Asked by David
Prove (A1 + A2 +...+ An)^T = A1^T + A2^T +...+ An^T using Mathimatical Induction.
Answers
Answered by
Steve
clearly false, since
(a+b)^2 ≠ a^2+b^2
I suspect there's more to this besides the misspelling ...
(a+b)^2 ≠ a^2+b^2
I suspect there's more to this besides the misspelling ...
Answered by
David
I probably should have elaborated. This is a proof based on the Transposition of Matrices, and it requires the use of the theorem "(A + B)^T = A^T + B^T". It is not a series of integers being raised to a power.
Also, yes. I accidentally misspelled Mathematics.
Also, yes. I accidentally misspelled Mathematics.
Answered by
Steve
well, if you have (A + B)^T = A^T + B^T then it is a cinch.
A1^T = A1^T (true for n=1
(A1+A2)^T = A1^T+A2^T (by the theorem)
matrix addition is associative, so
(A1+A2+A3)^T = ((A1+A2)+A3)^T
= (A1+A2)^T + A3^T
= A1^T+A2^T+A3^T
You can see how this can go directly from n=k to n=k+1
A1^T = A1^T (true for n=1
(A1+A2)^T = A1^T+A2^T (by the theorem)
matrix addition is associative, so
(A1+A2+A3)^T = ((A1+A2)+A3)^T
= (A1+A2)^T + A3^T
= A1^T+A2^T+A3^T
You can see how this can go directly from n=k to n=k+1
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