Linear Algebra, Linear Transformations
A function f that maps one vector space into another vector space
with the same scalars,
and satisfies f(a*x+b*y) = a*f(x)+b*f(y),
is a linear transformation.
In other words,
f commutes with addition and scaling, or if you prefer,
f is a module homomorphism.
This definition is disturbing when R is noncommutative.
(You can skip this part if you know you are working over the reals,
the complex numbers, or other commutative fields.)
We would like the function cx, scaling all vectors by c, to be linear.
It certainly looks linear.
And indeed c(x+y) = cx+cy, so we're ok there.
But premultiplication by c does not respect scaling.
In other words, c(ax) is not the same as (ac)x, because a and c do not commute.
The workaround is to multiply by c on the right.
Now (ax+by)c = a(xc)+b(xc).
Postmultiplication by c is indeed a linear map,
or a left module homomorphism if you prefer.
This assumes the vectors can be scaled on either side.
If you want postmultiplication to be a linear map,
our vector space is more than a left R module,
it is a bimodule,
with R on either side.
Of course the vector space could be a right R module, whence premultiplication by a constant implements a linear map.
Ok, that's enough of noncommutative rings.
Just to confuse you,
a linear transformation is also called a linear function,
or a linear map,
or a linear operator,
or a linear transform.
However, the terms operator and transform are usually used when the vectors are functions.
Yes, a function can map functions to functions.
Consider the set of continuous real valued functions on the closed interval [0,1].
These functions can be added and scaled, hence we have a vector space.
The operator 2f takes a function and doubles it.
Since 2f commutes with addition and scaling, it is a linear operator.
We will explore this topic in another section.
For now, let's concentrate on linear transformations.
Lines to Lines
Since the image of k times a vector is k times the image of the vector,
a linear transformation maps lines to lines in real space.
This is clear for lines that pass through the origin,
but it holds for all lines.
An off-origin line consists of an offset vector,
plus all scalar multiples of another vector.
For instance, the line y = 1 is [0,1] plus all multiples of [1,0].
The image of this line is a new offset vector plus multiples of another vector, hence another line.
Also, there is a fixed scale, a level of magnification,
for any given line.
If an inch of line is expanded to a mile under the transformation,
every inch of that line expands to a mile.
Of course other lines may expand by a different amount,
or they may be squashed down to a point (expanded by 0).
A simple transformation squashes 3 space down onto the xy plane.
If a line rises up from the plane at an angle, it is pushed back down into the plane,
and distance is compressed by a ratio that is a function of the angle of inclination.
The entire z axis is squashed into the origin, and distance is scaled by 0.
The kernel of a transformation is the set of points that are mapped to 0.
Verify that the kernel is a subspace.
In the previous example, the kernel is the z axis,
and the quotient space, or image space, is the xy plane.
Relationships Among Dimensions
Since the kernel is a subspace, assign it a basis x.
Expand this into a larger basis for the entire space.
Let y be the set of additional basis elements, not present in x.
Let w be the images of y under the transformation f.
Recall that everything in x, and everything spanned by x, maps to 0 under f.
If a linear combination of elements in w becomes 0,
then the same linear combination of preimages from y maps to something in the kernel,
which is spanned by x.
This means x∪y is not an independent set; not a basis.
Hence w is an independent set in the image space.
since the domain is spanned by x∪y, the image is spanned by w,
and w is a basis for the image space.
When the dimensions are finite,
the dimension of a space equals the dimension of its kernel plus the dimension of the image space.
Return to our example, f(x,y,z) = x,y,0.
Recall that the z axis is the kernel of this squashing transformation.
Let [0,0,1] be a basis for this kernel.
Extend this basis by [1,1,1] and [-1,1,1] to cover all of 3 space.
The image of these two basis elements is [1,1] and [-1,1], which is a basis for the xy plane.
A 3 dimensional space maps onto a 2 dimensional space with a 1 dimensional kernel.
Assume two subspaces are combined to span a larger space.
Think of two planes that combine to span 3 space, and intersect in a line.
Let x be a basis for the intersection.
Extend this by the vectors in y to build a basis for one of the subspaces,
and by the vectors in z to build a basis for the other subspace.
Now x∪y∪z spans the entire space.
Suppose a linear combination of vectors from this union produces 0.
Since x∪y is an independent set,
and x∪z is an independent set,
our linear combination must include vectors from both y and z.
combine the z vectors together to get a new nonzero vector that is spanned by x∪y.
However, if this vector, spanned by z, is contained in the space spanned by x and y,
it is in the intersection, and should be spanned by x alone.
Therefore x∪y∪z is an independent set,
and is a basis for the entire space.
The dimension of the space is the sum of the dimensions of the two subspaces,
minus the dimension of their intersection.