Hi, Thanks! This is great stuff. At the beginning of the video you said you "made it easy" and the system had a unique solution where [x,y] = [1,1]. Then, you changed the right side so there was no solution. In explaining this geometrically, you analogized to R3 with 2 vectors and asked, "What are the chances that a 3rd, randomly selected vector will lie in the same plane spanned by the first 2 vectors? Very unlikely." (I paraphrased that a bit.) So, is it correct to interpret the earlier statement that you 'made it easy' to mean that you CAREFULLY picked a solution (NOT random at all) so that, even though it is unlikely that a RANDOMLY selected vector would lie in the same space spanned by the first two vectors, in this case, it did because of your set up? And, once you picked that special vector, we are then left with only one solution (in this case [1,1]) but, if you had carefully picked another vector that lies in the space, perhaps [2,2,18,36] which is twice the first column and twice the second column, then the system would have the unique solution [2,2]. If this is correct, then there are infinitely many such vectors in the space spanned by the first two and, once we pick the 'solution vector' we lock down [x,y]. Is that correct? Thanks again for these outstanding videos! Steve
your probably long gone ...but I had to pause for the same reason I think ....... I understood it as ....that RANDOM ['arbitrary'] solution is unlikely to BE a solution [Pavel said HAVE a solution which I can't make sense of] as it's unlikely to be in the 'plane' of the 1st 2 vectors .....but there are infinitely many vectors as solutions in that 'plane' we're just unlikely to get one picking at random.
@@redberries8039 Both of you: no, Pavel very much meant "HAVE as solution". Remember, the vector on the right is not a solution, it's part of the question. The vector you get for [x,y] is the solution (if it exists). Not having a solution means that no combination of (in this case) x and y will cause the 4d vectors on the left to reach the 4d vector on the right.
You are confusing infinite solutions here. Infinite solutions in previous videos meant you could have infinite values of vector X = *ONE PARTICULAR VECTOR* but if you change the RHS, you'll definitely be getting a different solution but you can't call it the case of infinite solutions because it will never give two solutions for (I'm repeating again) *ONE PARTICULAR VECTOR* in the column space of coefficient matrix.
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
super
Hi,
Thanks! This is great stuff. At the beginning of the video you said you "made it easy" and the system had a unique solution where [x,y] = [1,1]. Then, you changed the right side so there was no solution. In explaining this geometrically, you analogized to R3 with 2 vectors and asked, "What are the chances that a 3rd, randomly selected vector will lie in the same plane spanned by the first 2 vectors? Very unlikely." (I paraphrased that a bit.)
So, is it correct to interpret the earlier statement that you 'made it easy' to mean that you CAREFULLY picked a solution (NOT random at all) so that, even though it is unlikely that a RANDOMLY selected vector would lie in the same space spanned by the first two vectors, in this case, it did because of your set up?
And, once you picked that special vector, we are then left with only one solution (in this case [1,1]) but, if you had carefully picked another vector that lies in the space, perhaps [2,2,18,36] which is twice the first column and twice the second column, then the system would have the unique solution [2,2].
If this is correct, then there are infinitely many such vectors in the space spanned by the first two and, once we pick the 'solution vector' we lock down [x,y]. Is that correct?
Thanks again for these outstanding videos!
Steve
your probably long gone ...but I had to pause for the same reason I think ....... I understood it as ....that RANDOM ['arbitrary'] solution is unlikely to BE a solution [Pavel said HAVE a solution which I can't make sense of] as it's unlikely to be in the 'plane' of the 1st 2 vectors .....but there are infinitely many vectors as solutions in that 'plane' we're just unlikely to get one picking at random.
@@redberries8039 Both of you: no, Pavel very much meant "HAVE as solution". Remember, the vector on the right is not a solution, it's part of the question. The vector you get for [x,y] is the solution (if it exists). Not having a solution means that no combination of (in this case) x and y will cause the 4d vectors on the left to reach the 4d vector on the right.
You are confusing infinite solutions here. Infinite solutions in previous videos meant you could have infinite values of vector X = *ONE PARTICULAR VECTOR* but if you change the RHS, you'll definitely be getting a different solution but you can't call it the case of infinite solutions because it will never give two solutions for (I'm repeating again) *ONE PARTICULAR VECTOR* in the column space of coefficient matrix.