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Linear Algebra: kernel and range of translation confusion
Jimmy King
Registered User regular
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I have no idea what is going on. This all made sense in class, what I am doing matches my notes and it matches what the book says to do, yet I don't get the same answers as the book. Most of the time I can see what the book did to get the answer and it's not doing what it says to when explaining how to work these problems. Please take a look at these and explain to me what the hell I am misunderstanding about what the book is doing here.
Now, maybe there is some difference between the range of T and a basis for the range of T, but I wouldn't know since the book doesn't explain it or show any examples of anything except finding a basis for the range of T.
My notes from class and the one example of doing this in the book says to take that reduced form of A, find the pivot columns, then take those same columns from A as part of the basis of the range, so in this case we have a row with 1 in column 1 as a leading value, and a row with a 1 in column 2 as a leading value, columns 1 and 2 from A are the basis of the range.
The example given is T(v) = Av where A is
[ 1 2 0 1 -1]
[ 2 1 3 1 0]
[-1 0 2 0 1]
[ 0 0 0 2 8]
They then reduce that as far as possible to
[1 0 2 0 -1]
[0 1 -1 0 -2]
[0 0 0 1 4]
[0 0 0 0 0]
And so the basis for the range is
{(1, 2, -1, 0), (2, 1, 0, 0), (1, 1, 0, 2)}
because there is a row with a 1 as the leading value in column 1, a row with a 1 as the leading value in column 2, and a row with 1 as the leading value in column 3.
So to the homework -
For the first couple: T(v) = Av represents the linear transformation T. Find a bases for (a) the kernel of T and (b) the range of T.
First problem:
A = [1 2]
[3 4]
a) So I'm looking to do
[1 2][X1] = [0]
[3 4][X2] = [0]
get A to reduced row echelon form of [1 0]
[0 1]
and we can see that
X1 = 0, X2 = 0, so ker(t) = {(0, 0)}, which is what the book says, so that's cool.
b) From the example above and my notes, the answer should be {(1, 3), (2, 4)}. The book's answer is {(1, 0), (0, 1)}. Why?
There are a couple more where they use the reduced form of A instead of referring back to the actual columns of A and I have no idea why.
Then there are these gems
Define the linear transformation T by T(x) = Ax. Find (a) ker(T), (b) nullity(T), (c) range(T), and (d) rank(T)
A =
[5 -3]
[1 1]
[1 -1]
A) ker(T) = {(0, 0)}
nullity(T) = 0 (that's fine... rank(T) - dim(T))
D) rank(T) = 2
but C...
range(T) = {4s, 4t, s-t)} according to the book... what?
Based on the only example of finding anything related to the range above we take A and get it reduced as far as possible, which gives us
[1 0]
[0 1]
[0 0]
We throw out that last row and range = {(5, 1, 1), (-3, 1, -1)}
There are a couple more with similar issues, but maybe if I can figure out what the hell is going on with these, I can figure those out.
Now, maybe there is some difference between the range of T and a basis for the range of T, but I wouldn't know since the book doesn't explain it or show any examples of anything except finding a basis for the range of T.
My notes from class and the one example of doing this in the book says to take that reduced form of A, find the pivot columns, then take those same columns from A as part of the basis of the range, so in this case we have a row with 1 in column 1 as a leading value, and a row with a 1 in column 2 as a leading value, columns 1 and 2 from A are the basis of the range.
The example given is T(v) = Av where A is
[ 1 2 0 1 -1]
[ 2 1 3 1 0]
[-1 0 2 0 1]
[ 0 0 0 2 8]
They then reduce that as far as possible to
[1 0 2 0 -1]
[0 1 -1 0 -2]
[0 0 0 1 4]
[0 0 0 0 0]
And so the basis for the range is
{(1, 2, -1, 0), (2, 1, 0, 0), (1, 1, 0, 2)}
because there is a row with a 1 as the leading value in column 1, a row with a 1 as the leading value in column 2, and a row with 1 as the leading value in column 3.
So to the homework -
For the first couple: T(v) = Av represents the linear transformation T. Find a bases for (a) the kernel of T and (b) the range of T.
First problem:
A = [1 2]
[3 4]
a) So I'm looking to do
[1 2][X1] = [0]
[3 4][X2] = [0]
get A to reduced row echelon form of [1 0]
[0 1]
and we can see that
X1 = 0, X2 = 0, so ker(t) = {(0, 0)}, which is what the book says, so that's cool.
b) From the example above and my notes, the answer should be {(1, 3), (2, 4)}. The book's answer is {(1, 0), (0, 1)}. Why?
There are a couple more where they use the reduced form of A instead of referring back to the actual columns of A and I have no idea why.
Then there are these gems
Define the linear transformation T by T(x) = Ax. Find (a) ker(T), (b) nullity(T), (c) range(T), and (d) rank(T)
A =
[5 -3]
[1 1]
[1 -1]
A) ker(T) = {(0, 0)}
nullity(T) = 0 (that's fine... rank(T) - dim(T))D) rank(T) = 2
but C...
range(T) = {4s, 4t, s-t)} according to the book... what?
Based on the only example of finding anything related to the range above we take A and get it reduced as far as possible, which gives us
[1 0]
[0 1]
[0 0]
We throw out that last row and range = {(5, 1, 1), (-3, 1, -1)}
There are a couple more with similar issues, but maybe if I can figure out what the hell is going on with these, I can figure those out.
Jimmy King on
0
Posts
A =
[1 2]
[3 4]
to the identity, then [1 3] and [2 4] span R^n. But an equivalent pair of vectors that are way easier to work with in a practical sense are [1 0] and [0 1] so let's go with that instead. You did it right, and your answer is also correct, but it is more correct to use the columns of the n x n identity matrix when this happens. So that should solve all of those kinds of problems where you're getting the columns of A and the book's answers are the columns of the equivalent identity matrix.
For your second question:
I genuinely don't know why they phrased it like that. However, what a basis means is that you can form any vector v in the space V by taking sums of n vectors vn, i.e. for all v in V there exist a v1, v2, ..., vn such that v = c1v1 +c2v2 + ... + cnvn. Then we call v1, v2, ... vn a basis.
So let's call c1 = s and c2 = t (because it's a transformation we use those as variables instead of x and y, conventions are weird) in your question. Then what you're saying is that the set of vectors that is the range of the linear transformation T is:
{5s - 3t, s + t, s - t} and now that looks somewhat like the answer they're getting. And in fact, if you play with those, you'll find with some quick row operations you can make it look like {4s, 4t, s - t}.
Basically, if you can transition from one set of elements to another in a matrix via the elementary row operations, then those are equivalent from the standpoint of basis vectors.
So [1 3] and [2 4] is equivalent to [1,0] and [0,1] or [5, 1, 1] and [-3, 1, -1] is equivalent to [4, 0, 1] and [0, 4, -1]. Two different ways of establishing the same space.
At least I'm pretty sure I remember that right. I know my explanation to your first question is right, not 100% sure about the second. If anyone has a suggestion on why they established the book answer like they did, I'd love to hear it. Like those are prettier, but why not reduce further, you know?
The basis of a vector space Q is a minimal set of vectors such that you can get any vector in the space Q by performing linear operations on the basis vectors. E.g. {(1,0,0),(0,0,1)} is a basis for the X-Z plane; the vectors which lie on X-Z plane form a vector space. There are an infinite number of bases for any vector space. E.g. {(2, 0, 0),(0,0,6)}, or {(1, 0, 1), (1, 0, -1)} are also bases for the X-Z plane. Usually when asked for a basis of a vector space, any basis is correct (unless you're specifically asked for an ortho-normal basis, for example).
When you're trying to find the range of a matrix, you're (usually) interested in the space, not any particular set of basis vectors. So in this case {(1,3), (2,4)} and {(1,0), (0,1)} are both sets of basis vectors for the same space (the whole plane), so both are correct.
For the last question, I'm going to guess (and I haven't worked it out), that {(4, 0, 1),(0, 4, -1)} is a basis for range(T), in which case every vector in range(T) can be written as (4s, 4t, s-t) for an appropriate choice of s and t.
For the first one, is whatever rows are pivots from the reduced form of the matrix always going to be valid as an answer or is that only in certain cases, like when it's able to be reduced all the way down to an identity matrix?
For the second one, why would I ever try to get it to that {(4, 0, 1), (0, 4, -1)} form and then why would it write it like they did using S and T as they did instead of the specific vectors? To get it to that {(4, 0 ,1), (0, 4, -1)} form you have to go out of your way when reducing it. In just a couple of steps I get it to
[1 0]
[0 1]
[5 -3]
And then you can obviously go two more to get it to
[1 0]
[0 1]
[4 -4]
Then multiply R3 by 1/4 R3 to get it to [1 -1] in R3. And then completely pointlessly multiple R1 and R2 each by 4 for no apparent reason other than to confuse students who are using this book. I guess maybe there's some other technique for finding this which comes up with that answer more naturally.
And then I guess when being very specific and correct writing the solutions out with the variables makes it "The range" whereas writing out the specific columns makes it "A basis for the range" because it's just one specific answer. Which is fine and basically goes back to whoever came up with the answers not actually reading the chapter to make sure their answers aligned with what the book says.
On the second question, I genuinely don't know why they chose to write it like that. I mean the [1 -1] appeals to me, but otherwise, nope. And yeah, you're right that technically that's the range with the s and t and I was choosing to write the basis, because a basis for a space defines the space. Basically: whatever, book. I think you know what you're doing and you should be getting full credit if you handed in what you had answered in your OP.
Based on other books in the same series my school used for my calculus classes and comments those teachers made about some of the answers in those, I'm going to have to guess some grad student was given the problem and told "here, solve this", but not given the chapter to read (or just didn't read it) to solve it in the way the book explains.
Thanks for the help on this stuff. Now I've got to go finish this nonsense.
I took linear algebra in college. I absolutely hated it but i managed to get through it. I tried to read your post and my vision is literally blurring every time i try to read the problem. It's like my brain is rejecting even looking at the shit anymore.
I'm beginning to think the class literally traumatized me.
So defining the linear transformation T by T(x) = Ax and given:
A =
[5 -3]
[1 1]
[1 -1]
find Range(T)
The key to this is to realize that the Range(T) is the Column Space of A.
So take the transpose of A which yields:
[5 1 1]
[-3 1 -1]
then row reduce and get:
[1 0 1/4]
[0 1 -1/4]
The author doesn't like to have fractions so he multiplies this matrix by 4 yielding:
[4 0 1]
[0 4 -1]
Since this represents the Column Space these are column vectors:
[4] [0]
[0] [4]
[1] [-1]
If each of these are parameterized then:
multiply t by the first vector and s by the second vector yielding:
[4t] [0]
[0] [4s]
[t] [-s]
Since the sum of these vectors are equal to:
[x1]
[x2]
[x3]
then
4t + 0 = x1
0 + 4s = x2
t - s = x3
and that's how we get Range(T) = {(x1, x2, x3)} = {(4t, 4s, t - s)}
The posts here helped me bigtime in finding this out and I'll use the same technique for other problems. Hope this helps others.