As anon points you out, try to use $\overline{z\cdot w} = \overline{z}\cdot \overline{w}$ and simplify notation.
For instance, you could just write vectors in $\mathbb{C}^2$ like this: $(z_1, z_2)$, with $z_1, z_2 \in \mathbb{C}$.
Second, I would try to convince myself that that product/conjugacy rule works too for products of matrices and vectors. That is:
$$
\overline{A\cdot v} = \overline{A}\cdot \overline{v} \ ,
$$
for, say, $A$ a $2\times 2$ complex matrix and $v\in \mathbb{C}^2$. What would happen if $A$ had real coefficitients?
Finally, I would write the equality I already know, namely
$$
A \cdot Y_0 = \lambda Y_0 \ .
$$
Staring at it should force inspiration to come. :-)
No, there are plenty of matrices with repeated eigenvalues which are diagonalizable. The easiest example is
$$
A=\begin{bmatrix}1&0\\0&1\end{bmatrix}.
$$
The identity matrix has $1$ as a double eigenvalue and is (already) diagonal. If you want to write this in diagonalized form, you can write
$$
A=I^{-1}AI=I^{-1}II
$$
since $A$ is a diagonal matrix.
In general, $2\times 2$ matrices with repeated eigenvalue $\lambda$ are diagonalizable if and only if the eigenspace corresponding to $\lambda$ is two dimensional. In other words, if
$$
A-\lambda I=\begin{bmatrix}a-\lambda&b\\c&d-\lambda\end{bmatrix}
$$
has a two-dimensional null space. Using the rank-nullity theorem, we get that this happens exactly when the matrix has $0$ pivots. If $A-\lambda I$ has any nonzero entries, then it will have a pivot. Therefore, a $2\times 2$ matrix with repeated eigenvalues is diagonalizable if and only if it is $\lambda I$.
If $B$ is an $n\times n$ matrix, all of whose eigenvalues are $\lambda$, a similar result holds. A sneakier way to prove this is that if $B$ is diagonalizable, then
$$
B=P^{-1}(\lambda I)P=\lambda P^{-1}IP=\lambda I,
$$
where P is an invertible (basis changing) matrix.
Therefore, the only $n\times n$ matrices with all eigenvalues the same and are diagonalizable are multiples of the identity.
If only some of $B$'s eigenvalues have multiplicity, then the situation becomes more complicated and you really need to compute the dimensions of all the eigenspaces.
As the other posters comment, there are diagonal matrices which are not multiples of the identity, for example
$$
\begin{bmatrix}1&0\\0&2\end{bmatrix}
$$
and if all the eigenvalues of a matrix are distinct, then the matrix is automatically diagonalizable, but there are plenty of cases where a matrix is diagonalizable, but has repeated eigenvalues.
Best Answer
We indeed find that if a real matrix $A$ has a complex eigenvalue $\lambda$, then the conjugate eigenvalue $\bar \lambda$ has the same algebraic and geometric multiplicity. In fact, we can say a bit more: $$ (A - \lambda I)|_{\ker(A - \lambda I)} ``=" (A - \bar \lambda I)|_{\ker(A - \bar \lambda I)}, $$ which is to say that all structures associated with the eigenvalues are the same. That is, the Jordan form of $A$ has the same number and sizes of blocks for $\lambda$ and $\bar \lambda$.
As far as proving multiplicities goes, we have the following: the algebraic multiplicity is the multiplicity of the root $\lambda$ in the characteristic polynomial $p(x) = \det(xI - A)$. As is true for any polynomial with real coefficients, the multiplicity of the root $\lambda$ is this the same as the multiplicity of the root $\bar \lambda$.
For geometric multiplicity, one approach is as follows: we note that matrices satisfy $\overline{A B} = \bar A \bar B$. It follows that if $v$ is a (complex) eigenvector associated with eigenvalue $\lambda$, then we have $$ Av = \lambda v \implies \overline{Av} = \overline{\lambda v} \implies A \bar v = \bar \lambda \bar v. $$ In other words, the map $v \mapsto \bar v$ is an invertible $\Bbb R$-linear map between the eigenspaces of $A$ associated with $\lambda$ and $\bar \lambda$. It follows that these spaces have the same dimension.