342: Set of n x n Quaternion Matrices Is 'Rings - Homomorphism Morphisms' Isomorphic to Set of Corresponding 2n x 2n Complex Matrices

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A description/proof of that set of n x n quaternion matrices is 'rings - homomorphism morphisms' isomorphic to set of corresponding 2n x 2n complex matrices

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of quaternion.
• The reader knows a definition of ring.
• The reader knows a definition of %structure kind% homomorphism.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that the set of the quaternions is 'rings - homomorphism morphisms' isomorphic to the set of the corresponding 2 x 2 complex matrices.
• The reader admits the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise.

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Target Context

• The reader will have a description and a proof of the proposition that the set of the n x n quaternion matrices is 'rings - homomorphism morphisms' isomorphic to the set of the corresponding 2n x 2n complex matrices, via the between-the-set-of-the-quaternions-and-the-set-of-the-corresponding-2-x-2-complex-matrices 'rings - homomorphism morphisms' isomorphism.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

Let us define the map, $f^{″}:M_n(\mathbb{H})\to M_{2n}(\mathbb{C})$, by replacing each element of the quaternion matrix with the corresponding 2 x 2 complex matrix by the 'rings - homomorphism morphisms' isomorphism, $f^{\prime }$. Then, the restriction of $f^{″}$ on the codomain, $f:M_n(\mathbb{H})\to f^{″}(M_n(\mathbb{H}))$ is a 'rings - homomorphism morphisms' isomorphism.

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2: Proof

Let us prove that $f$ is a rings homomorphism. Let $M_1,M_2$ be any $M_1,M_2\in M_n(\mathbb{H})$.

As for addition, $f(M_1+M_2)=f(M_1)+f(M_2)$, because any 2 x 2 complex matrix in $f(M_1+M_2)$ is $f^{\prime }(M_{1,i,j}+M_{2,i,j})=f^{\prime }(M_{1,i,j})+f^{\prime }(M_{2,i,j})$, because $f^{\prime }$ is rings homomorphic, and $f^{\prime }(M_{1,i,j})+f^{\prime }(M_{2,i,j})$ is the corresponding 2 x 2 complex matrix in $f(M_1)+f(M_2)$.

As for multiplication, for each 2 x 2 complex matrix, $M_3$, in $f(M_1{M}_2)$, $M_3=f^{\prime }(\sum _j{M}_{1,i,j}{M}_{2,j,k})=\sum _j{f}^{\prime }(M_{1,i,j})f^{\prime }(M_{2,j,k})$, because $f^{\prime }$ is a rings homomorphism, and $\sum _j{f}^{\prime }(M_{1,i,j})f^{\prime }(M_{2,j,k})$ is the corresponding 2 x 2 complex matrix in $f(M_1)f(M_2)$, because $f(M_1)f(M_2)$ is the blocks-wise multiplication where each 2 x 2 complex matrix is a block, by the the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise. So, $f(M_1{M}_2)=f(M_1)f(M_2)$.

Let us prove that $f$ is bijective.

If $M_1\ne M_2$, $f(M_1)\ne f(M_2)$, because $f^{\prime }$ is bijective. So, $f$ is injective. $f$ is obviously surjective, because the codomain is restricted to the image.

So, there is the inverse, $f^{-1}$.

Let us prove that $f^{-1}$ is a rings homomorphism. Let $f(M_1),f(M_2)$ be any $f(M_1),f(M_2)\in f(M_n(\mathbb{H}))$.

As for addition, $f^{-1}(f(M_1)+f(M_2))=f^{-1}(f(M_1))+f^{-1}(f(M_2))$, because $f^{-1}(f(M_1)+f(M_2))=f^{-1}(f(M_1+M_2))=M_1+M_2=f^{-1}(f(M_1))+f^{-1}(f(M_2))$.

As for multiplication, $f^{-1}(f(M_1)f(M_2))=f^{-1}(f(M_1{M}_2))=M_1{M}_2=f^{\prime -1}(f(M_1))f^{-1}(f(M_2))$.

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References

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