One extra fact that helped me with the proof at 4:55 (3.108): the annihilator of an entire vector space V is always {0} in V' (similarly, the annihilator of {0} in V is the whole dual space V').
Hi, Prof Axler! May I ask a question? When proving the second half of "range T' = (null T)^0" in the book (page 108), why it's sufficient to prove that they have the same dimension instead of proving "(null T)^0 ⊆ range T'"? Thank you!
In a finite-dimensional vector space, if one subspace is contained in another subspace and the two subspaces have the same dimension, then the two subspaces are equal (see Exercise 1 on page 48; the idea here is that a basis of the smaller subspace can be extended to a basis of the larger subspace, but because the two subspaces have the same dimension, the extension does not add any new vectors to the basis).
Thank you, Prof. Axler, for taking your precious time to answer my elementary question! Your book and videos are godsends. I took only some calculation-based math courses in college more than a dozen years ago, have rather limited proof-related experiences, and only begin to get a glimpse of what real math looks like because of your wonderful works.
@@sheldonaxler5197 By the way, professor, would you be by any chance planning to make a similar video series on your new book Measure, Integration & Real Analysis? That would be really phenomenal! ❤️
@@andyl.5998 I have not yet definitely decided, but I am leaning toward making a series of videos to accompany my new book Measure, Integration & Real Analysis. If I do make those videos, you will be able to find links to them on the book's website (measure.axler.net/) around September 2020. That website also contains a link to download the pdf file for the book legally and free.
I have a question. Is it possible to prove rangeT=W, if (rangeT)^{0}={0}, without using 3.107(b)? Because by using rangeT=W and 3.107(b), we can imply nullT'={0}, and we don't need the intermediate step (rangeT)^{0}={0}. So why is it mentioned?
At 11:00, the row rank of A is 2 instead of 4 because the 2 rows are linearly independent and all linear combinations of them should span a plane in R^4?
One extra fact that helped me with the proof at 4:55 (3.108): the annihilator of an entire vector space V is always {0} in V' (similarly, the annihilator of {0} in V is the whole dual space V').
thanks
I think you've misspoke at 5:52, it should probably be "well, if it's easier than proving that something is surjective".
7:07 May I prove dim range T= dim range T' this way?dim rangeT'=dim (kerT)^0=dimV-dim nullT=dim rangeT.
Certainly! I actually prefer this proof, as it doesn't require Result 3.95
yes, but i suppose the point was to simultaneously utilize the previous If ye know one is a subspace of the other
好好学习，天天向上
Hi, Prof Axler! May I ask a question? When proving the second half of "range T' = (null T)^0" in the book (page 108), why it's sufficient to prove that they have the same dimension instead of proving "(null T)^0 ⊆ range T'"? Thank you!
In a finite-dimensional vector space, if one subspace is contained in another subspace and the two subspaces have the same dimension, then the two subspaces are equal (see Exercise 1 on page 48; the idea here is that a basis of the smaller subspace can be extended to a basis of the larger subspace, but because the two subspaces have the same dimension, the extension does not add any new vectors to the basis).
Thank you, Prof. Axler, for taking your precious time to answer my elementary question!
Your book and videos are godsends. I took only some calculation-based math courses in college more than a dozen years ago, have rather limited proof-related experiences, and only begin to get a glimpse of what real math looks like because of your wonderful works.
@@sheldonaxler5197 By the way, professor, would you be by any chance planning to make a similar video series on your new book Measure, Integration & Real Analysis? That would be really phenomenal! ❤️
@@andyl.5998 I have not yet definitely decided, but I am leaning toward making a series of videos to accompany my new book Measure, Integration & Real Analysis. If I do make those videos, you will be able to find links to them on the book's website (measure.axler.net/) around September 2020. That website also contains a link to download the pdf file for the book legally and free.
Fantastic!
I have a question. Is it possible to prove rangeT=W, if (rangeT)^{0}={0}, without using 3.107(b)? Because by using rangeT=W and 3.107(b), we can imply nullT'={0}, and we don't need the intermediate step (rangeT)^{0}={0}. So why is it mentioned?
At 11:00, the row rank of A is 2 instead of 4 because the 2 rows are linearly independent and all linear combinations of them should span a plane in R^4?
The statement in the video is correct: the row rank of A equals 2. The row rank of A cannot equal 4 because a list of length 2 cannot span R^4.
The video is fine
Helpfull