Show a matrix satisfying $A^2 − 8A + 15I = 0$ is diagonalisable. 
A square matrix $A$ (of some size $n × n$) satisfies the condition $A^2 − 8A + 15I = 0$. Show that this matrix is similar to a diagonal matrix.

I know that we must show that 5 and 3 are the eigenvalues of this matrix, and that they yield n linearly independent eigenvectors, but I have no idea how.

Furthermore: Show that for every positive integer $k ≥ 8$ there exists a matrix $A$ satisfying the above condition with $tr(A) = k$.

I think there is some formula that the sum of the eigenvalues is equal to the trace, but I'm not entirely sure.
 A: You can rewrite the equation as
$$(A-5I)(A-3I)=0$$
This shows that $A-5I$ and $A-3I$ are divisors of zero and therefore singular matrices and $3$ and $5$ are eigenvalues of $A$
Let's now prove that
$$\operatorname{ker}(A-5I)\oplus \operatorname{ker}(A-3I)=\operatorname{ker}0=V$$
where $V$ is the whole vector space.
Let $x\in \operatorname{ker}(A-5I)\cap\operatorname{ker}(A-3I)$. This means $Ax=5x=3x$ or $2x=0$ so $\operatorname{ker}(A-5I)\cap\operatorname{ker}(A-3I)=\{0\}$
Now we have $\frac{1}{2}(A-3I)-\frac{1}{2}(A-5I)=I$ and so $\forall x\in V$ we have $x=\frac{1}{2}(A-3I)x-\frac{1}{2}(A-5I)x=x_1+x_2$ where $x_1\in\operatorname{im}(A-3I)$ and $x_2\in\operatorname{im}(A-5I)$
But $(A-5I)x_1=\frac{1}{2}(A-5I)(A-3I)x=0$ and similarly $(A-3I)x_2=-\frac{1}{2}(A-3I)(A-5I)x=0$ so we have proven our claim.
This means $A$ is diagonalisable and its trace is $\operatorname{tr}{A}=3\alpha+5\beta$ where $\alpha$ is the dimension of $\operatorname{ker}(A-3I)$ and $\beta$ the dimension of $\operatorname{ker}(A-5I)$ and of course $\alpha+\beta=\operatorname{dim}{V}$
A: *

*Clearly the minimal polynomial of $A$ is $x^2-8x+15 = (x-5)(x-3)$, so $5$ and $3$ are eigen-values.

*If $x,y\neq 0$ such that $Ax = 5x$ and $Ay = 3y$, then $\alpha x + \beta y =0$ implies that (applying $A$) gives
$$
5\alpha x + 3\beta y = 0 
$$
Since
$$
3\alpha x + 3\beta y =0 \Rightarrow 2\alpha x = 0 \Rightarrow \alpha = 0
$$
Similarly, $\beta = 0$ and so $\{x,y\}$ is linearly independent.

*Let $g_1(x) = (x-3)$ and $g_2(x) = (x-5)$, then $(g_1,g_2) = 1$ as polynomials, so $\exists$ polynomials $h_1, h_2$ such that
$$
h_1g_1 + h_2g_2 = 1
$$
So for any $v \in \mathbb{C}^n$, write
$$
v = h_1(A)g_1(A)(v) + h_2(A)g_2(A)v
$$
and note that $h_1(A)g_1(A)v$ lies in $\ker(A-5I)$ and $h_2(A)g_2(A)v$ lies in $\ker(A-3I)$. Hence, we have shown that
$$
\mathbb{C}^n = \ker(A-5I) + \ker(A-3I)
$$
By part 2, the sum is a direct sum and you get what you want by choosing a basis of $\ker(A-5I)$ together with a basis of $\ker(A-3I)$

