end{array} } \right].$ 
Case 3: $\lambda=-\sqrt{2}$ Eigen vector, $\epsilon_3$ is $\left[ {\begin{array}{cc}
   1\\
   -\sqrt{2}-1\\
   1-\sqrt{2}\\ 
  \end{array} } \right].$  There are three vectors $\epsilon_1$,$\epsilon_2$ and $\epsilon_3$ obtained as Eigen vectors corresponding to Eigen values $\lambda=1,\sqrt{2} and -\sqrt{2}.$ To verify, these eigen vectors are orthogonal, the necessary condition for orthogonality, based on the notation used here is,$$\epsilon_i .\epsilon_j =0$$ here dot denotes dot product or inner product and further i and j can take values from 1 to 3, such that $i\neq j. $ i.e, $\epsilon_1.\epsilon_2 =0$ shows $\epsilon_1$ is orthogonal to $\epsilon_2$ and $\epsilon_1.\epsilon_3 = 0$ shows $\epsilon_1$ is orthogonal to $\epsilon_3.$ and $\epsilon_2.\epsilon_3 = 0$ shows $\epsilon_2$ is orthogonal to $\epsilon_3.$ If $$\epsilon_1.\epsilon_2 =\epsilon_1.\epsilon_3 =\epsilon_2.\epsilon_3 = 0.$$ then, $\epsilon_1$,$\epsilon_2$ and$\epsilon_3$ are mutually orthogonal to each other. From the vectors obtained, it is found that they are not orthogonal to each other.
