My problem:

Suppose

1)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B \text{ is}: A$

2)Matrix for linear transformation $\displaystyle T$ relative to basis $\displaystyle B^\prime \text{ is}: A^\prime$

3)Transition matrix from basis $\displaystyle B^\prime$ to basis $\displaystyle B$ is: $\displaystyle P$

4)Transition matrix from basis $\displaystyle B$ to basis $\displaystyle B^\prime$ is : $\displaystyle P^{-1}$

Now we know that:

$\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $

And

$\displaystyle P^{-1}AP[\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime} $

where $\displaystyle [T(\mathbf{v})]_{B^\prime}$ is coordinate matrix of $\displaystyle T(\mathbf{v})$ relative to basis $\displaystyle B^\prime$ and $\displaystyle [\mathbf{v}]_{B^\prime}$ is coordinate matrix of $\displaystyle \mathbf{v}$ relative to $\displaystyle B^\prime$

Now the book(Elementary Linear Algebra by Larson 6th edition) I am reading states that:

"But by the definition of the matrix of a linear transformation relative to a basis, this implies that:

$\displaystyle A^\prime = P^{-1}AP $

"

My question:

My question is: why's that?

I know from above that:

$\displaystyle [T(\mathbf{v})]_{B^\prime} = A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

which makes:

$\displaystyle A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime} $

But what reasoning makes the statement $\displaystyle A^\prime = P^{-1}AP$ true? How did the author deduced this to be true?