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.