# Thread: clearification on Isomorphism question

1. ## clearification on Isomorphism question

Let $B$ be an $n \times n$ invertible matrix. Define $\Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB$ Prove that $\Phi$ is an isomorphism.

From the answer provided, it shows:

Let $A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C$

where do the values in blue come from?

2. Originally Posted by lllll
Let $B$ be an $n \times n$ invertible matrix. Define $\Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB$ Prove that $\Phi$ is an isomorphism.

From the answer provided, it shows:

Let $A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C$

where do the values in blue come from?
Hmm... I am stuck too...
Can you write it completely? What are you trying to prove?

Is $\Phi (A) =B^{-1}AB$ or $\Phi (A) =BAB^{-1}$?

3. The first thing that was shown was $\Phi$ was a linear

$\Phi$ was shown to be 1-1 by:

$\Phi (A) = 0 \therefore B^{-1}AB = 0 \Rightarrow N(\Phi) = 0$

$\Phi$ was shown to be onto by:

for any $C \in M_{n\times n}(\mathbb{R})$ Find an $A$ to show that $\Phi (A) = C$

then $A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C$

That's everything I have shown.

4. Originally Posted by lllll
Let $B$ be an $n \times n$ invertible matrix. Define $\Phi: M_{n \times n}(F) \ \mbox{by} \ \Phi (A) =B^{-1}AB$ Prove that $\Phi$ is an isomorphism.

From the answer provided, it shows:

Let $A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C$

where do the values in blue come from?
On second thought, I think I will just prove it my way. Probably the book has done something similar.

Proving $\Phi$ is a homomorphism is not hard. Proving 1-1 is even easier! So I dont understand why the textbook has done the mentioned manipulations...

If $\Phi(A) = \Phi(C) \Rightarrow B^{-1}AB = B^{-1}CB \Rightarrow A=C$

To prove homomorphism: Let W = XY, the let us prove $\Phi(W) = \Phi(X)\Phi(Y)$.

$\Phi(W) = B^{-1}WB = B^{-1}(XY)B$

$\Rightarrow B^{-1}(XY)B = B^{-1}(X{\color{blue}BB^{-1}}Y)B$
$\Rightarrow(B^{-1}XB)(B^{-1}YB) = \Phi(X)\Phi(Y)$

5. Originally Posted by lllll
The first thing that was shown was $\Phi$ was a linear

$\Phi$ was shown to be 1-1 by:

$\Phi (A) = 0 \therefore B^{-1}AB = 0 \Rightarrow N(\Phi) = 0$

$\Phi$ was shown to be onto by:

for any $C \in M_{n\times n}(\mathbb{R})$ Find an $A$ to show that $\Phi (A) = C$

then $A =B^{-1}CB \Rightarrow \Phi A = \Phi B^{-1}CB \Rightarrow {\color{blue} B}(B^{-1}CB){\color{blue} B^{-1}} = C$

That's everything I have shown.
I am sure they meant A =BCB^{-1}, its a typo
OR they wanted to show $\Phi (C) = A$

$\Phi (A) = C \Rightarrow B^{-1}AB = C \Rightarrow A = BCB^{-1}$
Since $C \in M_{n \times n} , \,det(C) \neq 0$ and hence $det(A) \neq 0$, thus $A \in M_{n \times n}$

So the map is onto.

6. Is there a way that this can be shown without using determinants? Other then solving numerical problems, we haven't looked at theorems governing them.

7. Originally Posted by lllll
Is there a way that this can be shown without using determinants? Other then solving numerical problems, we haven't looked at theorems governing them.
Well, than how will you establish that A is invertible? Without that you cannot be sure that $A \in M_{n \times n}$.

8. There is another way to show the map is onto. A consequence of the rank nullity theorem says that if you have a one-to-one linear transformation $f: V \mapsto U$ and the dimension of V and U are both equal and finite then the map is also onto. In this case $V=U = M_{n\times n}(\mathbb{R})$ which have the same finite dimension.