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If we know that some homomorphism $\phi: GL_{2}(\mathbb{R})\to G$ has a non-trivial kernel, consisting of invertible square real matrices, and we know that these matrices are similar to diagonal matrices by conjugation, then for some matrix $B \in GL_{2}(\mathbb{R})$ there exists some matrix $P$, such that $PBP^{-1} = A$, and thus $A$ is also in the kernel of $\phi$. However, by Proposition 7 on page 82 in Abstract Algebra by Dummit and Foote, we can read that "A subgroup $N$ of the group $G$ is normal if and only if it is the kernel of some homomorphism". But if $N$ is normal in our case, then this implies that $gNg^{-1} \subset N$ for all $g \in G$.

But does that imply that $PBP^{-1} = A$ for any $P \in GL_{2}(\mathbb{R})$. Am I mistaken or is this really possible? From my Linear Algebra experience, I thought there is a unique matrix $P$ for every matrix $B$, such that $PBP^{-1} = A$. Please help me clarify this interesting moment.

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  • $\begingroup$ OK I see where my confusion came from. We know that $PBP^{-1} \neq A$ for all $P \in M_{2\times2}(\mathbb{R})$, but $PBP^{-1}$ is still in the kernel of $\phi$ for any $P \in M_{2\times2}(\mathbb{R})$. Is this correct? $\endgroup$
    – sequence
    Oct 6, 2015 at 16:15
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    $\begingroup$ Whoa! There's a ton of confusion here. What does $M_{2 \times 2}(\mathbb{R})$ mean? All $2\times 2$ matrices (a commutative ring or specifically an abelian group under matrix addition)? Or just invertible matrices (i.e. $\mathrm{GL}_2(\mathbb{R})$) under matrix multiplication? Group homomorphism? Ring homomorphism? $\endgroup$
    – Bill Cook
    Oct 6, 2015 at 16:22
  • $\begingroup$ Group homomorphism, $GL_2(\mathbb{R})$. Sorry for the confusion, let me make some corrections to the post. $\endgroup$
    – sequence
    Oct 6, 2015 at 16:24
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    $\begingroup$ If you mean all $2 \times 2$ matrices, then the set of invertible matrices cannot be a kernel of a homomorphism. $\mathrm{GL}_2(\mathbb{R})$ (all invertible matrices) is a group under matrix multiplication, but it is not closed under addition: invertible matrices aren't a subgroup of all matrices (different operations). $\endgroup$
    – Bill Cook
    Oct 6, 2015 at 16:24
  • $\begingroup$ Ok. Next, issue: If the kernel is all invertible square matrices, then the kernel is the entire domain and so $\phi$ is the trivial homomorphism. $\endgroup$
    – Bill Cook
    Oct 6, 2015 at 16:25

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First, not all matrices can be diagonalized. So given a random $B \in \mathrm{GL}_2(\mathbb{R})$, we could have that $B$ is diagonalizable, but only over the complex numbers (this occurs when $B$ has complex eigenvalues). Or even worse, $B$ might not be diagonalizable over any field of scalars (it has a non-trivial Jordan block).

Next, given $PBP^{-1}=A$, yes, if $B$ is in the kernel, then $PBP^{-1}=A$ must be in the kernel as well (kernels are normal subgroups and normal subgroups are closed under conjugation). However, there's no reason to assume that $QBQ^{-1}=C$ is the same matrix (if we swap out $P$ for some other invertible matrix $Q$). However, $C=QBQ^{-1}$ would still be in the kernel (it just may not be the same element of the kernel).

Finally, IF a matrix $B$ is diagonalizable, then any matrix whose columns form a basis of eigenvectors, $P$, will diagonalize $B$.

Take the first column of $P$ (some eigenvector of $B$), rescale it (just not by zero) and you'll still have an eigenvector of $B$. Rescaling a vector won't change linear independence, so $Q$ (where $Q$ is $P$ with its first column rescaled) still diagonalizes $B$. Therefore, there are infinitely many matrices that diagonalize $B$!

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