Basis and Dimension
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- Опубліковано 23 кві 2019
- Now we know about vector spaces, so it's time to learn how to form something called a basis for that vector space. This is a set of linearly independent vectors that can be used as building blocks to make any other vector in the space. Let's take a closer look at this, as well as the dimension of a vector space, and what that means.
Script by Howard Whittle
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0:27 A Basis
1:31 Check Linear Combination
2:26 Span
2:52 Satisfying Linear Independence
3:21 A more complicated example (R 2x2)
3:54 Span check
4:23 Distribute the Scalars. Add up the new matrix.
4:45 Make sure a solution exists
5:43 Check Linear Independence
6:57 Row Echelon Form:
- No Free Variables
- All Scalars must be = 0
7:34 Both conditions verified ✔️
Basis With N Elements= Dimension N
9:05 Check Comprehension
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Basis vectors/Matrices seemed so far out of reach even after trying to understand them for a couple of weeks but after this video, which make them seem easy, I think I finally understand them. Thanks Dave! :))
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The reduced row echelon form isn't finished yet at 7:22, you can still do R3+R4, R2-R4 and after that R1-R3 which doesn't require you to solve the remaining set of equations.
You don't need to since you can see c4 is equal to zero which would then make the rest zero.
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When n (number of columns) is not equal to m (number of rows) then the set is always not a basis (if the question ask if a specific set of vectors is a basis or not)
but when n = m, then you have two possibilities depending on the det(A), in other words:
det(A) is equal to 0 ==> the set is not a basis
det(A) isn't equal to 0 ==> the set is a basis
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Can we expect a subspace who span vector space but vectors (elements ) in that subspace are linearly dependent?
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5:05 since you’re taking the determinate of the square matrix and it’s a none zero number, isn’t also linearly independent too?
that was what i was thinking too :)
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I already gave a like as soon as I saw the intro
I think the first 3 vectors is because for a R2, one need only 2 vectors for creating a base for R2. Plus, 3,2 could be 2 times the 1,0. Right?
Yeah, the could be built using 2 x + 1 x
Do free variables effect whether or not the basi can be linearly independent?
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Thx
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but the main question is-why the canonical basis is indexed by natural numbers?And can we describe canonaical basis in terms of matrices?
superb ....
Sir can we find the null space of set of vectors from M2x2 like we do for vectors in R^n
For the first question in the comprehension part, 0 is the determinant, so that should be linearly dependent right?
Independent
|A|=0 =linearly independent
At 5:26 how is the determinant 1? Cause multiplying the 4 brackets above from the formula (ad-bc) gets: 0, 0, 1, then the last one is 0-1 which is -1
doesnt matter regardless, if the det isnt equal to 0 we can proceed
Linear independent vectors means we cant take linear combination of them...on the other hand span is all the linear combination of those vectors. Basis is the vectors will be linearly independent + they will span. I am confused...how these 2 can be true at the same time?
What if such cases when determinant is zero yet it has infinitely many solutions?
2 v + 3 w.. In this v and w are vectors and these are basis as well?
If they are linearly independent, it means there are no linear combinations among the vectors. So, how can a basis have two conditions where (1) they are linearly independent and (2) they span the vector space V (by a linear combination of the vectors), don't the conditions contradict each other? Please clarify, and let me know if I'm missing something here.
Dude ! Linearly independent does not mean that they will have no linear combination its actually satisfies that they can have linear combination coz we check that the vectors are not dependant so that we can have all the possible linear combination!
Linear combination and linear dependent is two different things. Whether given two vector elements within the vector space are linearly dependent or not has to do nothing with the linear combination. Any set of vector elements can be written as linear combination (with respect to their coefficients)
what if instead of all leading ones we had a leading 2 in some position. thats okay right ?. since its not Reduced row echelon form
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How to apply curl to higher dimensional vector field
But Im not getting the determinant value as 1 in example 2 while checking for spaning 5.23 ...
Can somebody please help.... please...
How can you have more than a dimension of 3 in a 3D space? Wouldnt any more vectors are just repetitive and therefore be linearly dependent?
Mathematics isn't limited to the three spatial dimensions we are familiar with, it can utilize many more. We just are incapable of visualizing it.
1. The dimension of the matrix is
2. In the matrix , the entry in the third row and second column is _____.
3. For what values of and , the two matrices are equal?______
4. Write a diagonal matrix of order two, such that the entries on the diagonal zero._______
5. Given the following matrices and , then compute
a) b) c) d) e) f)
6. Find the products of a row matrix and the column matrix ; that is and YX.
7. A manufacturer produces three products: A, B, and C, which he can sell in two markets. Annual sales volume is indicted as follows.
Product A B C
Market I 10,000 units 2,000 units 80,000 units
Market II 6,000 units 20,000 units 8,000 units
a) If unit sales of A, B, and C are 2.50 Birr, 1.25 Birr, and 1.50 Birr, respectively, find the total revenue as a product of matrices in each market.
b) If unit sales of A, B, and C are 1.80 Birr, 1.20 Birr, and 0.80 Birr, respectively, find the gross profit as a product of matrices in each market
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For R 2X2 matix, can't we just say that the matices are linearly independent as their determinant is not equal to zero.
We created the matrix 4X4 which is a square matrix and its determinant is 1, so it satisfies that they are linearly independent.!!!
I thought the same thing, but idk
Doesn't the det of matrix being 1 (not 0) means its elements are linearly independent (so we don't need to form row echelon form)
I agree with this.
also, can you confirm that there is no point in ever checking both conditions for basis, i.e., condition 1: spanning, condition 2: linearly independent?
if you know it spans and number of vectors > number of dimensions, it can't be a basis.
if you know it spans and number of vectors = number of dimensions, it MUST be a basis.
if you know number of vectors < number of dimensions, it can't span?
you might as well just manually look, saves you work
Can somebody please help me? In the previous video where we had to check whether a matrix is linear independent or not by row operation, we didn't get all 1's in the main diagonal. But in this video why do I have to get all one's in the main diagonal?
works either way, i think
because all we aim is to reduce the variables
Isnt the set of vectors rank 4 ? How can a rank 4 span R2
6:36
how can we multiply R3 by -1
wont it change the equation??
Found this comment 6 months later and now i know the answer
if we multiply both sides of an equation it will still remain the same equation and is valid
here the other side of the equation(right of =) is 0 and 0 multiplied by anything gives 0 so we dont include it
in the comprehension, how the first one is not linearly independent ?
Can anyone provide me solution of last 2 questions?
4:54
thank you, can you give me a vector space with infinite dimensions?
Professor, vector space with the only vector "zero vector" has dimension 1. 8:23
Nope, it has dim 0 or none
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why is the first one not a basis in the comprehension
😮
Your explanation is good but you didnt explain the type of questions in the check comprehension
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No. The pacing is deliberate for those who need time to process what's on the screen. Teaching math is not a conversation.
The pacing is all good to me
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