📂Linear Algebra

Transformations on the Quotient Space of Diagonalizable Linear Transformations are also Diagonalizable

Theorem¹

Let $V$ be a dimension vector space, $T : V \to V$ be a linear transformation, and $W$ be an $T$-invariant subspace. If $T$ is diagonalizable, then $\overline{T} : V/W \to V/W$ is also diagonalizable. In this case, $V/W$ is the quotient space of $V$.

Proof

If $T$ is diagonalizable, so is $T|_{W}$, there exists a basis $\gamma = \left\{ v_{1}, v_{2}, \dots, v_{k} \right\}$ of $W$ such that $\begin{bmatrix}T|_{W} \end{bmatrix}_{\gamma}$ is a diagonal matrix. Remember that $\gamma$ is a set of eigenvectors of $T$. For each eigenvalue $\lambda$ of $T$, if there exists a corresponding eigenvector in $\gamma$, select it. If the set of selected eigenvectors is not a basis for the eigenspace $E_{\lambda}$ corresponding to the eigenvalue $\lambda$, it can be expanded to become a basis for $E_{\lambda}$. Let’s call the union of these bases for $E_{\lambda}$, $\beta$.

Lemma

If $T$ is diagonalizable and $\beta_{i}$ is a ordered basis for $E_{\lambda_{i}}$, then $\beta = \beta_{1} \cup \cdots \cup \beta_{k}$ is an ordered basis for $V$ containing the eigenvectors of $T$.

Then, $\beta$ is an ordered basis for $V$ consisting of its eigenvectors. Therefore, $\begin{bmatrix} T \end{bmatrix}_{\beta}$ is a diagonal matrix. However, considering the basis made in this way, if we call it $\beta = \gamma \cup \alpha$, the following holds:

$$\begin{bmatrix} T \end{bmatrix}_{\beta} = \begin{bmatrix} \begin{bmatrix} T|_{W} \end{bmatrix}_{\gamma} & A \\ O & \begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha} \end{bmatrix}$$

Therefore, since $\begin{bmatrix} T \end{bmatrix}_{\beta}$ is a diagonal matrix, $\begin{bmatrix}\ \overline{T}\ \end{bmatrix}_{\alpha}$ is a diagonal matrix, and $\overline{T}$ is diagonalizable.

■