Power of a Diagonalized Matrix?

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SUMMARY

The discussion centers on proving that if \(D\) is a diagonal matrix, then \(D^n\) is also a diagonal matrix with elements equal to the nth power of the elements of \(D\). Participants suggest using index notation and induction as effective methods for the proof. Key points include the observation that the trace and determinant do not suffice to establish the equality of matrices, and the necessity of formalizing the argument using proper summation notation. The conversation highlights the importance of clarity in mathematical proofs, particularly when dealing with indices.

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  • Understanding of diagonal matrices and their properties
  • Familiarity with matrix multiplication and summation notation
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  • Experience with Kronecker delta notation in matrix operations
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  • Study the properties of diagonal matrices in linear algebra
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  • #31
Yes, I see. The sum is gone altogether, so $$ { \left[ { { A }^{ 2 } } \right] }_{ ij }={ A }_{ ii }{ A }_{ ii }\quad When\quad i=j,\quad and\quad { \left[ { { A }^{ 2 } } \right] }_{ ij }=0\quad when\quad i\neq j $$.

Chris
 
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  • #32
kq6up said:
Yes, I see. The sum is gone altogether, so $$ { \left[ { { A }^{ 2 } } \right] }_{ ij }={ A }_{ ii }{ A }_{ ii }\quad When\quad i=j,\quad and\quad { \left[ { { A }^{ 2 } } \right] }_{ ij }=0\quad when\quad i\neq j $$.

Chris

That's it! The nth power goes exactly the same way. There is at most one nonzero term in that big summation.
 
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