Help with Linear Algebra Problem | Mathematics | Page 1

scenic pivot May 5, 2023, 2:02 AM

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Is my work for part a correct and how do I do part b?

keen tartanBOT May 5, 2023, 2:02 AM

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orchid chasm May 5, 2023, 2:05 AM

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To find if a vector is in the span of another set of vectors, you use an augmented matrix

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if the linear system is consistent after doing row reductions, then the vector is in the span of the vectors in your set

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Part a looks correct. If you put all of your vectors in a matrix then reduce it to RREF, the basis of the column space will form the basis of the span of your vectors

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As a linear algebra teacher, I would prefer you show your row operations.... hopefully your professor doesn't care

scenic pivot May 5, 2023, 2:19 AM

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orchid chasm To find if a vector is in the span of another set of vectors, you use an augment...

would this mean setting A = to
[a]
[b]
[c]
[d]

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is there not another way to determine if the vector belongs to the set with the information provided?

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orchid chasm May 5, 2023, 2:22 AM

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scenic pivot would this mean setting A = to [a] [b] [c] [d]

so part a actually helps us with part b

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by finding a basis, we can set up a simpler linear system, since we now have less vectors describing the same span

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so really we only need [a,b | v] where v is the vector we want to check is in the span

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actually, your work for a is incorrect

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what you found was a basis for the null space

scenic pivot May 5, 2023, 2:26 AM

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so how would I find a basis for the span

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would it be putting it equal to a, b, c, and d?

orchid chasm May 5, 2023, 2:26 AM

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your pivot columns in your RREF matrix correspond to your basis of the columns space in your original matrix

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so in the RREF form, your first and second column are pivot columns

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so your basis for the span is the first and second columns in your original matrix (before you reduced it)

scenic pivot May 5, 2023, 2:28 AM

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oh i see

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so it would be 1 2 2 0 and 2 5 2 1 correct?

orchid chasm May 5, 2023, 2:30 AM

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right!

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so then, instead of having a 5 column augmented matrix in part b, you only need to look at

scenic pivot May 5, 2023, 2:31 AM

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scenic pivot so it would be 1 2 2 0 and 2 5 2 1 correct?

so then what would be the justification behind this?

orchid chasm May 5, 2023, 2:32 AM

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$\begin{bmatrix} 1 & 2&1\ 2&5&1\2&2&0\0&1&1\\end{bmatrix}

slim ibexBOT May 5, 2023, 2:32 AM

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Cheunk

$\begin{bmatrix} 1 & 2&1\\ 2&5&1\\2&2&0\\0&1&1\\\end{bmatrix}
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.58 \end{document}
                   
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

orchid chasm May 5, 2023, 2:32 AM

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the above should be an augmented matrix

orchid chasm May 5, 2023, 2:33 AM

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scenic pivot so then what would be the justification behind this?

the pivot columns by definition are linearly independent, and row reductions preserves the independence relationship of the columns

scenic pivot May 5, 2023, 2:34 AM

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so would x3 and x4 be unconstrained variables here?

orchid chasm May 5, 2023, 2:35 AM

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okay so i don’t think you’re understanding what i mean

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look at your RREF matrix

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columns 3 and 4 can be written as linear combinations of columns 1 and 2

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so we can remove them, and the resultant set will still have the same span

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thus we have a linearly independent spanning set aka a basis

scenic pivot May 5, 2023, 2:37 AM

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ohhhhh i get it now ty

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so by linear combination that means that the columns can be multiplied by x3 and x4 right?

orchid chasm May 5, 2023, 2:39 AM

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a linear combination is a sum of any scalar multiple of the vectors in the set

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so in your RREF matrix column 3 is a linear combination of columns 1 and 2 because if we multiply column 1 by -1 and add it to column 2, we get column 3

scenic pivot May 5, 2023, 2:42 AM

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So do you have to do those calculations to determine whether they are linear combinations or can you figure it out looking at the equations i wrote down below the matrices?

orchid chasm May 5, 2023, 2:43 AM

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the equations are irrelevant

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you only need to look at the RREF

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i would suggest you look over these definitions in your textbook

scenic pivot May 5, 2023, 2:45 AM

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does that mean the only way to determine if there are linear combinations is to look at the RREF and check each vector?

orchid chasm May 5, 2023, 2:46 AM

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Kind of

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by reducing the matrix, we can find all of the pivot columns

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those are the columns that have a ‘1’ in one spot, and ‘0’ at every other spot in the column

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if we can identify the pivots, we know all of the other columns are in the span of the pivots

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so we can remove them without changing the span

scenic pivot May 5, 2023, 2:48 AM

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orchid chasm if we can identify the pivots, we know all of the other columns are in the span ...

does this hold true everytime?

orchid chasm May 5, 2023, 2:48 AM

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if you have a textbook, you should look at the definitions for pivot, linear independence, span, and linear combination

orchid chasm May 5, 2023, 2:49 AM

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scenic pivot does this hold true everytime?

yes. if you have a set of vectors, put them in a matrix, then reduce it to RREF, the columns in the ORIGINAL matrix, which are in the same spot as the pivot columns in the RREF will always be a linearly independently spanning set

scenic pivot May 5, 2023, 2:50 AM

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i think i get it now

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so basically just reduce to RREF and take the columns from the original matrix from the same positiion as the pivot columns

orchid chasm May 5, 2023, 2:51 AM

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right! then when you do part b, you use an augmented matrix with those columns

scenic pivot May 5, 2023, 2:52 AM

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the basis columns or the pivot columns?

orchid chasm May 5, 2023, 2:52 AM

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if the matrix is consistent, then the answer is yes, else that vector is not in the span

scenic pivot May 5, 2023, 2:52 AM

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or just A

orchid chasm May 5, 2023, 2:52 AM

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the basis columns

scenic pivot May 5, 2023, 2:52 AM

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oh

orchid chasm May 5, 2023, 2:52 AM

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slim ibex **Cheunk** ```latex $\begin{bmatrix} 1 & 2&1\\ 2&5&1\\2&2&0\\0&1&1\\\end{bmatrix...

see here

scenic pivot May 5, 2023, 2:53 AM

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whre does 1 1 0 1 come from?

orchid chasm May 5, 2023, 2:53 AM

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that is the one the problem is asking you to see is in the span!

scenic pivot May 5, 2023, 2:53 AM

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oh i see

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I got
1 0 0
0 1 0
0 0 1
0 0 0

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so does it belong to the span?

orchid chasm May 5, 2023, 2:54 AM

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so what does that mean as a linear system?

scenic pivot May 5, 2023, 2:54 AM

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its consistent

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oh wait

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is it inconsistent cus in the 3rd row 0 = 1

orchid chasm May 5, 2023, 2:55 AM

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exactly!

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so that vector cannot be in the span of the others

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you could also have done that by putting all of the vectors in a matrix and augmenting the final vector

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but part a allows us to simplify the row operations by getting rid of 2 of the columns

scenic pivot May 5, 2023, 2:57 AM

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so putting A equal to 1 1 0 1?

orchid chasm May 5, 2023, 2:58 AM

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yes

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you should check and see that that also is inconsistent

scenic pivot May 5, 2023, 2:58 AM

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yes that also gave 0=1

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but couldnt you also check by putting x1=1, x2=1, x3=0, x4=1 into the equations x1-x3+2x4=0 and x2+x3-x4=0?

orchid chasm May 5, 2023, 3:00 AM

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sure, but this does not use any linear algebra techniques

scenic pivot May 5, 2023, 3:01 AM

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so its a valid way to check but it just doesnt use linear algebra

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but it gives the same results correct?

orchid chasm May 5, 2023, 3:02 AM

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yes. your professor would probably take points off if you did it that way though 😂

scenic pivot May 5, 2023, 3:02 AM

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oh ok lol

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would multiplying the matrix A and the given vector and seeing if it equals zero also work?

orchid chasm May 5, 2023, 3:03 AM

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no. if you multiply Av you will just get another vector. it would not tell you if the vector was in the span of the columns of the matrix

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if you multiply Av and get zero it would tell you that v is in the null space of A though

scenic pivot May 5, 2023, 3:04 AM

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Oh i see

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Ok then what would be the justification behind augmenting matrix A and the given vector to see if it is consistent?

orchid chasm May 5, 2023, 3:05 AM

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If it is consistent after you do the row reductions

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then you can solve the linear system to find a solution

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this solution will be a set of scalars that you scale your vectors by to make the vector as a linear combination

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https://math.stackexchange.com/questions/1949704/how-to-determine-if-vector-b-is-in-the-span-of-matrix-a

Mathematics Stack Exchange

How to determine if vector b is in the span of matrix A?

Given a matrix A = \begin{bmatrix}
1 &2 &3 \
4 &5 &6 \
7 &8 &9
\end{bmatrix}

Determine if vector $b$ is in $span(A)$
where
$$
b = \begin{bmatrix}
1 \
2 \
4 ...

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this example explains it well

scenic pivot May 5, 2023, 3:09 AM

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So could you say that because the system is inconsistent after doing row operations, there are no scalars that can give you the vector provided

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i dont know how to word it

orchid chasm May 5, 2023, 3:11 AM

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yeah that’s close enough

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or you could say

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since the matrix is inconsistent, the is no c1,c2 such that

v = c1 y1 + c2 y2

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where v is (1 1 0 1)

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and y1 and y2 are your vectors from part a

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remember, the span is the set of ALL linear combinations

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so if there is no such linear combination, the vector cannot be in the span

scenic pivot May 5, 2023, 3:16 AM

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orchid chasm remember, the span is the set of ALL linear combinations

linear combinations of y1 and y2?

orchid chasm May 5, 2023, 3:17 AM

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so

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if you have a set of vectors

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say S = {y1, y2, y3}

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then the span of S is all of the linear combinations of y1, y2, y3

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for example
-2 y1 + 3y2 -πy3 is in the span

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so is
2y1 - y3

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it’s ANY sum of ANY scalar multiple of the vectors in the set

scenic pivot May 5, 2023, 3:19 AM

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so for the span of S it would by any scalar multiples of the 4 vectors in S?

orchid chasm May 5, 2023, 3:19 AM

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right! that’s the span of S

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so what you’ve done is ask

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can i find scalars multiples of these vectors which add to (1 1 0 1)

scenic pivot May 5, 2023, 3:20 AM

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and so if you set S equal to the vector you want if you dont get a solution then there are no scalar multiples that can get you to that vector

orchid chasm May 5, 2023, 3:20 AM

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exactly!!

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and what part A did was get rid of the extra vectors in the set that weren’t “making the span bigger”

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the set we have from the problem statement and the set we got as an answer for part a have the same span

scenic pivot May 5, 2023, 3:22 AM

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ok that makes much more sense now

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so the justification would be the same for the basis vectors and all 4 vectors right?

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since they are basically the same

orchid chasm May 5, 2023, 3:23 AM

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right. the benefit of removing the vectors we did in part a though is that we have a lot less row operations to do in part b

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there are 2 less columns to worry about

scenic pivot May 5, 2023, 3:23 AM

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so its basically just to make it easier

orchid chasm May 5, 2023, 3:24 AM

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yes

scenic pivot May 5, 2023, 3:24 AM

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So for this would I just do the same thing and put -1 0 1 1 equal to A?

orchid chasm May 5, 2023, 3:25 AM

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hm

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so you could do that

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but what do you know about a vector in the null space?

scenic pivot May 5, 2023, 3:26 AM

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its equal to zero?

orchid chasm May 5, 2023, 3:27 AM

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well

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Av = 0

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so if you multiply v by A and get 0, then v is in the null space

scenic pivot May 5, 2023, 3:27 AM

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oh right

orchid chasm May 5, 2023, 3:27 AM

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so that way you don’t need row reductions, just matrix vector multiply

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also, to help you understand span

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i want you to pick any x3 and x4 from this part and calculate what vector you get from that

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then multiply that vector by A on the left

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if you did that problem right, you will ALWAYS get 0

scenic pivot May 5, 2023, 3:31 AM

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Yep I got zero

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so just to be sure my answer for c was correct?

orchid chasm May 5, 2023, 3:31 AM

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looks good to me!

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the benefit of a basis is

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any vector in the span of the basis vectors can be written as a unique linear combination of the basis vectors

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if you have a set that spans but isn’t linearly independent, you can represent vectors in a lot of different ways

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but there’s only one linear combination for each vector if you have a basis

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so your intuition was correct

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to do part d, you could either use an augmented matrix with your basis vectors as columns, or multiply the vector by the matrix A and see if you get 0

scenic pivot May 5, 2023, 3:34 AM

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just to make sure, part c is asking for a spanning set for N(A)

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spanning set and basis are different right?

orchid chasm May 5, 2023, 3:35 AM

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ah good point! a basis is a spanning set which is linearly independent

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so a basis is a spanning set

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but not necessarily the other way around

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remember a basis of a space H is a set that:
a. is linearly independent
b. spans H

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if you get rid of a. you only have a spanning set

scenic pivot May 5, 2023, 3:37 AM

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does that mean a spanning set is only linearly dependent?

orchid chasm May 5, 2023, 3:39 AM

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no it only spans

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If it spans the space, it’s a spanning set

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if it is linearly independent, it is a linearly independent set

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if it is both, it’s a basis

scenic pivot May 5, 2023, 3:41 AM

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ah i see

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is it possible to have multiple spanning sets or is there only one

orchid chasm May 5, 2023, 3:43 AM

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no there can be many!

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you can have an infinite number of bases of ℝ² for instance

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{(1 0),(0 1)} is the “standard basis”

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{(-1 2) , (-4 -7) } is also a basis

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(and therefore also both spanning sets)

scenic pivot May 5, 2023, 3:45 AM

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so would {(-1, 2), (-2, 4)} be a spanning set but not a basis because its linearly dependent

orchid chasm May 5, 2023, 3:46 AM

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hmm well not quite

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it would span something, but not ℝ²

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{(-1 2), (-2 4), (-1 -1)} would be a spanning set of ℝ² but not linearly independent

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notice that you can only make the span bigger or keep it the same by adding vectors

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so a better way to find a spanning set that isn’t linearly independent would be to add a vector that is already in the span of the stuff we have in the basis (like what i did)

scenic pivot May 5, 2023, 3:49 AM

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ah ok

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wait sho how did you get {(-1 2) , (-4 -7) } is also a basis

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did you just choose two random linearly independent vectors?

orchid chasm May 5, 2023, 3:50 AM

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yes! we have a shortcut for sets of 2 vectors

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if one is not a scalar multiple of the other, the set is linearly independent

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if i multiply the first vector by 4 to make them agree in the first coordinate, that forces my second coordinate to be 8

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8 is not equal to -7 so they are linearly independent

scenic pivot May 5, 2023, 3:51 AM

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so what made {(-1 2) (-2, 4)} not a spanning set?

orchid chasm May 5, 2023, 3:52 AM

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well (-2,4) is in the span of (-1 2)

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so without decreasing the span, we can consider the set containing only (-1 2)

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now the vector (1 1) is in ℝ²

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but it is not in the span of (-1 2)

scenic pivot May 5, 2023, 3:53 AM

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what about (-1 2) (-2 4) and (-3 6)?

orchid chasm May 5, 2023, 3:54 AM

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ah! these are good questions

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both (-2 4) and (-3 6) are in the span of (-1 2)

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so we can throw then both out without decreasing the span! still not a spanning set of ℝ²

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so we have a nice theorem here

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if we have a linearly independent set of n many vectors

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it will always be a basis of ℝ^n

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so if we have 3 linearly independent vectors from ℝ³

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they will form a basis of ℝ³

scenic pivot May 5, 2023, 3:56 AM

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so then what defines a spanning set of ℝ²?

orchid chasm May 5, 2023, 4:01 AM

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A set which can make any vector in ℝ²

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as a linear combination

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think of the set of all linear combinations of (1 0) and (0 1)

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they look like

a(1 0) + b(0 1)

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if we put those together

a(1 0) + b(0 1) = (a b)

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well since i can choose any a and any b, i can get anything in ℝ²!

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now lets add a new vector (1 1)

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i can make that already out of my basis

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so now my set is linearly dependent

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but now the linear combinations look like

a(1 0) + b(0 1) + c (1 1) = (a + c b + c)

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but if i just make c = 0, i have the same thing as before!

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so this set is still a spanning set, but no linearly independent (so not a basis)

scenic pivot May 5, 2023, 4:07 AM

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ok i think i kinda get it but im still confused as to why {(-1, 2), (-2, 4)} isnt a spanning set

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is it because one of the vectors spans the other

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and by spanning the other vector does that mean they are scalar multiples?

orchid chasm May 5, 2023, 4:08 AM

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so it is a spanning set for the span of either of the individual vectors

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but it is not a spanning set of ℝ²

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because there are vectors that we can’t make out of those 2 vectors

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everything in the span of those vectors looks like (-a 2a)

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and there are definitely things that don’t look like that

scenic pivot May 5, 2023, 4:10 AM

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what would be an example of a vector that we cant make?

orchid chasm May 5, 2023, 4:10 AM

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(1 1)

scenic pivot May 5, 2023, 4:10 AM

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oh i see now

orchid chasm May 5, 2023, 4:10 AM

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you can check using an augmented matrix. it will be inconsistent

scenic pivot May 5, 2023, 4:11 AM

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so then how do you know if two vectors will be a spanning set of ℝ²

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if they arent scalar multiples are they always a spanning set of ℝ²?

orchid chasm May 5, 2023, 4:12 AM

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yeah that theorem i told you earlier applies

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any 2 linearly independent vectors will be a basis

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and in ℝ³ any linearly independent set of 3 vectors will form a basis, etc etc

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the fastest way to check if a set of vectors is a basis is to first count that you have enough

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then put them all in a matrix

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if the matrix reduces to the identity, they form a basis

scenic pivot May 5, 2023, 4:14 AM

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ah i see now

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and just to be sure you can have more than 2 vectors that are a spanning set of ℝ² correct?

orchid chasm May 5, 2023, 4:15 AM

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yes!

orchid chasm May 5, 2023, 4:15 AM

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orchid chasm but now the linear combinations look like a(1 0) + b(0 1) + c (1 1) = (a + c b...

this example is a spanning set that

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has more than 2

scenic pivot May 5, 2023, 4:16 AM

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ok perfect

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also back to part d, you said that i can check if the vector belongs to the null space by augmenting it with the basis vectors

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i did that but got an inconsistent system even though i multiplied A and x and got zero

orchid chasm May 5, 2023, 4:17 AM

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scenic pivot also back to part d, you said that i can check if the vector belongs to the null...

did you do this with the basis vectors of the null space?

scenic pivot May 5, 2023, 4:17 AM

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wouldnt the basis vectors be the same for part a and c?

orchid chasm May 5, 2023, 4:18 AM

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no! the basis from c gives you the basis for all of the vectors that get mapped to 0 by A

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the basis from a gives you a basis for all of the vectors that are linear combinations of the columns of A

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so basically, the basis from a gives you a basis to find all b such that Ax = b

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the basis from c gives you a basis to find all x such that Ax = 0

scenic pivot May 5, 2023, 4:20 AM

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so in this case is the spanning set the same as the basis for part c?

orchid chasm May 5, 2023, 4:21 AM

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yes

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remember a basis is a spanning set

scenic pivot May 5, 2023, 4:23 AM

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i was just confused because using the equations for the null space gives you the basis vectors but for part a the equations gave the x3 and x4 which were removed to get the basis vectors

orchid chasm May 5, 2023, 4:24 AM

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yes they are different processes

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it comes from the fact that the null space is a subspace of the domain but the column space is a subspace of the codomain

scenic pivot May 5, 2023, 4:25 AM

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Ok and very last question, can I check if the vector belongs to the null space by checking if it matches with the equations I derived like I could do in part a and b?

orchid chasm May 5, 2023, 4:26 AM

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so if you wanted to check with an augmented matrix

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you can use

orchid chasm May 5, 2023, 4:27 AM

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slim ibex **Cheunk** ```latex $\begin{bmatrix} 1 & 2&1\\ 2&5&1\\2&2&0\\0&1&1\\\end{bmatrix...

this setup

scenic pivot May 5, 2023, 4:27 AM

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what about for the equations x1-x3+2x4=0 and x2+x3-x4=0?

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i just want to know if it works or not

orchid chasm May 5, 2023, 4:28 AM

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you should avoid doing that in general

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but you could do it with your basis vectors from the null space instead of your vectors from part a

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so as you’ve written it that would not check if it’s in the null space

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instead it would be

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using the numbers from the vectors in the basis you found for part c

scenic pivot May 5, 2023, 4:30 AM

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I get
1 0 1
0 1 1
0 0 0
0 0 0

orchid chasm May 5, 2023, 4:31 AM

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right! so that means it is in the null space

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because that is a consistent system

scenic pivot May 5, 2023, 4:32 AM

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so as long as its consistent even if its something like
1239 9412 2
0 241 429849
0 0 0
0 0 0
it would belong to the null space?

orchid chasm May 5, 2023, 4:32 AM

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yup

scenic pivot May 5, 2023, 4:32 AM

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ok awesome

orchid chasm May 5, 2023, 4:32 AM

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but you should always reduce to RREF

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i think you can mark as solved btw

scenic pivot May 5, 2023, 4:37 AM

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Alright then tysm for all your help, I learned a lot 😄

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how do i mark as solved?

orchid chasm May 5, 2023, 4:37 AM

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you're welcome! you were asking good questions so it was nice to help you

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i have no idea

scenic pivot May 5, 2023, 4:37 AM

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o lol

orchid chasm May 5, 2023, 4:38 AM

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look under the #❓how-to-get-help channel maybe?

scenic pivot May 5, 2023, 4:38 AM

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i think i figured it out

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before i close it is it cool if i add u if i have anymore questions? its cool if u dont want to tho

orchid chasm May 5, 2023, 4:39 AM

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cool

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sure go ahead. i can't guarantee i can help you quickly lol i was at the gym this whole time and answering between sets lol

scenic pivot May 5, 2023, 4:40 AM

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o dang lol, mb if i bothered u

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thanks for the help anyways have a good rest of your night

orchid chasm May 5, 2023, 4:41 AM

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wasn't a bother lol it helps pass the time and helps me to understand it better too

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you too! good luck if you have finals or anything

scenic pivot May 5, 2023, 4:42 AM

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thank you! 🙂

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.close

#Help with Linear Algebra Problem