Linear Transformation; Geometric Representation

[srfriggen]

Homework Statement

(note; all column vectors will be represented as transposed row vectors, and matrices will be look like that on a Ti-83 or similar)

L: R^3 -> R^2 is given by,

L([x1, x2, x3]) = [2x1 + x2 - x3
x1 + 3x2 +2x3]*

*Matrix



Relevant Equations:

2x1 + x2 - x3 = b1 (a)
x1 + 3x2 + 2x3 = b2 (b)

Determine whether L is one-to-one

The Attempt at a Solution

I set L(x) = b for a particular vector b = [b1, b2] in R^2 and solved for the x's in terms of b1 and b2.

after using rref I find x1 and x2 are the leading variables, and x3 is arbitrary, which I call r. Solving the equations leads to;

x1 = r + (1/5)(3b1 - b2)

x2 = -r + (1/5)(2b2 - b1)

Since the nonleading variable x3 = r is arbitrary, there are infinitely many solutions for x for a particular b. thus, L is not one-to-one.


My question is, what is happening geometrically in this problem?

Could this be represented as (1) two planes intersecting for form a line, where the line is the solution set?

(2) A plane lying on top of a plane? (I don't think this is true for if it were I could choose any x1 x2 and x3 and would get the same solution in the equations (a) and (b)... right?

(3) A plane in R^3 intersecting the plane that makes up the subspace R^2 but not going "through" the subspace... just touching it to form a line.

(4) A plane in R^3 intersecting the R^2 plane and passing through it (is that possible)?

I suspect either (3) or (4) is happening.

Any insight would be greatly appreciated.

 

[jbunniii]

Just a side note: the fact that L maps R^3 to R^2 tells you immediately that L cannot be one-to-one, because the target space doesn't have high enough dimension. (It would have to be at least 3.)

 

[jbunniii]

Geometrically, I like to think of linear maps in terms of how they transform a unit ball in the domain.

All linear maps change balls into ellipsoids by stretching or squashing along certain axes. (There's a decomposition, called the singular value decomposition, that tells you exactly which axes, and what the stretch or squash factors are.)

A map might squash the sphere so much along one or more axes that it "flattens" the sphere into a lower-dimensional object, e.g., a 3-dimensional ball becomes a 2-dimensional disc. This happens if and only if the map is not one-to-one.

 

[srfriggen]

Geometrically, I like to think of linear maps in terms of how they transform a unit ball in the domain.

All linear maps change balls into ellipsoids by stretching or squashing along certain axes. (There's a decomposition, called the singular value decomposition, that tells you exactly which axes, and what the stretch or squash factors are.)

A map might squash the sphere so much along one or more axes that it "flattens" the sphere into a lower-dimensional object, e.g., a 3-dimensional ball becomes a 2-dimensional disc. This happens if and only if the map is not one-to-one.

That is an extremely interesting idea I must say. This topic gets more and more interesting as I go on.

But can the geometric representation be thought of in one of the ways I described? A week ago it would have seemed obvious that it was the intersection of 2 planes... now when learning about what a Linear Transformation is the idea is becoming a bit more abstract.

Also, the solution set is one dimensional, correct? So using your analogy wouldn't the sphere be squished and stretched to a 1 dimensional object?

 

[jbunniii]

That is an extremely interesting idea I must say. This topic gets more and more interesting as I go on.

But can the geometric representation be thought of in one of the ways I described? A week ago it would have seemed obvious that it was the intersection of 2 planes... now when learning about what a Linear Transformation is the idea is becoming a bit more abstract.

Also, the solution set is one dimensional, correct? So using your analogy wouldn't the sphere be squished and stretched to a 1 dimensional object?

Your map L takes vectors in R^3 and maps them to vectors in R^2, so you already know some kind of "collapse" is occurring.

To understand exactly what is collapsing, look at the null space of the map. This is the set of (x1,x2,x3) such that L(x1,x2,x3) = (0,0).

The null space is a subspace of R^3. What does this mean? All subspaces of R^3 are of one of the following types:

0-dimensional subspace: just the point (0,0,0)
1-dimensional subspace: a line that passes through (0,0,0)
2-dimensional subspace: a plane that passes through (0,0,0)
3-dimensional subspace: all of R^3

For your specific map, the null space is 1-dimensional. This means that there is a certain line in R^3, passing through the origin, which your map collapses to the origin.

For points that aren't on the line, imagine "tiling" all of R^3 with parallel copies of the null space line. These parallel copies of the null space lines are called "cosets" of the null space. Every point of R^3 lies on exactly one such coset line. All points on that coset line are mapped by L to the same point of R^2.

P.S. You can think of each "row" of the matrix of L as a linear map in its own right (from R^3 to R^1), each of which has a null space that is 2-dimensional, hence a plane passing through the origin. The null space of L is then the intersection of these planes, which forms a line. So maybe that's the geometric interpretation you are looking for?

 

[srfriggen]

Your map L takes vectors in R^3 and maps them to vectors in R^2, so you already know some kind of "collapse" is occurring.

To understand exactly what is collapsing, look at the null space of the map. This is the set of (x1,x2,x3) such that L(x1,x2,x3) = (0,0).

The null space is a subspace of R^3. What does this mean? All subspaces of R^3 are of one of the following types:

0-dimensional subspace: just the point (0,0,0)
1-dimensional subspace: a line that passes through (0,0,0)
2-dimensional subspace: a plane that passes through (0,0,0)
3-dimensional subspace: all of R^3

For your specific map, the null space is 1-dimensional. This means that there is a certain line in R^3, passing through the origin, which your map collapses to the origin.

For points that aren't on the line, imagine "tiling" all of R^3 with parallel copies of the null space line. These parallel copies of the null space lines are called "cosets" of the null space. Every point of R^3 lies on exactly one such coset line. All points on that coset line are mapped by L to the same point of R^2.

P.S. You can think of each "row" of the matrix of L as a linear map in its own right (from R^3 to R^1), each of which has a null space that is 2-dimensional, hence a plane passing through the origin. The null space of L is then the intersection of these planes, which forms a line. So maybe that's the geometric interpretation you are looking for?

holy hell that makes things so much clearer now! So it seems the image I had in my head (actually, bad word choice here)... the geometrical interpretation I had was of the null space(s)... two planes intersecting planes forming the line. I was trying to find a geometric interpretation for the images, but I now see that would be a plane in R^2... after you collapse all the co-sets of the null space, along with the null space, you would be left with a plane in R^2, correct? I don't have my book out, but it seems pretty intuitive that plane would be all of R^2...hence the range of L is R^2.

So since my map takes any of those co-sets and "collapses" them down to a point in R^2, it is easy to see that L is not one-to-one (infinitely many points on a line to single point in R^2). I now also see why it's more practical to look at the nullspace to check if L is one-to-one.

If I've misinterpreted anything please let me know, but I feel like I got it, so thank you!

 

[jbunniii]

holy hell that makes things so much clearer now! So it seems the image I had in my head (actually, bad word choice here)... the geometrical interpretation I had was of the null space(s)... two planes intersecting planes forming the line. I was trying to find a geometric interpretation for the images, but I now see that would be a plane in R^2... after you collapse all the co-sets of the null space, along with the null space, you would be left with a plane in R^2, correct? I don't have my book out, but it seems pretty intuitive that plane would be all of R^2...hence the range of L is R^2.

So since my map takes any of those co-sets and "collapses" them down to a point in R^2, it is easy to see that L is not one-to-one (infinitely many points on a line to single point in R^2). I now also see why it's more practical to look at the nullspace to check if L is one-to-one.

If I've misinterpreted anything please let me know, but I feel like I got it, so thank you!

Yes, everything you wrote is correct.

Regarding the image of a linear map, it is always a subspace of the target vector space. In this case your target vector space is R^2. That means the image has to be either all of R^2, or a single line passing through the origin, or just the origin. Those are the only three possibilities.

Moreover, once you know the dimension of the null space, you immediately know the dimension of the image as well, through this important theorem:

If V and W are (finite-dimensional) vector spaces and L is a linear map from V to W, then

dim(V) = dim(image(L)) + dim(null(L))

Since the null space of your particular L has dimension 1, and dim(V) = dim(R^3) = 3, it follows that dim(image(L)) = 2. Therefore the image of L is all of R^2.

If the null space were larger, say an entire plane in R^3, then the image of the map gets smaller: a single line in R^2.

And of course if the null space is all of R^3, then every point gets mapped to the origin of R^2, so the image is zero-dimensional: just the origin.

 

[srfriggen]

Yes, everything you wrote is correct.

Regarding the image of a linear map, it is always a subspace of the target vector space. In this case your target vector space is R^2. That means the image has to be either all of R^2, or a single line passing through the origin, or just the origin. Those are the only three possibilities.

Moreover, once you know the dimension of the null space, you immediately know the dimension of the image as well, through this important theorem:

If V and W are (finite-dimensional) vector spaces and L is a linear map from V to W, then

dim(V) = dim(image(L)) + dim(null(L))

Since the null space of your particular L has dimension 1, and dim(V) = dim(R^3) = 3, it follows that dim(image(L)) = 2. Therefore the image of L is all of R^2.

If the null space were larger, say an entire plane in R^3, then the image of the map gets smaller: a single line in R^2.

And of course if the null space is all of R^3, then every point gets mapped to the origin of R^2, so the image is zero-dimensional: just the origin.

that's a great theorem... can I use it to say that if if L is one-to-one then dim(V)=dim(image) ? (I don't think i could say "iff" in that case).

 

[jbunniii]

that's a great theorem... can I use it to say that if if L is one-to-one then dim(V)=dim(image) ? (I don't think i could say "iff" in that case).

Yes, and in fact you CAN say "iff", provided that dim(V) is finite:

For any linear map, L is one to one iff dim(null(L)) = 0

Also, if dim(V) is finite, then

dim(null(L)) = 0 iff dim(V) = dim(image(L))

If dim(V) is infinite, then the ==> direction is still true, but the <== direction need not be true.

 

[srfriggen]

Yes, and in fact you CAN say "iff", provided that dim(V) is finite:

For any linear map, L is one to one iff dim(null(L)) = 0

Also, if dim(V) is finite, then

dim(null(L)) = 0 iff dim(V) = dim(image(L))

If dim(V) is infinite, then the ==> direction is still true, but the <== direction need not be true.

I see... wow this really is a great course! Starting to have a lot of fun with it, thanks for the help!

 

[jbunniii]

I see... wow this really is a great course! Starting to have a lot of fun with it, thanks for the help!

Glad you're enjoying it - linear algebra is a really interesting and deep subject once you start thinking about it at a bit higher level than just matrix manipulations. Hopefully your course will get around to proving all the assertions I made in this thread. But meanwhile your intuition seems good and you are asking the right sorts of questions to get a deeper understanding.