I have been given a matrix P where the columns represent a basis in B. I have also been given a matrix A which is the standard matrix for T. I am then supposed to calculate the matrix for T relative to the basis B, "i.e calculate $[T]_{\cal BB}$".
I have also been given that $T(x)=Ax$
I am very confused about what this means, honestly, I have no clue whatsoever about where to start. Can anyone help me out?
Thanks in advance!
/Nick
On
Knowing that the new components of a vector $v$ are given by multiplying $P^{-1}v$, it is easy to solve you exercise.
For the linear transformation $T$ you calculate $Av=APP^{-1}v$ from you can see how the matrix $AP$ relates the old version $Tv$ with the new of $v$, but, to change everything in terms of the new basis you do $P^{-1}Av=P^{-1}APP^{-1}v$ and associate as $$P^{-1}Av=(P^{-1}AP)P^{-1}v$$ where you can see how the new components of $Tv$ are related with the new components of $v$ through the multiplication $P^{-1}AP$.
Short answer:
$[T]_{\mathcal{BB}}=P^{-1}AP$.
Long answer:
$\Bbb R^3$ is a vector space of finite dimension, so it has a basis. In fact, there exists a canonical basis, which we can call $(e_1,e_2,e_3)$. This basis is canonical in the sense that a triple $(x_1,x_2,x_3)\in\Bbb R^3$ corresponds to the vector $x_1e_1+x_2e_x+x_3e_3$. We represent this vector by $$\left[\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right].$$
However, we can give $\Bbb R^3$ another basis, say $$\begin{aligned} f_1&=(1,1,1)\\ f_2&=(1,2,2)\\ f_3&=(-1,-2,-1) \end{aligned}$$ (by the way, these are the columns of $P$). Then any $(x_1,x_2,x_3)$ in $\Bbb R^3$ can be written as $\lambda_1f_1+\lambda_2f_2+\lambda_3f_3$ for some $\lambda_1,\lambda_2,\lambda_3$, but takes a bit more effort to calculate these coefficients. Let's start with something easier: if we know $(\lambda_1,\lambda_2,\lambda_3)$, then how can we recover $(x_1,x_2,x_3)$? We can remark that if $(\lambda_1,\lambda_2,\lambda_3)=(1,0,0)$, then $(x_1,x_2,x_3)=(1,1,1)$. Doing the same for $(\lambda_1,\lambda_2,\lambda_3)=(0,1,0)$ and $(0,0,1)$, we find that, for any $(\lambda_1,\lambda_2,\lambda_3)$, $$\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=P\cdot\begin{bmatrix}\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}.$$ We then conclude that $$\begin{bmatrix}\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}=P^{-1}\cdot\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}.$$
Back to the operator $T$. $T:\Bbb R^3\rightarrow\Bbb R^3$ is a linear operator in $\Bbb R^3$, and therefore can be represented by a matrix: however, this representation depends on a choice of bases, both for the input and for the output. Its representation in the canonical basis is $A$. We can understand this step by step. Take $x$ a vector in $\Bbb R^3$. We can write it in the canonical basis as $x_1e_1+x_2e_2+x_3e_3$, which gives us a vector $\left[\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}\right]$. $T(x)$ is also a vector in $\Bbb R^3$, so we can also write it out in the canonical basis. This is where $A$ comes in: the expression of $T(x)$ in the canonical basis is $A\cdot\left[\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}\right]$.
Now, instead of representing $x$ in the canonical basis, and getting an output in the canonical basis, we want to represent $x$ in the basis $\mathcal B$, and get the output also in the basis $\mathcal B$. We are therefore looking for a matrix $[T]_{\mathcal{BB}}$ that takes as input a vector written in basis $\mathcal B$, i.e. a triple $(\lambda_1,\lambda_2,\lambda_3)$, and returns another triple in the basis $\mathcal B$. But we only have a matrix that takes as input a vector written in the canonical basis, i.e. a triple $(x_1,x_2,x_3)$, and likewise returns a triple written in the canonical basis.
How can we go from the canonical basis to $\mathcal B$ and back? you might ask. This is where $P$ comes in. $P$ takes a vector written in the basis $\mathcal B$ and returns a matrix in the canonical basis. Similarly, $P^{-1}$ takes a vector written in the canonical basis and returns a matrix in the basis $\mathcal B$. We now have all the tools required to calculate $[T]_\mathcal{BB}$: $$[T]_{\mathcal{BB}}=P^{-1}AP.$$ To understand why we multiply in this order, imagine a vector being multiplied on the right: $$[T]_{\mathcal{BB}}x=P^{-1}APx.$$ $x$ is written in the basis $\mathcal B$. $Px$ transforms this into a vector written in the canonical basis. We can then apply $A$, which gives $APx$. Finally, we want to come back to the basis $\mathcal B$, so we multiply by $P^{-1}$, giving $P^{-1}APx$.
I'm not too sure if I explained myself clearly, but I hope it helps you understand changes of bases and matrix representation in general!