📂Linear Algebra

Basis Transformation (Coordinate Transformation) of Linear Transformations

Overview¹

Let $V$ to $n$ dimensional vector space, and call it $\mathbf{v} \in V$. Let $\beta, \beta^{\prime}$ be the ordered basis of $V$. Then, the two coordinates $[\mathbf{v}]_{\beta}$ and $[\mathbf{v}]_{\beta^{\prime}}$ of $\mathbf{v}$ are transformed by the coordinate transformation matrix $Q$ as follows.

$$[\mathbf{v}]_{\beta} = Q [\mathbf{v}]_{\beta^{\prime}}$$

Now, suppose a linear transformation $T : V \to V$ is given. Then, for each ordered basis, there exist matrix representations $\begin{bmatrix} T \end{bmatrix}_{\beta}$ and $\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}}$. Just like the coordinates of vector $\mathbf{v}$ of $V$ are transformed by $Q$, these two matrices are also transformed by $Q$.

Theorem

Let $\beta, \beta^{\prime}$ be the ordered basis of the $n$-dimensional vector space $V$, and $T : V \to V$ be a linear transformation. Let $Q = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}$ be the coordinate transformation matrix converting $\beta^{\prime}$-coordinates to $\beta$-coordinates. Then, the following holds.

$$\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}} = Q^{-1} \begin{bmatrix} T \end{bmatrix}_{\beta} Q$$

Explanation

Such two matrices $\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}}$ and $\begin{bmatrix} T \end{bmatrix}$ are called similar.

Proof

Lemma

Let $V, W, Z$ be a finite-dimensional vector space, and let $\alpha, \beta, \gamma$ be their respective ordered bases. Also, let $T : V \to W$ and $U : W \to Z$ be linear transformations. Then,

$$[UT]_{\alpha}^{\gamma} = [U]_{\beta}^{\gamma}[T]_{\alpha}^{\beta}$$

By the lemma,

$$Q\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}} = \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta}\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}}^{\beta^{\prime}} = \begin{bmatrix} IT \end{bmatrix}_{\beta^{\prime}}^{\beta} = \begin{bmatrix} TI \end{bmatrix}_{\beta^{\prime}}^{\beta} = \begin{bmatrix} T \end{bmatrix}_{\beta}^{\beta} \begin{bmatrix} I \end{bmatrix}_{\beta^{\prime}}^{\beta} = \begin{bmatrix} T \end{bmatrix}_{\beta}Q$$

Since [$Q$ is invertible]

$$\begin{bmatrix} T \end{bmatrix}_{\beta^{\prime}} = Q^{-1} \begin{bmatrix} T \end{bmatrix}_{\beta} Q$$

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