Finding Eigenvectors using different basis of V - Linear Algebra Done Right by Sheldon Axler Section 5.C (Eigenspaces and Diagonal Matrices) page 159 Finding Eigenvectors using different basis of V - Linear Algebra Done Right by Sheldon Axler Section 5.C (Eigenspaces and Diagonal Matrices) page 159 Example 5.45. Given the below matrix , how do I find a different basis for V so that the matrix will be diagonal. If the eigenvalues are 2,5 and 8. Sheldon Axler said this is a simple equation and we have to find a nonzero solution to the equation 
$$T(x,y,z) = λ(x,y,z)$$
$$ \begin{pmatrix} 2 & 1 & 0 \\ 0 & 5 & 3\\0 &0 & 8 \end{pmatrix} $$
The answer was the list
$$(1,0,0), (1,3,0) , (1,6,6)$$ 
Please how did he come by this answer using the eigenvalues 2,5 and 8
 A: Once a matrix is upper-triangular our life is pretty easy; for example you can read the eigenvalues off the diagonal.
Then to figure out the eigenspace associated with e.g. the eigenvalue 8, you want to look at $$\begin{bmatrix}2&1&0\\0&5&3\\0&0&8\end{bmatrix} - 
\begin{bmatrix}8&0&0\\0&8&0\\0&0&8\end{bmatrix}=\begin{bmatrix}-6&1&0\\0&-3&3\\0&0&0\end{bmatrix}$$The upper-triangular form then helps us in another way, which is that we can then start from the bottom up, so if we assume a form $[a, b, 1]^T$ for our eigenvector, we find from the bottom line that $0=0$ and we have some clear degeneracy to work with; then the next line tells us that $b=1$,  then the next line tells us that $a=1/6$. The span of all such vectors is the eigenspace; there are no other eigenvectors to be concerned with here: all vectors of the form $[u/6, u, u]^T$ are eigenvectors with eigenvalue 8.
Similarly with $5$ although there is more calculation needed, we start from  $$\begin{bmatrix}2&1&0\\0&5&3\\0&0&8\end{bmatrix} - 
\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}=\begin{bmatrix}-3&1&0\\0&0&3\\0&0&3\end{bmatrix},$$
now if we assume $[a, 1, c]^T$ as our form, and start again from the bottom up, the last equation says $3c=0$ so therefore $c=0$ and the second equation now says $0=0$ giving our necessary degeneracy, and finally the first equation says that $-3a + 1 = 0$ and therefore that $a=1/3$, so the desired eigenspace is the vectors $[u/3, u, 0]^T$.
When things are no longer upper-triangular life gets harder, but in general if you can find those eigenvalues and then subtract $\lambda I_n$ you should be able to see some sort of linear dependence among the rows which you can use to find the non-trivial vectors mapped to $[0,0,\dots 0]^T.$
A: 
How did he come by this answer using the eigenvalues $2, 5$, and $8$?

If $$\begin{pmatrix} 2 & 1 & 0 \\ 0 & 5 & 3 \\ 0 & 0 & 8 \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} =\lambda \begin{pmatrix}x \\ y \\ z\end{pmatrix}$$
then $2x+y=\lambda x, \tag1$
$5y+3z=\lambda y,\tag2$ 
and $ 8z=\lambda z. \tag3$
From (3), we have $\lambda=8, $ in which case, from (2), $z=y,$ and, from (1), $y=6x$, yielding the eigenvector $(1,6,6)$; or $z=0$, in which case, from (2), $\lambda y=5y$, so $\lambda=5,$ in which case from the first equation $y=3x$, yielding the eigenvector $(1,3,0)$; or $y=0$, in which case from the first equation $\lambda =2$ and we have the eigenvector $(1,0,0)$.
