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I'll try to summarize here what I understand so far about the concepts of change of basis matrix etc.

Let $\beta$ and $\gamma$ be two different ordered bases for the vectorspace $V$, and let $P =[Id_V]_{\beta}^{\gamma}$ be the matrixrepresentation of the identical transformation, with respect to $\beta$ and $\gamma$. It can be easily shown that $P$ is invertible; the inverse is $[Id_V]_{\gamma}^{\beta}$. When an arbitrary vector $x \in V$ has coordinate vector $[x]_{\beta}$ with respect to $\beta$, then its coordinate vector with respect to $\gamma$ is given as \begin{align*} [x]_{\gamma} = P [x]_{\beta}. \end{align*} The matrix $P = [Id_V]_{\beta}^{\gamma}$ is called the change of basis matrix from $\beta$ to $\gamma$.

Example:

If we work with $\mathbb{R}^2$, then two possible bases are $\gamma = \left\{(1,1), (1,-1)\right\}$ and $\beta = \left\{(2,4),(3,1)\right\}$. Now, since $(2,4) = 3(1,1) - 1(1,-1)$ and $(3,1)=2(1,1)+1(1,-1)$ we have that \begin{align*} P = [Id_V]_{\beta}^{\gamma} = \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix}, \end{align*} whereby $P$ the $\beta$-coordinates in $\gamma$-coordinates changes. Let $x = (7,9) \in \mathbb{R}^2$ be an arbitrary vector that has a coordinate vector $[x]_{\beta} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ with respect to the basis $\beta$. Then its coordinate vector with respect to $\gamma$ is given as \begin{align*} \begin{pmatrix} 3 & 2 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 8 \\ -1 \end{pmatrix}. \end{align*} We notice that in the columns of this matrix $P$ are the coordinates of vectors $\beta$ with respect to the basis $\gamma$.

So far so good, now I'm having trouble relating this information to the next two theorems.

Theorem 1.: Let $T$ be a linear operator over a finite dimensional vectorspace $V$, and let $\beta$ and $\gamma$ be two ordered bases for $V$. Let $[T]_{\beta}$ and $[T]_{\gamma}$ be two matrixrepresentations of this transformation. If $P$ is the change of basis matrix from $\beta$ to $\gamma$, then we have that \begin{align*} [T]_{\beta} = P^{-1} [T]_{\gamma} P. \end{align*} Proof: Let $Id: V \rightarrow V$ be the identical transformation. Then we have that $T = Id \circ T = T \circ Id$. Now, since $P = [Id]_{\beta}^{\gamma}$, we have that \begin{align*} P[T]_{\beta} = [Id]_{\beta}^{\gamma} [T]_{\beta}^{\beta} = [Id \circ T]_{\beta}^{\gamma} = [T \circ Id]_{\beta}^{\gamma} = [T]_{\gamma}^{\gamma} [Id]_{\beta}^{\gamma} = [T]_{\gamma} P, \end{align*} from where it follows that $[T]_{\beta} = P^{-1} [T]_{\gamma} P$. Q.E.D.

I'm not sure what's the connection between this theorem and the information above about coordinate vectors.

I could also make this theorem more general, expanding it to a linear map (not necessarily an operator).

Theorem 2.: Let $T : V \rightarrow W$ be a linear map, and let $\beta, \beta'$ be two different ordered bases for $V$, and $\gamma, \gamma'$ two different ordered bases for $W$. Let $A = [T]_{\beta}^{\gamma}$ and $B = [T]_{\beta'}^{\gamma'}$ be two matrixrepresentations of $T$. If $P$ is the change of basis matrix from $\beta$ to $\beta'$, and $Q$ is the change of basis matrix from $\gamma$ to $\gamma'$, then the relationship between $A$ and $B$ is given as \begin{align*} B = QAP^{-1}. \end{align*}

Would this be correct? Also, what do $P$ and $Q$ represent now? Are they still representing the identical transformation? If this is correct, could someone give me an example?

Thanks in advance for clarifying up any misconceptions I might have.

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Let $T$ be reflection over $y=x$. It is a linear transformation. We have

$$T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\x\end{pmatrix}$$

Let $\beta=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}, \gamma=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}\}$ be two basis. We can find $P =[Id_V]_{\beta}^{\gamma}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$

The matrices of the linear transformation $T$ in terms of two basis are

$$[T]_{\beta}=\begin{pmatrix}0&1\\1&0\end{pmatrix}, [T]_{\gamma}=\begin{pmatrix}-1&0\\1&1\end{pmatrix}$$

Theorem 1 says

$$\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}^{-1}\begin{pmatrix}-1&0\\1&1\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}$$

which can be easily verified.

An example for the second theorem can be constructed from a linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^1$. For example, projection onto $x$-axis. $P$ and $Q$ are still the identical transformation. This time we have

$$T\begin{pmatrix}x\\y\end{pmatrix}=x$$

Let $\beta=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}, \beta'=\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}1\\1\end{pmatrix}\}$ be two basis for $\mathbb{R}^2$. We still have $P =[Id_V]_{\beta}^{\beta'}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$.

Let $\gamma=1, \gamma'=-1$. Then $Q=[Id_V]_{\gamma}^{\gamma'}=-1$.

$$[T]_{\beta}^{\gamma}=\begin{pmatrix}1&0\end{pmatrix}, [T]_{\beta'}^{\gamma'}=\begin{pmatrix}-1&-1\end{pmatrix}$$

To get $[T]_{\beta'}^{\gamma'}$, compute transformation of $T$ on the basis $\beta'$, and write the results in terms of $\gamma'$:

$$T\begin{pmatrix}1\\0\end{pmatrix}=1=-1\cdot -1\\ T\begin{pmatrix}1\\1\end{pmatrix}=1=-1\cdot -1$$

It can be found that

$$\begin{pmatrix}-1&-1\end{pmatrix}=Q\begin{pmatrix}1&0\end{pmatrix}P^{-1}$$

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  • $\begingroup$ Thanks for the help. I got a question though. How did you find $Q = [Id_V]_{\gamma}^{\gamma'}$ in your second example? What did you enter as argument in the $T\begin{pmatrix} x \\ y \end{pmatrix}$? $\endgroup$ – Kamil Apr 11 '15 at 20:24
  • $\begingroup$ Also, in the last step: matrix multiplication doesn't work out (not defined for the sizes you say). We have that $ -1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$. But that's multiplication of a $2 \times 1$ matrix with a $2 \times 2$. $\endgroup$ – Kamil Apr 11 '15 at 20:34
  • $\begingroup$ @Kamil: I got $Q = [Id_V]_{\gamma}^{\gamma'}$ exactly in the same way as you got $P$. FInd $-1$ in terms of $1$: $-1=-1\cdot 1$. The last step is $-1 \begin{pmatrix} 1& 0 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 0 & -1 \end{pmatrix}$, multiplication of $1\times 2$ with $2 \times 2$. I added how I got $[T]^{\gamma'}_{\beta'}$ in the answer. $\endgroup$ – KittyL Apr 13 '15 at 9:27

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