Hello Prof. Axler Example 2 of Page No 61 of the book and 5:00 of this video mentions that (2,0,1) and (0,5,1) form a basis for the Range of T (which is basically R^3) My question is how can a vector list of length 2 span or form a basis of R^3
@@sheldonaxler5197 Thanks for the reply Sir However this leads us further to a question ‘Does there exist any Linear Map T from R^2 to R^3 where the Range of T equals whole of the R^3?’
In the proof of a linear map T from V to W not being injective when dim V > dim W, it is stated (similar to the way it is stated in pp. 64 of the book) that: dim V > dim W ≥ dim range T. It is explained in the video that what is meant exactly here is dim V > dim W ≥ dim range T. But without the explanation in the video, these inequalities are easily interpreted as dim V > dim W and dim V ≥ dim range T instead and this leads to some head scratching! I wish what is meant were expressed in the related inequalities more clearly!
@dragonzito mojado {0} is an infinite-dimensional vector space since its basis can be => {v1,v2.....vn} for all n in N And {0}={0.v1, 0.v2......0.vn} So for the good of all, dim{0} is taken as zero
@@rahuldalkari4285 basis of a vector space should be a subset ,vector space {0} only contains one vector 0, you can take a non zero scalar a and show a.0=0 implies 0 is linearly dependent..and so its dimension is zero
@@rahuldalkari4285 the span of the empty list is defined to be {0}. the empty list is also declared to be linearly independent. hence the empty list is a basis of {0}. thus dim{0} = the length of the empty list = 0.
Not necessarily. Suppose e_1, ..., e_n is a basis of V. Define linear maps T_j (j going from 1 to n) from V to V by the action T_j e_k = \delta_jk e_1 on basis vectors. Here \delta_jk = 1 if j=k and =0 otherwise. It is easy to check that the maps T_1, .. T_n are linearly independent, but the range of each is just span(e_1).
A very interesting question, this, in my p.o.v requires a lot to disprove. =>It is to be noted that phi (empty set) cannot be obtained, since all subspaces have the zero element in them: = It is to be noted that "as you should verify", implies that the deduction could be done by common sense, as it is left for the reader to do so. Let's say Hypothetically, that V is a vector space of dim = n, and for the sake of argument, that L(V,W) is the set of linear transformations of V on W, where W is A space of dim=m. Make-up Definition: Two semi-Disjoint spaces are two spaces which are defined by V intersection W = {0} Goal: if a Set of linear maps is linearly independent, then the intersection of ranges of all these maps = {0) Deduction I: => We know that L(V,W) is a vector space, so let us, for sake of _possible_ necessity, find it's dim: i- We need to do some cheating here and activate our basic understanding of matrices, we know that a linear transformation, can be expressed in the form of A matrix multiplication (Revise the preceding chapters), so this reveals the nature of L(V,W), which is a Matrix space. ii- the dim, henceforth, of L, could be deduced like this, by the law of matrix multiplication, for any matrix multiplication, the number of columns of the first matrix, must be equal to the row of the second one, the second matrix is a vector of Dim n, therefore the transformation has n columns, and since the resulting vector is a m-vector, therefore, the matrix has m rows, this implies that the dim of L(V,W) is nxm, as you should verify. => then the number of bases of L(V,W) is nxm, therefore, set of linearly independent transformations is equal to mxn. Deduction II: => we know that the range of a transformation is a subspace of W. => the set of ranges of each basis of L, form a set of subspaces on W. Question: Does this set of ranges form a subspace in W? Common sense: there are no possible Linear transformation From V to W that will give a vector outside W => the union of all those subspaces form W. The union of all those subspaces are the union of all the ranges, right? the union of all the ranges form a vector space of dim M, but the number of those ranges are mxn>m. therefore, as you should verify, they are not linearly independent.
Dear Prof. Axler, on p.61 of the textbook you mention that the differentiation map D:P(R) -> P(R) is surjective. If I am not mistaken, P(R) is the vector space of polynomials with degree m with coefficients in R. However, given that applying D to a polynomial reduces its degree by 1, it seems to me the range of D is of dimension "m" rather than "m+1", the latter being the dimension of P(R). If that is the case, then the range and vector space do not coincide, and D is not surjective. Could you please point out what I am missing. Thank you.
P(R) is the vector space of polynomials with coefficients in R. This is an infinite-dimensional vector space, not a vector space of dimension m+1 as written in the comment above.
@@sheldonaxler5197 Thank you so much Prof. Axler. I had completely misunderstood the definition of P(R) and missed Example 2.16 on p.32. I just want to say your text is excellent and I'm looking forward to buying your Measure Theory text too.
Thank you for video, it reminds, stresses concepts and useful when doing exercises.
Hello Prof. Axler
Example 2 of Page No 61 of the book and 5:00 of this video mentions that (2,0,1) and (0,5,1) form a basis for the Range of T (which is basically R^3)
My question is how can a vector list of length 2 span or form a basis of R^3
The range of that operator T is not equal to R^3. Instead, the range is a two-dimensional subspace of R^3.
@@sheldonaxler5197 Thanks for the reply Sir
However this leads us further to a question ‘Does there exist any Linear Map T from R^2 to R^3 where the Range of T equals whole of the R^3?’
@@monislamba3036 The answer to your question is "no", as shown by 3.24 in the third edition of Linear Algebra Done Right.
In the proof of a linear map T from V to W not being injective when dim V > dim W, it is stated (similar to the way it is stated in pp. 64 of the book) that:
dim V > dim W
≥ dim range T.
It is explained in the video that what is meant exactly here is dim V > dim W ≥ dim range T.
But without the explanation in the video, these inequalities are easily interpreted as
dim V > dim W and dim V ≥ dim range T
instead and this leads to some head scratching!
I wish what is meant were expressed in the related inequalities more clearly!
So I think for proof at 8:50 to make sense for me, the dim of {0} should be zero.
@dragonzito mojado {0} is an infinite-dimensional vector space since its basis can be => {v1,v2.....vn} for all n in N
And {0}={0.v1, 0.v2......0.vn}
So for the good of all, dim{0} is taken as zero
@@rahuldalkari4285 basis of a vector space should be a subset ,vector space {0} only contains one vector 0, you can take a non zero scalar a and show a.0=0 implies 0 is linearly dependent..and so its dimension is zero
@@rahuldalkari4285 the span of the empty list is defined to be {0}. the empty list is also declared to be linearly independent. hence the empty list is a basis of {0}. thus dim{0} = the length of the empty list = 0.
Dear prof., I have one doubt. If I have a set of linearly independent linear maps then will the ranges of all these maps be disjoint subspaces?
Not necessarily. Suppose e_1, ..., e_n is a basis of V. Define linear maps T_j (j going from 1 to n) from V to V by the action T_j e_k = \delta_jk e_1 on basis vectors. Here \delta_jk = 1 if j=k and =0 otherwise. It is easy to check that the maps T_1, .. T_n are linearly independent, but the range of each is just span(e_1).
A very interesting question, this, in my p.o.v requires a lot to disprove.
=>It is to be noted that phi (empty set) cannot be obtained, since all subspaces have the zero element in them:
= It is to be noted that "as you should verify", implies that the deduction could be done by common sense, as it is left for the reader to do so.
Let's say Hypothetically, that V is a vector space of dim = n, and for the sake of argument, that L(V,W) is the set of linear transformations of V on W, where W is A space of dim=m.
Make-up Definition: Two semi-Disjoint spaces are two spaces which are defined by V intersection W = {0}
Goal: if a Set of linear maps is linearly independent, then the intersection of ranges of all these maps = {0)
Deduction I:
=> We know that L(V,W) is a vector space, so let us, for sake of _possible_ necessity, find it's dim:
i- We need to do some cheating here and activate our basic understanding of matrices, we know that a linear transformation, can be expressed in the form of A matrix multiplication (Revise the preceding chapters), so this reveals the nature of L(V,W), which is a Matrix space.
ii- the dim, henceforth, of L, could be deduced like this, by the law of matrix multiplication, for any matrix multiplication, the number of columns of the first matrix, must be equal to the row of the second one, the second matrix is a vector of Dim n, therefore the transformation has n columns, and since the resulting vector is a m-vector, therefore, the matrix has m rows, this implies that the dim of L(V,W) is nxm, as you should verify.
=> then the number of bases of L(V,W) is nxm, therefore, set of linearly independent transformations is equal to mxn.
Deduction II:
=> we know that the range of a transformation is a subspace of W.
=> the set of ranges of each basis of L, form a set of subspaces on W.
Question: Does this set of ranges form a subspace in W?
Common sense: there are no possible Linear transformation From V to W that will give a vector outside W
=> the union of all those subspaces form W.
The union of all those subspaces are the union of all the ranges, right? the union of all the ranges form a vector space of dim M, but the number of those ranges are mxn>m.
therefore, as you should verify, they are not linearly independent.
Dear Prof. Axler, on p.61 of the textbook you mention that the differentiation map D:P(R) -> P(R) is surjective. If I am not mistaken, P(R) is the vector space of polynomials with degree m with coefficients in R. However, given that applying D to a polynomial reduces its degree by 1, it seems to me the range of D is of dimension "m" rather than "m+1", the latter being the dimension of P(R). If that is the case, then the range and vector space do not coincide, and D is not surjective. Could you please point out what I am missing. Thank you.
P(R) is the vector space of polynomials with coefficients in R. This is an infinite-dimensional vector space, not a vector space of dimension m+1 as written in the comment above.
@@sheldonaxler5197 Thank you so much Prof. Axler. I had completely misunderstood the definition of P(R) and missed Example 2.16 on p.32. I just want to say your text is excellent and I'm looking forward to buying your Measure Theory text too.