Let A and B benxn matrices. Recall that the trace of A is defined by tr(A) = Aii. i=1 Prove that tr(AB) = tr(BA) and tr(A) = tr(A'). %3D

Advanced Engineering Mathematics
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ISBN:9780470458365
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Chapter2: Second-order Linear Odes
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Please see attached Advanced Linear Algebra question below. 

 

How to fully prove that tr(AB) = tr(BA) and tr(A) = tr(A^t)?

 

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Let \( A \) and \( B \) be \( n \times n \) matrices. Recall that the trace of \( A \) is defined by

\[
\text{tr}(A) = \sum_{i=1}^{n} A_{ii}.
\]

Prove that \(\text{tr}(AB) = \text{tr}(BA)\) and \(\text{tr}(A) = \text{tr}(A^t)\).
Transcribed Image Text:Let \( A \) and \( B \) be \( n \times n \) matrices. Recall that the trace of \( A \) is defined by \[ \text{tr}(A) = \sum_{i=1}^{n} A_{ii}. \] Prove that \(\text{tr}(AB) = \text{tr}(BA)\) and \(\text{tr}(A) = \text{tr}(A^t)\).
It's natural that we have \( \text{tr}(A) = \text{tr}(A^t) \) since \( A_{ii} = A^t_{ii} \) for all \( i \). On the other hand, we have

\[
\text{tr}(AB) = \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{ik}B_{ki}
\]

\[
= \sum_{k=1}^{n} \sum_{i=1}^{n} B_{ki}A_{ik} = \sum_{k=1}^{n} (BA)_{kk}
\]

\[
= \text{tr}(BA).
\]
Transcribed Image Text:It's natural that we have \( \text{tr}(A) = \text{tr}(A^t) \) since \( A_{ii} = A^t_{ii} \) for all \( i \). On the other hand, we have \[ \text{tr}(AB) = \sum_{i=1}^{n} (AB)_{ii} = \sum_{i=1}^{n} \sum_{k=1}^{n} A_{ik}B_{ki} \] \[ = \sum_{k=1}^{n} \sum_{i=1}^{n} B_{ki}A_{ik} = \sum_{k=1}^{n} (BA)_{kk} \] \[ = \text{tr}(BA). \]
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