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

Question

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 \to 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

(\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix})

But then when I read several linear algebra book, it said if

T (v_{i}) = \sum_{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})

?

Answer 1

Your book has a typo (as did I); it should be

T (v_{i}) = \sum_{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

(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) .

Applying the algorithm for matrix multiplication,

(\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}) (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} a_{11} \\ a_{21} \\ a_{31} \end{matrix}) = a_{11} v_{1} + a_{21} v_{2} + a_{31} v_{3} = \sum_{j = 1}^{3} a_{j 1} v_{j} .

Answer 2

They're written it wrong. They should've written

T (v_{j}) = \sum_{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.

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

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