Relation between diagonalizibilty of linear operator and its characteristic polynomial? In Hoffman-Kunze (Linear Algebra) there is this theorem :

The first two statements are equivalent, does this means if characteristic polynomial can be factored in the above way then T is diagonalizable?
But consider : 

Characteristic Polynomial of A satisfies statement 2 but matrix A is not diagonalizable, So statement 1 and statement 2 are not equivalent. Am I missing something, Please Help.
 A: First: The second statement has two clauses. It asserts that two things happen:


*

*The characteristic polynomial splits; and

*The dimension of the eigenspace of $\lambda$ equals the multiplicity of $\lambda$ as a root of the characteristic polynomial for every eigenvalue $\lambda$.


So the second statement is not just about being able to factor the characteristic polynomial.
Second: by my calculations, your matrix has characteristic polynomial (expanding along the second row)
$$\begin{align*}
\det(A-tI) &= \det\left(\begin{array}{ccc}
6-t & 3 & -8\\
0 & -2-t & 0\\
1 & 0 & -3-t
\end{array}\right)\\
&= (-2-t)\det\left(\begin{array}{cc}
6-t & -8\\
1 & -3-t
\end{array}\right)\\
&= (-2-t)\Bigl( (6-t)(-3-t) + 8\Bigr)\\
&= (-2-t)(t^2 -3t - 10) = -(t+2)^2(t-5).
\end{align*}$$
Let us write then $c_1=-2$, $d_1=2$, $c_2=5$, $d_2=1$, using the notation of the theorem. 
It is true that it is not diagonalizable, because $W_1$, the nullspace of $A+2I$, has dimension $1$. But this means that it does not satisfy the second statement, because $\dim(W_1) = 1 \lt 2=d_1$. That is, $A$ does not satisfy the second statement in the theorem, because it has two satisfy two clauses and it only satisfies one of them.
A: $(\textrm{ii})$ In other words what it says is that $\textsf T$ is diagonalizable if and only if the dimension of each eigenspace associated with $c_i$ is equal to the multiplicity of the same eigenvalue $c_i$.
In your case, you forgot to check the dimension.
