(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 \(\displaystyle T: \mathbb{R}^3 \longrightarrow \mathbb{R}^2\) , say ...

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

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

... ... ...

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

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

Is the expression for the linear transformation

\(\displaystyle T: \mathbb{R}^3 \longrightarrow \mathbb{R}^2\)

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

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

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

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

in terms of the basis \(\displaystyle B'\) or in terms of the standard basis?[By the way, I think it is, by convention, that linear transformations from \(\displaystyle \mathbb{R}^n\) to \(\displaystyle \mathbb{R}^m\) 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 \(\displaystyle B\) and \(\displaystyle B'\), I am not sure ... ... ]
Question 2

Does the matrix of the transformation

\(\displaystyle A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 5 \end{bmatrix}\)

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

or

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

where \(\displaystyle S_1\) is the standard basis for \(\displaystyle \mathbb{R}^3\)

and \(\displaystyle S_2\) is the standard basis for \(\displaystyle \mathbb{R}^2\)
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