Linear Algebra 4.2.2 Column Spaces

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Kimberly Brehm

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15:50 is a mistake guys she was confused for some reasons, the reasoning does not add up. V is in column space of A

resoluation

Wow, this video is an absolute gem for anyone diving into the world of linear algebra! The explanations are clear and concise, making complex concepts seem so much more approachable. I appreciate how the instructor breaks down each topic, from vectors to matrices and beyond. The visual aids and examples provided really help solidify the understanding. Linear algebra can be a challenging subject, but this video makes it feel like a breeze. Kudos to the creator for making such an enlightening and well-crafted tutorial! 🌟

mohammadshujauddin

Studying for finals and have been super lost this whole semester, these videos have cleared up so much for me. Thanks

jollyroger

In 15:50, v could be linear combination of columns of the matrix instead of column itself. You might have made a mistake there.

zhihaojia

Why in the last practice of the video vector v is not in the Column Space of A? Vector v can be written as a linear combination of columns of matrix A: v=(-3, 5, 1) = 1*(1, 0, 0) + 7*(0, 1, 0) +(-1)*(4, 2, 3) + 4*(0, 0, 1) = 1*Col1 + 7*Col2 + (-1)*Col3 + 4*Col4. Where am I wrong?

antongorelov

Ideally for last question, we should consider the augmented matrix of A|V and see if it is consistent, to see if V belongs to Col A. ( According to Lay 5th Edition section 4.2 Question 7 b )

sayanbaidya

@ 5:05 the negative on the 3 was dropped from the first row? so the system should have been consistent?

ruinationmkII

15:52, I have a question here, what if v is a linear combination of the columns of A? is the statement would be true?

-Mohamed_bayan

Is Col A the same thing as span(A)? If so, why do we need all these deffinitions of the same things?

kassymchurchman

IN 15:44 could as you say vector v isnt a column of matrix A . could i also reson -> 'AND also because vector v isnt a linear combination of any column vectors of matrix A ? ' because if it was a scalar multiple of a vector in A then it could be in the span of the vectors of A .

lets say vector v was [ 0 0 2] then it would be in the column space of A as its a scalar multiple of the last vector a4 in a ( ie [0 0 1] . scalar = 2 ) ? all other scalars being 0 for the remaining column vectors in A.

mohnish

thank you very much for this useful course
I watched MIT open course as supplement I felt that in this unit you went a bit too quick it would be perfect if you provided another very fundamental example so that we really understand what is it about

ahmedmusa

In 6:00 you call the system inconsistent, but the system could still have infinitely many solutions, doesn't that make the system consistent at that moment?So you would have to do more row operations to decide if it's consistent or not?

inaksiv

Thank you very much for this. I just wanted to know why you preferred to do 2R2 - R3 instead of R3 - 2R2 at 9:12. I tend to think of it in its original linear form and it gets sometimes confusing. I love the course!

brilliantstarss

At 15:24, There is a linear combination of the columns of A that give v. One of column 1 plus 7 of column 2 minus one of column 3 plus 4 of column 4. Is that anything?

gyorgyo

IIITD'27 ATTENDANCE, LIKE MY COMMENT

darshgupta

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