Difference between revisions of "Aufgaben:Problem 5"
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| − | + | Follow the link! (Original Solution Sheldon) | |
| − | + | [https://polybox.ethz.ch/public.php?service=files&t=3cd27159ac0fca5b53c53fe4cff98dcc] | |
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| − | + | An alternative solution can be found here (3.). (MuLö Linalg) | |
| + | [http://www.math.ethz.ch/education/bachelor/lectures/hs2013/math/linalg1/l12] | ||
| − | = | + | == Task == |
| − | + | Determine the center of the algebra of complex d × d-matrices. | |
| − | + | Hint: the center of an algebra \(A\) is defined as \(Z(A) = \{\phi \in A | \phi \psi = \psi \phi \; \forall \psi \in A \} \). | |
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| + | == Problem 5 (Sheldon) == | ||
| + | Let big bold letters denote elements of \(\mathbb{C}^{d \times d}\) and small Greek letters complex numbers. | ||
| − | + | Let \(M \in Z\) | |
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| − | + | We make the following observations and conclude that \( Z = \{\lambda\mathbb{I}\}\) | |
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| − | + | Obviously, the elementes of the center are closed under scalar multiplication (they even form a vector space over \(\mathbb{C}\)). \(\mathbb{I} \in Z \Rightarrow \{\lambda\mathbb{I}\} \subseteq Z\) | |
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| − | + | The theorem about Jordan decomposition states that \(M\) is conjugate to a matrix in Jordan normal form \(J\) and \(J = CMC^{-1}=MCC^{-1}=M\). | |
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| − | + | \(M\) is a diagonal matrix. | |
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| − | + | Suppose to the contrary that \(M\) has a '1' in the superdiagonal, then \(2M\) is supposed to be in the center as well, but it can't, because it is not in Jordan normal form. | |
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| − | + | All diagonal entries of \(M\) are equal. | |
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| − | + | Let \( M =\begin{pmatrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & &\ddots& \\ 0 & \ldots & 0 & \lambda_d \end{pmatrix} \) and \( E_{ij} \) be the matrix with zero entries except for a '1' in the i-th row and j-th column. | |
| − | + | \(\lambda_i E_{ij} = ME_{ij} = E_{ij}M = \lambda_j E_{ij} \Rightarrow \lambda_i = \lambda_j \) | |
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| − | + | Because \( 1 \leq i \leq j \leq d \) can be chosen arbitrary, this proves the other inclusion \(M \in \{\lambda\mathbb{I}\}\). | |
| − | + | == Problem 5 (Musterloesung Illmanen) == | |
| − | + | Let \(E_{ij} \in \mathbb{K}^{n\times n} \) be the matrix which has the entry 1 at \((i,j)\) and 0 else. From proposition \(AE_{ij} = E_{ij}A ~~ \forall 1 \leq i, j \leq n\). Let \(1 \leq i, j \leq n \) be fixed: \(AE_{ij} \) consists solely out of zeros except for the j-th column which is equal to \((a_{1i}, \ldots , a_{ni})^T\). So we get \( a_{ii} = a_{jj} ~~ \wedge ~~ a_{ki} = 0 ~~\forall k \neq j\) | |
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| − | + | We can do this for all indices and this leads to \(a_{ij} = 0 \forall i \neq j ~~\wedge ~~ a_{11} = a_{22} = \ldots = a_{nn} \). Therefor \(A\) has to have the form \(\lambda E_n, \lambda \in \mathbb{K}\) | |
| − | + | ===Ilmanen's solution componentwise=== | |
| − | $$ | + | |
| − | + | Let \(A \in Z(\mathrm{Mat}_d(\mathbb{C}))\). | |
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| + | Let \(E_{ij}\) be the \(d \times d\)-matrix with \((E_{ij})_{kl} = \delta_{ik} \delta_{jl} \). | ||
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| + | Now let \(1 \leq i, j \leq d\) with \(i \neq j\) be fixed and consider | ||
| + | $$(E_{ij} A)_{kl} = \sum \limits_{m}(E_{ij})_{km} A_{ml} = \sum \limits_{m}\delta_{ki} \delta_{jm} A_{ml} = \delta_{ki} A_{jl}.$$ | ||
| + | Since \(A\) commutes with all compex \(d \times d\)-matrices, this is the same as | ||
| + | $$(A E_{ij})_{kl} = \sum \limits_{m}A_{km} (E_{ij})_{ml} = \sum \limits_{m} A_{km} \delta_{mi} \delta_{jl} = \delta_{jl} A_{ki}.$$ | ||
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| + | Thus, we have that | ||
| + | $$\delta_{ki} A_{jl} = \delta_{jl} A_{ki}$$ | ||
| + | Now set \(k=i, l=j\). We then find | ||
| + | $$A_{ii} = A_{jj}$$ | ||
| + | Next, set \(k=l=i\). Still, \(j \neq i\). It follows that | ||
| + | $$A_{ji} = 0.$$ | ||
| + | As this holds for an arbitrary choice of \(i \neq j\), the required form for \(A\) is | ||
| + | $$A = \lambda \mathbb{I}_d$$ | ||
| + | for some \(\lambda \in \mathbb{C}\). And obviously \(\forall \lambda \in \mathbb{C}: \lambda \mathbb{I} \in Z(\mathrm{Mat}_d(\mathbb{C}))\). | ||
Latest revision as of 14:01, 30 July 2015
Follow the link! (Original Solution Sheldon) [1]
An alternative solution can be found here (3.). (MuLö Linalg) [2]
Contents
Task
Determine the center of the algebra of complex d × d-matrices. Hint: the center of an algebra \(A\) is defined as \(Z(A) = \{\phi \in A | \phi \psi = \psi \phi \; \forall \psi \in A \} \).
Problem 5 (Sheldon)
Let big bold letters denote elements of \(\mathbb{C}^{d \times d}\) and small Greek letters complex numbers.
Let \(M \in Z\)
We make the following observations and conclude that \( Z = \{\lambda\mathbb{I}\}\)
Obviously, the elementes of the center are closed under scalar multiplication (they even form a vector space over \(\mathbb{C}\)). \(\mathbb{I} \in Z \Rightarrow \{\lambda\mathbb{I}\} \subseteq Z\)
The theorem about Jordan decomposition states that \(M\) is conjugate to a matrix in Jordan normal form \(J\) and \(J = CMC^{-1}=MCC^{-1}=M\).
\(M\) is a diagonal matrix.
Suppose to the contrary that \(M\) has a '1' in the superdiagonal, then \(2M\) is supposed to be in the center as well, but it can't, because it is not in Jordan normal form.
All diagonal entries of \(M\) are equal.
Let \( M =\begin{pmatrix} \lambda_1 & 0 & \ldots & 0 \\ 0 & \lambda_2 & \ldots & 0 \\ \vdots & &\ddots& \\ 0 & \ldots & 0 & \lambda_d \end{pmatrix} \) and \( E_{ij} \) be the matrix with zero entries except for a '1' in the i-th row and j-th column.
\(\lambda_i E_{ij} = ME_{ij} = E_{ij}M = \lambda_j E_{ij} \Rightarrow \lambda_i = \lambda_j \)
Because \( 1 \leq i \leq j \leq d \) can be chosen arbitrary, this proves the other inclusion \(M \in \{\lambda\mathbb{I}\}\).
Problem 5 (Musterloesung Illmanen)
Let \(E_{ij} \in \mathbb{K}^{n\times n} \) be the matrix which has the entry 1 at \((i,j)\) and 0 else. From proposition \(AE_{ij} = E_{ij}A ~~ \forall 1 \leq i, j \leq n\). Let \(1 \leq i, j \leq n \) be fixed: \(AE_{ij} \) consists solely out of zeros except for the j-th column which is equal to \((a_{1i}, \ldots , a_{ni})^T\). So we get \( a_{ii} = a_{jj} ~~ \wedge ~~ a_{ki} = 0 ~~\forall k \neq j\)
We can do this for all indices and this leads to \(a_{ij} = 0 \forall i \neq j ~~\wedge ~~ a_{11} = a_{22} = \ldots = a_{nn} \). Therefor \(A\) has to have the form \(\lambda E_n, \lambda \in \mathbb{K}\)
Ilmanen's solution componentwise
Let \(A \in Z(\mathrm{Mat}_d(\mathbb{C}))\).
Let \(E_{ij}\) be the \(d \times d\)-matrix with \((E_{ij})_{kl} = \delta_{ik} \delta_{jl} \).
Now let \(1 \leq i, j \leq d\) with \(i \neq j\) be fixed and consider $$(E_{ij} A)_{kl} = \sum \limits_{m}(E_{ij})_{km} A_{ml} = \sum \limits_{m}\delta_{ki} \delta_{jm} A_{ml} = \delta_{ki} A_{jl}.$$ Since \(A\) commutes with all compex \(d \times d\)-matrices, this is the same as $$(A E_{ij})_{kl} = \sum \limits_{m}A_{km} (E_{ij})_{ml} = \sum \limits_{m} A_{km} \delta_{mi} \delta_{jl} = \delta_{jl} A_{ki}.$$
Thus, we have that $$\delta_{ki} A_{jl} = \delta_{jl} A_{ki}$$ Now set \(k=i, l=j\). We then find $$A_{ii} = A_{jj}$$ Next, set \(k=l=i\). Still, \(j \neq i\). It follows that $$A_{ji} = 0.$$ As this holds for an arbitrary choice of \(i \neq j\), the required form for \(A\) is $$A = \lambda \mathbb{I}_d$$ for some \(\lambda \in \mathbb{C}\). And obviously \(\forall \lambda \in \mathbb{C}: \lambda \mathbb{I} \in Z(\mathrm{Mat}_d(\mathbb{C}))\).
