I put the matrix into a row echelon form but I got the matrix without $a$-s. What should I do ?
M= $\pmatrix{1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0}$
It is not OK.
I need to find the eigenvalues (also their algebraic multiplicity) and eigenvectors of the matrix $M$: $$ \pmatrix{a&4&a\cr 0&-a&0\cr 1&0&1\cr}, ~~ a \in \mathbb R$$
Calculate the characteristic polynomial:
$$\det(tI-A)=\begin{vmatrix}t-a&-4&-a\\0&t+a&0\\-1&0&t-1\end{vmatrix}=(t+a)\left[(t-1)(t-a)-a\right]=$$
$$=(t+a)(t^2-(a+1)t)=t(t+a)(t-(a+1))$$
so if $\;a, a+1\neq0\;$ then the matrix has three different eigenvalues: $\;0, -a,-a-1\;$, and is thus diagonalizable. Now let us see what happens when $\;a=0\iff\text{ eigenvalue}=0\;$, by evaluating the resulting homogeneous system to get the matrix's eigenvectors:
$$\begin{cases}4y=0\\-x-z=0\end{cases}\implies \text{only one lin. independent eigenvector}\;\;\begin{pmatrix}1\\0\\\!\!-1\end{pmatrix}$$
and thus the matrix isn't diagonalizable, and now for $\;a=-1\iff\text{ eigenvalue}=0\;$ :
$$\begin{cases}x-4y+z=0\\-y=0\\-x-z=0\end{cases}\implies\text{only one lin. independent eigenvector again}: \begin{pmatrix}1\\0\\\!\!-1\end{pmatrix}$$
In any case thus, the matrix is diagonalizable iff $\;a,\,a+1\neq0\;$