a) Prove that Tr(AB) = Tr(BA). Hint: See proof of (9.13). c) If S is a symmetric matrix and A is an antisymmetric matrix, show that Tr(SA) = 0. Hint: Consider Tr(SA)" and prove that Tr(SA) = – Tr(SA).
a) Prove that Tr(AB) = Tr(BA). Hint: See proof of (9.13). c) If S is a symmetric matrix and A is an antisymmetric matrix, show that Tr(SA) = 0. Hint: Consider Tr(SA)" and prove that Tr(SA) = – Tr(SA).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Matrix Trace Properties and Cyclic Permutations**
The equation (9.13) states the cyclic property of the trace operation for matrices:
\[ \text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB). \]
### Proof:
We can prove this cyclic property as follows:
1. Start with the trace of the product of matrices \(ABC\):
\[
\text{Tr}(ABC) = \sum_i (ABC)_{ii} = \sum_i \sum_j \sum_k A_{ij} B_{jk} C_{ki}
\]
2. Rearrange to express in terms of matrix \(BCA\):
\[
= \sum_i \sum_j \sum_k B_{jk} C_{ki} A_{ij} = \text{Tr}(BCA)
\]
3. Further rearrange to express in terms of matrix \(CAB\):
\[
= \sum_i \sum_j \sum_k C_{ki} A_{ij} B_{jk} = \text{Tr}(CAB)
\]
### Important Note:
The trace of the permutation \( \text{Tr}(ABC) \) is **not** generally equal to \( \text{Tr}(ACB) \). This distinction is crucial when working with non-commutative matrices.
The illustration is a clear demonstration of the cyclic nature of the trace operation for matrices, and an important reminder of its limitations regarding non-cyclic permutations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b6d687b-9b3f-47af-8b4d-c251df44093c%2F5b7f870e-5ed8-46b8-ad65-63dde279938d%2Fr62ogwg_processed.png&w=3840&q=75)
Transcribed Image Text:**Matrix Trace Properties and Cyclic Permutations**
The equation (9.13) states the cyclic property of the trace operation for matrices:
\[ \text{Tr}(ABC) = \text{Tr}(BCA) = \text{Tr}(CAB). \]
### Proof:
We can prove this cyclic property as follows:
1. Start with the trace of the product of matrices \(ABC\):
\[
\text{Tr}(ABC) = \sum_i (ABC)_{ii} = \sum_i \sum_j \sum_k A_{ij} B_{jk} C_{ki}
\]
2. Rearrange to express in terms of matrix \(BCA\):
\[
= \sum_i \sum_j \sum_k B_{jk} C_{ki} A_{ij} = \text{Tr}(BCA)
\]
3. Further rearrange to express in terms of matrix \(CAB\):
\[
= \sum_i \sum_j \sum_k C_{ki} A_{ij} B_{jk} = \text{Tr}(CAB)
\]
### Important Note:
The trace of the permutation \( \text{Tr}(ABC) \) is **not** generally equal to \( \text{Tr}(ACB) \). This distinction is crucial when working with non-commutative matrices.
The illustration is a clear demonstration of the cyclic nature of the trace operation for matrices, and an important reminder of its limitations regarding non-cyclic permutations.
![(a) Prove that Tr(AB) = Tr(BA). *Hint:* See proof of (9.13).
(b) [Text is redacted]
(c) If S is a symmetric matrix and A is an antisymmetric matrix, show that Tr(SA) = 0. *Hint:* Consider Tr(SA)ᵀ and prove that Tr(SA) = -Tr(SA).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b6d687b-9b3f-47af-8b4d-c251df44093c%2F5b7f870e-5ed8-46b8-ad65-63dde279938d%2Fnpmgjgmk_processed.png&w=3840&q=75)
Transcribed Image Text:(a) Prove that Tr(AB) = Tr(BA). *Hint:* See proof of (9.13).
(b) [Text is redacted]
(c) If S is a symmetric matrix and A is an antisymmetric matrix, show that Tr(SA) = 0. *Hint:* Consider Tr(SA)ᵀ and prove that Tr(SA) = -Tr(SA).
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