Understanding this example of eigenspaces. This is example 5.37 from Axler Linear Algebra Done Right.
Suppose the matrix of an operator $T\in\mathcal{L}(V)$ with respect to a basis $v_1,v_2,v_3$ of $V$ is the matrix
\begin{bmatrix}
8&0&0\\0&5&0\\0&0&5
\end{bmatrix}
Then $E(8,T)=\text{span}(v_1)$ and $E(5,T)=\text{span}(v_2,v_3)$.
Here the notation $E(\lambda,T)$ represents the eigenspace of $T$ corresponding to $\lambda$.  I can't quite get how those two eigenspaces were calculated.  They seem obvious, but I want to know how they got it.
For the first one, I can verify how any scalar multiple of $v_1$ will land you in $E(8,t)$, since $\text{null}(T-8I)cv=c\cdot\text{null}(T-8I)v=cv_1=\text{span}(v_1)$.  But I cannot see how to get the result either for the first or second example or how to know the eigenspace is in fact not larger, containing vectors that we don't know about.  A little detail and walkthrough would be appreciated.  Thank you.
 A: For an arbitrary vector $v=\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 \in V$,
$$Tv=8\lambda_1 v_1 + 5 \lambda_2 v_2 + 5 \lambda_3 v_3.$$
and this vector $v \in E(5,T)$ if and only if 
$$Tv=5v=5\lambda_1 v_1 + 5 \lambda_2 v_2 + 5 \lambda_3 v_3.$$
This will be the case if and only if $\lambda_1 = 0$, since $v$ has a unique representation as a linear combination of base vectors $v_1,v_2,v3$. But $\lambda_1=0$ is equivalent to $v \in \operatorname{span}(v_1,v_2)$.
A: It might help to think about it in terms of matrix multiplication, rather than algebra:
$$ M.v = \lambda v$$
is the eigenvalue equation so put
$$\left(\begin{array}{ccc} 8 & 0& 0 \\0 & 5 & 0 \\0 & 0 & 5 \end{array}\right)\left(\begin{array}{c} a \\b \\ c \end{array}\right) = \lambda \left(\begin{array}{c} a \\b \\ c \end{array}\right)$$
Multiplying this out will give you three equations that are very simple. They can only be solved by particular values of $\lambda$, as you say, that is either $\lambda$ is 8 or 5. Each value has its own solution space as well. 
If you can write out those answers in terms of $a, b,$ and $c$, perhaps you can relate that to your original problem??
A: Let $\mathcal{B} = (v_1,v_2,v_3)$ be an ordered basis for a three dimensional vector space $\mathbb{V}$ over $\mathbb{F}$ and let us denote the matrix representing an operator $T \colon V \rightarrow V$ with respect to $\mathcal{B}$ by $[T]_{\mathcal{B}}$. Recall that by definition, the $i$-th column of $[T]_{\mathcal{B}}$ is the column vector $[T(v_i)]_{\mathcal{B}}$ (that is, the representation of the result of applying $T$ to $v_i$ with respect to the basis $\mathcal{B}$).
In your case, we immediately deduce that 
$$ T(v_1) = 8\cdot v_1 + 0 \cdot v_2 + 0 \cdot v_3 = 8v_1,\\
   T(v_2) = 0 \cdot v_1 + 5 \cdot v_2 + 0 \cdot v_3 = 5v_2, \\
   T(v_3) = 0 \cdot v_1 + 0 \cdot v_2 + 5 \cdot v_3 = 5v_3. $$
Thus, we immediately have $\mathrm{span} \{v_1 \} \subseteq E(8,T)$ and $\mathrm{span} \{v_2, v_3\} \subseteq E(5,T)$. Why do we have equality and why don't we have any other eigenvalues? Let $v \in V$ and write $v = a_1 v_1 + a_2 v_2 + a_3 v_3$ for $a_i \in \mathbb{F}$. Then,
$$ T(v) = T(a_1 v_1 + a_2 v_2 + a_3 v_3) = a_1 T(v_1) + a_2 T(v_2) + a_3 T(v_3) = 8 a_1 v_1 + 5 (a_2 v_2 + a_3 v_3). $$
This also follows from the basic and useful property that $[T]_{\mathcal{B}} [v]_{\mathcal{B}} = [Tv]_{\mathcal{B}}$. If $Tv = \lambda v$ for some $\lambda \in \mathbb{F}$ then we have
$$ T(v) = 8 a_1 v_1 + 5 a_2 v_2 + 5 a_3 v_3 = \lambda a_1 v_1 + \lambda a_2 v_2 + \lambda a_3 v_3. $$
Since $\mathcal{B}$ is a basis, we can compare coefficients and deduce that
$$ 8a_1 = \lambda a_1, \\ 5 a_2 = \lambda a_2, \\ 5 a_3 = \lambda a_3. $$
If $a_1 \neq 0$ then we must have $\lambda = 8$ and $a_2 = a_3 = 0$, showing that $E(8,T) = \mathrm{span} \{ v_1 \}$. Similarly, if $a_1 = 0$ and $a_2 \neq 0$ or $a_3 \neq 0$ we must have $\lambda = 5$ and also showing that $E(5, T) = \mathrm{span} \{ v_2, v_3 \}$. 
By the way, you can also deduce that $\mathrm{span} \{ v_1 \} = E(T,8)$ and $\mathrm{span} \{ v_2, v_3 \}$ from the fact that $(v_1, v_2, v_3)$ is a basis for $V$. If you had $v \in E(T,8)$ that is linearly independent of $v_1$ then by the fact that eigenvectors corresponding to different eigenvalues are linearly independent you would have a list $(v_1,v_2,v_3,v)$ of four linearly independent vectors which is not possible in a three dimensional space. Similarly you can't have an eigenvector in $E(T,5)$ independent of $v_1,v_2$ nor an eigenvector belonging to a completely different eigenspace. However, this argument uses the not-entirely trivial result that eigenvectors corresponding to different eigenvalues are linearly independent and you don't really need that to analyze $T$.
