1. **Problem Statement:** Explain the definition of a diagonalizable operator and provide an example. 2. **Definition:** A linear operator $T: V \to V$ on a vector space $V$ is called diagonalizable if there exists a basis $B$ of $V$ such that the matrix representation of $T$ with respect to $B$, denoted $[T]_B$, is a diagonal matrix. 3. **Explanation:** This means we can find a basis consisting entirely of eigenvectors of $T$. In this basis, the action of $T$ is simply scaling each basis vector by its eigenvalue, which corresponds to the diagonal entries of the matrix. 4. **Important Note:** Diagonalizability is equivalent to $V$ having a basis of eigenvectors of $T$. This simplifies many computations and helps understand the operator's structure. 5. **Example:** Consider the linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $$T(x,y) = (3x, 2y).$$ 6. The matrix of $T$ in the standard basis is $$[T] = \begin{pmatrix}3 & 0 \\ 0 & 2\end{pmatrix},$$ which is already diagonal. 7. The eigenvalues are $3$ and $2$, with eigenvectors $(1,0)$ and $(0,1)$ respectively. 8. Since there is a basis of eigenvectors, $T$ is diagonalizable. **Final answer:** A linear operator is diagonalizable if it has a basis of eigenvectors, making its matrix representation diagonal. The operator $T(x,y) = (3x, 2y)$ is an example of a diagonalizable operator.