MHB (Very) Basic Questions on Linear Transformations and Their Matrices

[Math Amateur]

Firstly, my apologies to Deveno in the event that he has already answered these questions in a previous post ...

Now ...

Suppose we have a linear transformation $$T: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$$ , say ...

Suppose also that $$\mathbb{R}^3$$ has basis $$B$$ and $$\mathbb{R}^2$$ has basis $$B'$$, neither of which is the standard basis ...

Suppose further that $$T(x, y, z) = ( x + 2y - z , 3x + 5z )$$ ... ...

... ... ...

Then (if I am right) we write the matrix $$A$$, of the transformation as follows:

$$A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 5 \end{bmatrix}$$... BUT ... questions ...Question 1

Is the expression for the linear transformation

$$T: \mathbb{R}^3 \longrightarrow \mathbb{R}^2$$

an expression in terms of the transformation of $$v = (x, y, z)$$ into $$w = T(v)$$ in terms of the bases $$B$$ and $$B'$$ ... ...

That is, when we input some vector $$v = ( 2, 1, -3 )$$ , say ... ... is that vector to be read as being in terms of the basis $$B$$ or in terms of the standard basis ... ...

... ... and is the output vector from applying T, namely

$$T(v) = ( 2, 1, 3 ) = ( x + 2y - z , 3x + 5z ) = ( 2 + 2(1) - (-3) , 3(2) + 5(-3) ) = ( 7, -9 )$$

in terms of the basis $$B'$$ or in terms of the standard basis?[By the way, I think it is, by convention, that linear transformations from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ are expressed as if they are from a standard basis to a standard basis ... but why they are not taken to be in the declared bases $$B$$ and $$B'$$, I am not sure ... ... ]
Question 2

Does the matrix of the transformation

$$A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 5 \end{bmatrix}$$

represent the transformation from $$[v]_B$$ to $$[T(v)]_{B'}$$

or

does it represent the transformation from $$[v]_{S_1}$$ to $$[T(v)]_{S_2}$$

where $$S_1$$ is the standard basis for $$\mathbb{R}^3$$

and $$S_2$$ is the standard basis for $$\mathbb{R}^2$$
Hope someone can help ...

Peter

 

[Deveno]

A matrix representation of a linear transformation DEPENDS on the basis chosen.

In the same vein, the COORDINATES of a vector ALSO depend on the basis chosen.

By itself, the triple $(2,0,5)$ means nothing-it is just three numbers separated by commas, and enclosed in parentheses.

By convention (and *solely* by convention), elements of $F^n$ (where $F$ is any field) are usually denoted by their representation in the standard basis:

$e_1 = (1,0,\dots,0)$
$e_2 = (0,1,\dots,0)$
$\vdots$
$e_n = (0,0,\dots,1)$so when we say, $(x,y,z) \in \Bbb R^3$, for example, what we really MEAN is, the linear combination:

$xe_1 + ye_2 + ze_3$.

Given an $n$-dimensional vector space, the only point unambiguously defined by an $n$-tuple is the $0$-vector, which is the same in any basis. For example, the point in $3$-space you may think of as the $x$-unit vector ($e_1$) might be what I think of as $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0)$, because my $3$-space has the $xy$-plane rotated 45 degrees.

Given a linear transformation, there isn't THE matrix representation, only *A* linear representation *relative to some choice of bases*. Indeed there is a linear isomorphism from:

$\text{Hom}_F(U,V) \to \text{Mat}_{\dim(V) \times \dim(U)}(F)$

but this isomorphism isn't unique, we get a different one for each pair of bases chosen for $U$ and $V$.

Bases are a great way to turn our calculations with vectors into calculations in the underlying field-but a vector space doesn't "come" with a basis supplied (it doesn't care what coordinate system you choose).

 

[Math Amateur]

A matrix representation of a linear transformation DEPENDS on the basis chosen.

In the same vein, the COORDINATES of a vector ALSO depend on the basis chosen.

By itself, the triple $(2,0,5)$ means nothing-it is just three numbers separated by commas, and enclosed in parentheses.

By convention (and *solely* by convention), elements of $F^n$ (where $F$ is any field) are usually denoted by their representation in the standard basis:

$e_1 = (1,0,\dots,0)$
$e_2 = (0,1,\dots,0)$
$\vdots$
$e_n = (0,0,\dots,1)$so when we say, $(x,y,z) \in \Bbb R^3$, for example, what we really MEAN is, the linear combination:

$xe_1 + ye_2 + ze_3$.

Given an $n$-dimensional vector space, the only point unambiguously defined by an $n$-tuple is the $0$-vector, which is the same in any basis. For example, the point in $3$-space you may think of as the $x$-unit vector ($e_1$) might be what I think of as $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},0)$, because my $3$-space has the $xy$-plane rotated 45 degrees.

Given a linear transformation, there isn't THE matrix representation, only *A* linear representation *relative to some choice of bases*. Indeed there is a linear isomorphism from:

$\text{Hom}_F(U,V) \to \text{Mat}_{\dim(V) \times \dim(U)}(F)$

but this isomorphism isn't unique, we get a different one for each pair of bases chosen for $U$ and $V$.

Bases are a great way to turn our calculations with vectors into calculations in the underlying field-but a vector space doesn't "come" with a basis supplied (it doesn't care what coordinate system you choose).

Well! ... ... That was REALLY HELPFUL!

Thanks Deveno ... that has cleared a few things up for me ...

Thanks again,

Peter