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
Topics
About: matrix
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
Starting Context
- The reader knows a definition of matrix.
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.
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.
Main Body
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 \((i, j)\).
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, \(M (k)_{i, j}\). \(M (k)_{i, j} = 2^{-1} (M (k)_{i, j} + M (k)_{j, i}) + 2^{-1} (M (k)_{i, j} - M (k)_{j, i})\). \({M (k)_{i, j}}' := 2^{-1} (M (k)_{i, j} + M (k)_{j, i})\) is symmetric, because \({M (k)_{j, i}}' = 2^{-1} (M (k)_{j, i} + M (k)_{i, j}) = 2^{-1} (M (k)_{i, j} + M (k)_{j, i}) = {M (k)_{i, j}}'\); \({M (k)_{i, j}}'' := 2^{-1} (M (k)_{i, j} - M (k)_{j, i})\) is antisymmetric, because \({M (k)_{j, i}}'' := 2^{-1} (M (k)_{j, i} - M (k)_{i, j}) = - 2^{-1} (M (k)_{i, j} - M (k)_{j, i}) = - {M (k)_{i, j}}''\).
3: Description 2
The separation of Description 1 is unique.
4: Proof 2
Let us think of only 3-dimensional cases, \(M_{i, j, k}\), as before.
According to Description 1, \(M (k)_{i, j} = {M (k)_{i, j}}' + {M (k)_{i, j}}''\). 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. \(M (k)_{i, j} + M (k)_{j, i} = {M (k)_{i, j}}' + {M (k)_{i, j}}'' + {M (k)_{j, i}}' + {M (k)_{j, i}}'' = 2 {M (k)_{i, j}}'\); likewise, \(M (k)_{i, j} + M (k)_{j, i} = 2 {M (k)_{i, j}}'''\). So, \(2 {M (k)_{i, j}}' = 2 {M (k)_{i, j}}'''\), and \({M (k)_{i, j}}' = {M (k)_{i, j}}'''\). \({M (k)_{i, j}}'' = M (k)_{i, j} - {M (k)_{i, j}}' = M (k)_{i, j} - {M (k)_{i, j}}''' = {M (k)_{i, j}}''''\).
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 \({M (k)_{i, j}}' = {M (k)_{i, j}}'_1 + {M (k)_{i, j}}'_2\), \({M (k)_{i, j}}'_i\) may not be symmetric.
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. \({M (k)_{1, 1}}' = 1, {M (k)_{1, 2}}' = 2, {M (k)_{2, 1}}' = 2, {M (k)_{2, 2}}' = 3\), which is indeed symmetric. \({M (k)_{1, 1}}'_1 = 1, {M (k)_{1, 2}}'_1 = 1, {M (k)_{2, 1}}'_1 = 0, {M (k)_{2, 2}}'_1 = 1\) and \({M (k)_{1, 1}}'_2 = 0, {M (k)_{1, 2}}'_2 = 1, {M (k)_{2, 1}}'_2 = 2, {M (k)_{2, 2}}'_2 = 2\), which indeed satisfy \({M (k)_{i, j}}' = {M (k)_{i, j}}'_1 + {M (k)_{i, j}}'_2\), but \({M (k)_{1, 2}}'_1 \neq {M (k)_{2, 1}}'_1\), non-symmetric.
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 \(M_{i, j, k} = {M_{i, j, k}}' + {M_{i, j, k}}'' + {M_{i, j, k}}''' + {M_{i, j, k}}''''\) where \({M_{i, j, k}}'\) 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 \(j, k\).
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, \(M_{i, j, k} = M (k)_{i, j} = {M (k)_{i, j}}'\). Let us separate it with respect to the \((j, k)\) pair: \({M (k)_{i, j}}' = 2^{-1} ({M (k)_{i, j}}' + {M (j)_{i, k}}') + 2^{-1} ({M (k)_{i, j}}' - {M (j)_{i, k}}')\). Although \(N_{i, j, k} := {M (k)_{i, j}}' + {M (j)_{i, k}}'\) is symmetric with respect to the \((j, k)\) pair, it is not symmetric with respect to the \((i, j)\) pair any more, as \({M (k)_{j, i}}' + {M (i)_{j, k}}' \neq {M (k)_{i, j}}' + {M (j)_{i, k}}'\) 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 \(N_{1, 2, 1} \neq N_{2, 1, 1}\). By Description 2, there is no other option for separating \({M (k)_{i, j}}'\), so, \(M_{i, j, k}\) cannot be separated as desired.
