Let V be a 3 dimensional vector space over a field F and fix ( <mi mathvariant="bold">v

Rebecca Villa 2022-07-15 Answered
Let V be a 3 dimensional vector space over a field F and fix ( v 1 , v 2 , v 3 ) as a basis. Consider a linear transformation T : V V. Then we have
T ( v 1 ) = a 11 v 1 + a 21 v 2 + a 31 v 3
T ( v 2 ) = a 12 v 1 + a 22 v 2 + a 32 v 3
T ( v 3 ) = a 13 v 1 + a 23 v 2 + a 33 v 3
So that we can identify T by the matrix
( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 )
But then when I read several linear algebra book, it said if T ( v i ) = j a i j v j , then we can identify T by the matrix ( a i j ). My problem is: isn't the matrix is ( a j i ) instead of ( a i j )?
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Answers (2)

kawiarkahh
Answered 2022-07-16 Author has 15 answers
Your book has a typo (as did I); it should be
T ( v i ) = j a j i v j .
Here is an example: in the basis { v 1 , v 2 , v 3 }, we know that v 1 corresponds to the column vector
( 1 0 0 ) .
Applying the algorithm for matrix multiplication,
( a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 ) ( 1 0 0 ) = ( a 11 a 21 a 31 ) = a 11 v 1 + a 21 v 2 + a 31 v 3 = j = 1 3 a j 1 v j .
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Patatiniuh
Answered 2022-07-17 Author has 5 answers
They're written it wrong. They should've written T ( v j ) = i a i j v i - edited so that this is correct now. I saw that i wasn't free in the original way it was written but not the transpose part. I think a form that doesn't transpose is inherently easier to work with.
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