Linear operators & mappings between vector spaces

["Don't panic!"]

Hi,

I'm having a bit of difficulty with the following definition of a linear mapping between two vector spaces:

Suppose we have two n-dimensional vector spaces V and W and a set of linearly independent vectors \mathcal{S} = \lbrace \mathbf{v}_{i}\rbrace_{i=1, \ldots , n} which forms a basis for V. We define the linear operator T which maps the basis vectors \mathbf{v}_{j} to their representations in W, i.e. T:V \rightarrow W, as T\left(\mathbf{v}_{j}\right)= \sum_{i=1}^{n} T_{ij}\mathbf{w}_{i} where \mathcal{B}= \lbrace\mathbf{w}_{i} \rbrace_{i=1, \ldots , n} is a basis for W.

What I'm struggling with is, why is the transformation expressed as \sum_{i=1}^{n} T_{ij}\mathbf{w}_{i} and not \sum_{j=1}^{n} T_{ij}\mathbf{w}_{j} ? Is it purely definition, or is there some deeper meaning behind it? (and if so, is there any way of deriving this expression?).

Sorry to ask a probably very trivial question, but it's been bugging me, and I can't seem to find a satisfactory answer from trawling the internet.

 

[Stephen Tashi]

What I'm struggling with is, why is the transformation expressed as \sum_{i=1}^{n} T_{ij}\mathbf{w}_{i} and not \sum_{j=1}^{n} T_{ij}\mathbf{w}_{j} ? Is it purely definition, or is there some deeper meaning behind it? (and if so, is there any way of deriving this expression?).

The left hand side T(v_j) has a j in it. If you summed over j on the right hand side you wouldn't have any j remaining on the right hand side. So the equation would make a claim that T(v_j) is a sum involving subscripts that are constant except for the index i. How would we interpret that?

Mathematics is traditionally written in a sloppy way from the viewpoint of logic. An equation with a with unknowns like i and j is used with the preceeding quantifiers "for each i and for each j" omitted. So the statement T(v_j) = T_{i\ 1} w_1 + T_{i\ 2} w_2 + ... T_{i\ n} w_n would be interpreted to be a condition that was true for each pair of indices i , j. However this wouldn't make sense as definition unless terms like T_{i\ 1} w_1 had the same value for all indices i. We don't want that restriction in the definition of a linear transformation.

You could also ask why we don't define the transformation by T(v_j) = \sum_{i=1}^n T_{j\ i} w_i.

It's just a matter of tradition. Linear transformations are associated with matrices. The traditional way to think of this is to think of a column vector v being transformed by being left multipled by a matrix T. If you wanted to think of a transformation as a row vector v being right multiplied by a matrix T, you'd change the order of T's indexes i,j in the definition.

 

[Fredrik]

What I'm struggling with is, why is the transformation expressed as \sum_{i=1}^{n} T_{ij}\mathbf{w}_{i} and not \sum_{j=1}^{n} T_{ij}\mathbf{w}_{j} ? Is it purely definition, or is there some deeper meaning behind it? (and if so, is there any way of deriving this expression?).

That first definition ensures that $(Tv_j)_i$ (the ith component of $Tv_j$ in the $\mathcal B$ basis) is $T_{ij}$.

Recall that when we're given a linear operator T, we define the matrix [T] corresponding to it by $[T]_{ij}=(Tv_j)_i$. (See the FAQ post I linked to in your other thread). The reason that we use this convention rather than $[T]_{ij}=(Tv_i)_j$ is that it's nice to have the matrix equation that corresponds to y=Tx be [y]=[T][x], rather than $[y]^T=[x]^T[T]$.

The convention you're asking about ensures that $[T]_{ij}=T_{ij}$.

 

["Don't panic!"]

The left hand side T(v_j) has a j in it. If you summed over j on the right hand side you wouldn't have any j remaining on the right hand side. So the equation would make a claim that T(v_j) is a sum involving subscripts that are constant except for the index i. How would we interpret that?You could also ask why we don't define the transformation by T(v_j) = \sum_{i=1}^n T_{j\ i} w_i.

Thanks. Sorry, when I wrote it with the sum over the other index, I had meant in the form that you've put above.

I guess I was just trying to make sense of it really, as usually, if one expresses a vector in terms of a column matrix, with respect to a given ordered basis, for example
\left[ \mathbf{v}\right]_{ \mathcal{B}} = \left( \begin{matrix} a_{1} \\ \vdots \\ a_{n} \end{matrix} \right)
where \mathcal{B} is an n-dimensional basis for V and \mathbf{v} \in V. The components a_{i} of the column matrix are the components of the vector \mathbf{v} with respect to the basis vectors \mathbf{v}_{i}. ( \left[\, \cdot \,\right]_{\mathcal{ B}} defines an isomorphism between V and \mathbb{R}^{n} ).

Then, for some linear operator \mathcal{A}, we have that, with respect to the basis \mathcal{B}: \left[\mathcal{ A} \left( \mathbf{v} \right) \right]_{\mathcal{ B}} = \left( \sum_{j=1}^{n} A_{ij} a_{j} \right)
where the a_{j} are the components of \mathbf{v} with respect to the basis vectors \mathbf{v}_{i} \in \mathcal{B} (the right-hand side of the equation denotes a column matrix whose components are \sum_{j=1}^{n} A_{ij} a_{j} ).
In my mind, I rationalised it by using that one can the basis vectors as the standard basis in the basis that they define, i.e. \left[ \mathbf{v}_{i}\right]_{ \mathcal{B}} = \left( \begin{matrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0 \end{matrix} \right)
where \mathbf{v}_{i} \in \mathcal{B}. And in that sense, for the j^{th} basis vector \mathbf{v}_{j} \in \mathcal{B}, the vector "picks off" the elements of the j^{th} column of the matrix representation of \mathcal{A} in the basis \mathcal{B}, such that \left[ \mathcal{A} \left( \mathbf{v}_{j} \right) \right]_{\mathcal{B} } = \left( \begin{matrix} A_{1j} \\ \vdots \\ A_{nj} \end{matrix} \right) = A_{1j} \left( \begin{matrix} 1 \\ 0 \\ \vdots \\ 0 \end{matrix} \right) + \cdots + A_{nj} \left( \begin{matrix} 0 \\ \vdots \\ 0 \\ 1 \end{matrix} \right) = A_{1j} \left[ \mathbf{v}_{1}\right]_{ \mathcal{B}} + \cdots + A_{nj} \left[ \mathbf{v}_{n}\right]_{ \mathcal{B}} = \left[ A_{1j} \mathbf{v}_{1} + \cdots + A_{nj} \mathbf{v}_{n} \right]_{ \mathcal{ B}} = \left[ \sum_{i=1}^{n} A_{ij} \mathbf{v}_{i} \right]_{\mathcal{B}}
from which one could imply that \mathcal{A} \left( \mathbf{v}_{j} \right) = \sum_{i=1}^{n} A_{ij} \mathbf{v}_{i} however, I'm not convinced that I've got this write and it didn't seem to be the most general way of doing it?!

 

["Don't panic!"]

Cheers Fredrik. My question actually arose after reading your FAQ, I was just a bit unsure. So, is the way that an operator acts on the basis vectors defined that way so that one recovers the matrix equation \left[ \mathbf{y}\right] = \left[\mathcal{T} \right] \left[\mathbf{x} \right]. instead of \left[ \mathbf{y}\right]^{T} = \left[\mathbf{x} \right]^{T} \left[\mathcal{T} \right]?
Also, is it correct to deduce it the way I did in the above post?

 

[Fredrik]

Then, for some linear operator \mathcal{A}, we have that, with respect to the basis \mathcal{B}: \left[\mathcal{ A} \left( \mathbf{v} \right) \right]_{\mathcal{ B}} = \left( \sum_{j=1}^{n} A_{ij} a_{j} \right)

Your calculation looks OK, but there are some inaccuracies. I see nothing in the calculation that indicates a misunderstanding, but the formula above is weird, because it says that a matrix is equal to a number. The right-hand side is the ith component of the left-hand side.

It's also a little confusing to use the same letter of the alphabet (v) for the basis vectors and the arbitrary vector. I will write $\mathbf x=\sum x_i\mathbf v_i$ instead.

If you want to prove that $A\mathbf v_j=\sum_{i=1}^n A_{ij}\mathbf v_i$, where $A_{ij}$ is the row i, column j component of the matrix representation of A, you really only have to realize the left-hand side is a vector, and can therefore be expressed as a linear combination of the basis vectors: $A\mathbf v_j = \sum_{i=1}^n (A\mathbf v_j)_i \mathbf v_i$. Then you recall that $A_{ij}$ is defined by $A_{ij}=(A\mathbf v_j)_i$.

Then you can use this result to evaluate $A\mathbf x$, where $\mathbf x$ is an arbitrary vector.
$$A\mathbf x =A\bigg(\sum_j x_j \mathbf v_j\bigg) =\sum_j x_j A\mathbf v_j =\sum_j x_j\bigg(\sum_i A_{ij}\mathbf v_i\bigg) =\sum_i\bigg(\sum_j A_{ij}x_j\bigg)\mathbf v_i.$$ This implies that the ith component of $A\mathbf x$ is
$$(A\mathbf x)_i =\sum_j A_{ij} x_j.$$

 

["Don't panic!"]

Cheers Fredrik, that's most helpful!

Yes, the formula \mathcal{A} \left[ \left(\mathbf{v} \right) \right]_{\mathcal{B}} = \left(\sum_{j=1}^{n} A_{ij} a_{j}\right) was just laziness on my part - I didn't want to write out all the components of the column vector, so I implied it, all be it a bit ambiguously, through the parentheses around the sum. Apologies for the confusion though, and thank you very much for your help once again.

 

[FactChecker]

Either way works. They are just picking one way. The notation is simpler.

 

[Incnis Mrsi]

The question features basis vectors, whereas a routine operation is calculation of components; one rarely refers to the basis directly. For components we have, in standard notation, an intuitive order of indices (T v)^k = T^k_ℓ v^ℓ, where k is the index of a row and ℓ enumerates both columns of the matrix T^k_ℓ and components v^ℓ.

Original poster wants to see how images of basis vectors of the first space are expressed from basis vectors of the second space. Yes, there is a reversion, because bases are a concept dual to components, namely (e_i)^k = δ^k_i.