240: Some Facts about Separating Possibly-Higher-than-2-Dimensional Matrix into Symmetric Part and Antisymmetric Part w.r.t. Indices Pair

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A description/proof of that some facts about separating possibly-higher-than-2-dimensional matrix into symmetric part and antisymmetric part w.r.t. indices pair

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Topics

About: matrix

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description 1
• 2: Proof 1
• 3: Description 2
• 4: Proof 2
• 5: Description 3
• 6: Proof 3
• 7: Description 4
• 8: Proof 4

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Starting Context

• The reader knows a definition of matrix.

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Target Context

• The reader will have a description and a proof of the proposition that some facts about separating any possibly-higher-than-2-dimensional matrix into the symmetric part and the antisymmetric part with respect to any indices pair.

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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 1

Any possibly higher-than-2-dimensional matrix, $M_{i,j,...,k}$, is the sum of a symmetric part and a antisymmetric part with respect to any indices pair, for example $(i,j)$.

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2: Proof 1

Let us think of only 3-dimensional cases, $M_{i,j,k}$, as higher-dimensional cases can be easily extrapolated.

After the pair, $(i,j)$, has been chosen, with $k$ fixed, $M_{i,j,k}$ is a 2-dimensional matrix, $M(k{)}_{i,j}$. $M(k{)}_{i,j}=2^{-1}(M(k{)}_{i,j}+M(k{)}_{j,i})+2^{-1}(M(k{)}_{i,j}-M(k{)}_{j,i})$. ${M(k{)}_{i,j}}^{\prime }:=2^{-1}(M(k{)}_{i,j}+M(k{)}_{j,i})$ is symmetric, because ${M(k{)}_{j,i}}^{\prime }=2^{-1}(M(k{)}_{j,i}+M(k{)}_{i,j})=2^{-1}(M(k{)}_{i,j}+M(k{)}_{j,i})={M(k{)}_{i,j}}^{\prime }$; ${M(k{)}_{i,j}}^{″}:=2^{-1}(M(k{)}_{i,j}-M(k{)}_{j,i})$ is antisymmetric, because ${M(k{)}_{j,i}}^{″}:=2^{-1}(M(k{)}_{j,i}-M(k{)}_{i,j})=-2^{-1}(M(k{)}_{i,j}-M(k{)}_{j,i})=-{M(k{)}_{i,j}}^{″}$.

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3: Description 2

The separation of Description 1 is unique.

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4: Proof 2

Let us think of only 3-dimensional cases, $M_{i,j,k}$, as before.

According to Description 1, $M(k{)}_{i,j}={M(k{)}_{i,j}}^{\prime }+{M(k{)}_{i,j}}^{″}$. Let us suppose that there was another separation, $M(k{)}_{i,j}={M(k{)}_{i,j}}^{‴}+{M(k{)}_{i,j}}^{⁗}$ where ${M(k{)}_{i,j}}^{‴}$ would be symmetric and ${M(k{)}_{i,j}}^{⁗}$ would be antisymmetric. $M(k{)}_{i,j}+M(k{)}_{j,i}={M(k{)}_{i,j}}^{\prime }+{M(k{)}_{i,j}}^{″}+{M(k{)}_{j,i}}^{\prime }+{M(k{)}_{j,i}}^{″}=2{M(k{)}_{i,j}}^{\prime }$; likewise, $M(k{)}_{i,j}+M(k{)}_{j,i}=2{M(k{)}_{i,j}}^{‴}$. So, $2{M(k{)}_{i,j}}^{\prime }=2{M(k{)}_{i,j}}^{‴}$, and ${M(k{)}_{i,j}}^{\prime }={M(k{)}_{i,j}}^{‴}$. ${M(k{)}_{i,j}}^{″}=M(k{)}_{i,j}-{M(k{)}_{i,j}}^{\prime }=M(k{)}_{i,j}-{M(k{)}_{i,j}}^{‴}={M(k{)}_{i,j}}^{⁗}$.

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5: Description 3

The symmetric part or the antisymmetric part of Description 1 could be separated into non-symmetric parts or into non-antisymmetric parts, which means that for example, when ${M(k{)}_{i,j}}^{\prime }={M(k{)}_{i,j}}_1^{\prime }+{M(k{)}_{i,j}}_2^{\prime }$, ${M(k{)}_{i,j}}_i^{\prime }$ may not be symmetric.

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6: Proof 3

This Description may be obvious, but let us not carelessly imagine that any part of the symmetric part was symmetric.

A counterexample suffices. ${M(k{)}_{1,1}}^{\prime }=1,{M(k{)}_{1,2}}^{\prime }=2,{M(k{)}_{2,1}}^{\prime }=2,{M(k{)}_{2,2}}^{\prime }=3$, which is indeed symmetric. ${M(k{)}_{1,1}}_1^{\prime }=1,{M(k{)}_{1,2}}_1^{\prime }=1,{M(k{)}_{2,1}}_1^{\prime }=0,{M(k{)}_{2,2}}_1^{\prime }=1$ and ${M(k{)}_{1,1}}_2^{\prime }=0,{M(k{)}_{1,2}}_2^{\prime }=1,{M(k{)}_{2,1}}_2^{\prime }=2,{M(k{)}_{2,2}}_2^{\prime }=2$, which indeed satisfy ${M(k{)}_{i,j}}^{\prime }={M(k{)}_{i,j}}_1^{\prime }+{M(k{)}_{i,j}}_2^{\prime }$, but ${M(k{)}_{1,2}}_1^{\prime }\ne {M(k{)}_{2,1}}_1^{\prime }$, non-symmetric.

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7: Description 4

A matrix cannot be necessarily separated into symmetric antisymmetric parts simultaneously with respect to multiple indices pairs, which means that for example, an $M_{i,j,k}$ cannot be $M_{i,j,k}={M_{i,j,k}}^{\prime }+{M_{i,j,k}}^{″}+{M_{i,j,k}}^{‴}+{M_{i,j,k}}^{⁗}$ where ${M_{i,j,k}}^{\prime }$ is symmetric with respect to $(i,j)$ and is symmetric with respect to $j,k$, ${M_{i,j,k}}^{″}$ is symmetric with respect to $(i,j)$ and is antisymmetric with respect to $j,k$, ${M_{i,j,k}}^{‴}$ is antisymmetric with respect to $(i,j)$ and is symmetric with respect to $j,k$, and ${M_{i,j,k}}^{⁗}$ is antisymmetric with respect to $(i,j)$ and is antisymmetric with respect to $j,k$.

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8: Proof 4

A counterexample suffices. Let us think of the case in which $M(k{)}_{i,j}$ is symmetric with respect to the $i,j$ pair for each $k$. So, $M_{i,j,k}=M(k{)}_{i,j}={M(k{)}_{i,j}}^{\prime }$. Let us separate it with respect to the $(j,k)$ pair: ${M(k{)}_{i,j}}^{\prime }=2^{-1}({M(k{)}_{i,j}}^{\prime }+{M(j{)}_{i,k}}^{\prime })+2^{-1}({M(k{)}_{i,j}}^{\prime }-{M(j{)}_{i,k}}^{\prime })$. Although $N_{i,j,k}:={M(k{)}_{i,j}}^{\prime }+{M(j{)}_{i,k}}^{\prime }$ is symmetric with respect to the $(j,k)$ pair, it is not symmetric with respect to the $(i,j)$ pair any more, as ${M(k{)}_{j,i}}^{\prime }+{M(i{)}_{j,k}}^{\prime }\ne {M(k{)}_{i,j}}^{\prime }+{M(j{)}_{i,k}}^{\prime }$ in general, which is what Description 3 states, for example, when $M_{1,1,1}=1;M_{1,1,2}=4;M_{1,2,1}=2;M_{1,2,2}=5;M_{2,1,1}=2;M_{2,1,2}=5;M_{2,2,1}=3;M_{2,2,2}=6$, $N_{1,1,1}=2;N_{1,1,2}=6;N_{1,2,1}=6;N_{1,2,2}=10;N_{2,1,1}=4;N_{2,1,2}=8;N_{2,2,1}=8;N_{2,2,2}=12$, which is not symmetric with respect to the $i,j$ pair, as $N_{1,2,1}\ne N_{2,1,1}$. By Description 2, there is no other option for separating ${M(k{)}_{i,j}}^{\prime }$, so, $M_{i,j,k}$ cannot be separated as desired.

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References

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