# Thread: Linear transformation relative to a basis:A question

1. ## Linear transformation relative to a basis:A question

My problem:
Suppose
1)Matrix for linear transformation $T$ relative to basis $B \text{ is}: A$
2)Matrix for linear transformation $T$ relative to basis $B^\prime \text{ is}: A^\prime$
3)Transition matrix from basis $B^\prime$ to basis $B$ is: $P$
4)Transition matrix from basis $B$ to basis $B^\prime$ is : $P^{-1}$

Now we know that:
$A^\prime [\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime}$

And
$P^{-1}AP[\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime}$
where $[T(\mathbf{v})]_{B^\prime}$ is coordinate matrix of $T(\mathbf{v})$ relative to basis $B^\prime$ and $[\mathbf{v}]_{B^\prime}$ is coordinate matrix of $\mathbf{v}$ relative to $B^\prime$

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

My question:
My question is: why's that?

I know from above that:
$[T(\mathbf{v})]_{B^\prime} = A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime}$

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

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

2. ## Re: Linear transformation relative to a basis:A question

Originally Posted by x3bnm
My problem:
Suppose
1)Matrix for linear transformation $T$ relative to basis $B \text{ is}: A$
2)Matrix for linear transformation $T$ relative to basis $B^\prime \text{ is}: A^\prime$
3)Transition matrix from basis $B^\prime$ to basis $B$ is: $P$
4)Transition matrix from basis $B$ to basis $B^\prime$ is : $P^{-1}$

Now we know that:
$A^\prime [\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime}$

And
$P^{-1}AP[\mathbf{v}]_{B^\prime} = [T(\mathbf{v})]_{B^\prime}$
where $[T(\mathbf{v})]_{B^\prime}$ is coordinate matrix of $T(\mathbf{v})$ relative to basis $B^\prime$ and $[\mathbf{v}]_{B^\prime}$ is coordinate matrix of $\mathbf{v}$ relative to $B^\prime$

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

My question:
My question is: why's that?

I know from above that:
$[T(\mathbf{v})]_{B^\prime} = A^\prime [\mathbf{v}]_{B^\prime} = P^{-1}AP[\mathbf{v}]_{B^\prime}$

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

But what reasoning makes the statement $A^\prime = P^{-1}AP$ true? How did the author deduced this to be true?
if two functions are the same for every input of the domain, they are the same function.

so if A' and $P^{-1}AP$ are the same for every $[\mathbf{v}]_{B'}$, they are the same function (and if they are both linear transformations, the same linear transformation).

3. ## Re: Linear transformation relative to a basis:A question

Originally Posted by Deveno
if two functions are the same for every input of the domain, they are the same function.

so if A' and $P^{-1}AP$ are the same for every $[\mathbf{v}]_{B'}$, they are the same function (and if they are both linear transformations, the same linear transformation).
Thanks for reply. But suppose three matrices and $[x]$ are such that:
$ABx = CBx$ where $AB = CB$ for all $[x]$

Then by the properties of matrix:
$A$ is not necessarily equal to $C$

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

why my reasoning above is not a problem for this situation?

Why still $A^\prime = P^{-1}AP$?

4. ## Re: Linear transformation relative to a basis:A question

there's no cancellation going on, we're saying A' and P^-1AP are the same function (the fact that they have the same matrix relative to the basis B' is a big clue).

5. ## Re: Linear transformation relative to a basis:A question

Originally Posted by Deveno
there's no cancellation going on, we're saying A' and P^-1AP are the same function (the fact that they have the same matrix relative to the basis B' is a big clue).
Thanks a lot. That I understand.