I had given one matrix .
$\begin{bmatrix}0&1&0 \\-1 & 0&0\\0&0&1\end{bmatrix}$
I had found its rational form as
$\begin{bmatrix}-1&0&0 \\0 & 1&0\\0&0&1\end{bmatrix}$
Can we deduce it is diagonalisable?
Or it there any way to say so form rational form
Any Help will be appreciated
Each block on the diagonal in the rational canonical form is the companion matrix of a polynomial, which is both the minimal and characteristic polynomial for that block. Now, a matrix is diagonalizable (over a field $F$) if and only if its minimal polynomial splits into a product of distinct linear factors (over $F$). The largest block in the RCF is associated with the minimal polynomial of the full matrix, so that's the key; we read off that polynomial and try to factor it. If it splits right, with linear factors and no repeats, the matrix is diagonalizable. If we can't factor it all the way or there's a repeated factor, then the matrix isn't diagonalizable.
Now, your example? Wrong, in multiple ways. First, those two matrices aren't similar; their determinants don't match. Second, your second matrix isn't the rational canonical form of anything; the only diagonal matrices that are are multiples of the identity. The actual RCF of this matrix is $$\begin{pmatrix}0&0&1\\1&0&-1\\0&1&1\end{pmatrix}$$ with associated polynomial $x^3-x^2+x-1=(x^2+1)(x-1)$. As that's only one block, this is the minimal polynomial of the matrix.
Is it diagonalizable? That depends; are we working in a field such as $\mathbb{C}$ with square roots of $-1$, or a field such as $\mathbb{R}$ without them? There are no repeated factors, but we can only fully factor the polynomial if we have $\pm i$.