so
#linear-algebra
slow scroll
np
merry shuttle
is there a way to figure out how many basic variables are in the general solution of a system with x linear equations in y variables that has z free variables?
merry shuttle
so the number of linear equations doesnt affect the number of free variables?
yeah that makes sense tbh
slow scroll
sure it does, and so does the linear dependence/independence of the equations. Have you ever heard of rank-nullity theorem by any chance?
slow scroll
do you know anything about fundamental subspaces?
slow scroll
ah ok then. Just a remark: you can't have more "basic variables" than there are equations. That may have been the relationship you were looking for
merry shuttle
oh okay
so in the general solution of a system with x linear equations in y variables that has z free variables, given that x > a (basic variables), a = y-z
slow scroll
right, and you don't need to be given x>a. That will always be the case :p
merry shuttle
okay
so in the case where x <= a, what would happen to the number of basic variables?
slow scroll
it would have to equal the number of equations then. It is NEVER the case that the number of basic variables exceed the number of equations
This is simply because if you are given equations that all have variables x y and z, then of course there is nowhere for a w to come from
Similarly the number of basic variables can’t exceed the total number of variables.
slow scroll
@topaz plover did you get it?
topaz plover
nope im not sure which one it is
slow scroll
you have a free variable (the last column doesn't have a pivot). do you know what that means?
topaz plover
infinite?
slow scroll
right. the solutions (x,y,z) are x=5-2z, y = -1+3z where the z coord can be anything
slow scroll
npnp
green garden
@north sierra math is a cycle of "i don't get it those are just words/symbols and they make no sense together" followed by "holy crap no way that's amazing" moments
strap in
(that said, a good professor can make a huge difference)
gray dust
show us your work @north sierra
north sierra
slow scroll
looks good
gray dust
get it into rref form
north sierra
whats that
gray dust
turn the main diagonal into 1's
slow scroll
or you could convert back to a normal system at this point and sub things in
thats what i normally do
gray dust
@slow scroll i think the hw wants him to rref all the way
@north sierra eliminate the 5 from R1
north sierra
the answer in the book is (-8,3)
gray dust
didn't ask for the answer, just asked you to eliminate the 5 in R1
slow scroll
lol
gray dust
you can add rows to each other
north sierra
what does that mean?
slow scroll
so its like the elimination you were doing, but in the opposite direction now
gray dust
for example you can take all the numbers of R2 and add them to their corresponding numbers in R1
green garden
@north sierra it might help to list out elementary operations
that's your toolbox
easier to work if you have the tools laid out in front of you
north sierra
the way they explain the three elementary operations in my boom is very confusing
book
gray dust
e
- you can multiply a row by a non zero constant
- you can multiply a row by a non zero constant and add the result to another row
- you can swap rows
gray dust
you goofed up
gray dust
by eliminating x_1 from the system
north sierra
im not allowed to do that?
slow scroll
it looks like you wrote R1*5R2 
north sierra
yeah
gray dust
lmao
slow scroll
dat aint an elementart row operation
royal storm
You can only multiply rows by a scalar, add rows, and add scalar multiples of rows
gray dust
e
- you can multiply a row by a non zero constant
- you can multiply a row by a non zero constant and add the result to another row
- you can swap rows
green garden
1 5 | 7
0 1 | 3
go back to this. you want to eliminate the 5 in the top right, because then you have 1x + 0y = ? and 0x + 1y = ? and the problem is finished.
no, how do you do get rid of the 5?
north sierra
i did option 2 right?
royal storm
you did none of them
north sierra
-5
green garden
1 * _ = -5
north sierra
-5
green garden
do you see a way out yet?
north sierra
nope
green garden
how do you get rid of 5
you add -5 to it
how do you add something w/ an elementary operation? add a row to it
royal storm
The general idea is that you want each column to contain one thing that's not 0, except for the last one, because that's your answer
green garden
well you dont see -5 anywhere... where are you gonna get a -5?
you've done this before, i can see because you did 3 steps prior
north sierra
row1 * -1?
green garden
that's one way to do it
but
then you have the -5 where you want 0
and you can't add the row to itself, that's not a valid elementary operation
so you need to get the -5 from elsewhere
there is only one other place you could possibly get it from
green garden
well, it would delete the 1 in the top left, which you need to keep
also i dont think that even makes sense as an operation
north sierra
Row 1 * 1?
green garden
what happens if you do Row2 * -1 + Row1?
green garden
thta's not what -1 + 5 equals
north sierra
a little bit
green garden
you can freely manipulate that 5 by messing with the 1 below it
north sierra
how do i know when to stop?
green garden
well.... what if you just use something else instead of -1?
1 * -1 + 5 = 4
1 * _ + 5 = 0
what is _
north sierra
-5
green garden
try it
north sierra
ok
i meant to put -8 sorry
not positive 8
wow its the same answer as the books
so how did you manage to get it fast?
any tricks?
gray dust
rref calculator
north sierra
?
gray dust
rref = reduced row echelon form btw
gray dust
it's not a method
you used gaussian elimination, the method, to turn the matrix into reduced row echelon form, the result
north sierra
oh
is there a way to do this fast?
or is it guess and check
and also how do i know when i'm done?
i only knew i was done cause i checked the answer from before
gray dust
you know you're done when you get your matrix into RREF
(or if you show the matrix is inconsistent or has free variables, whatever)
gray dust
when the main diagonal of the matrix is all 1's
and every other element is a 0 (besides the right column)
tbh RREF can take on slightly different forms but that's usually the way you want to get it
gray dust
gray dust
a good way to know is if you have a row of 0s with a nonzero at the right side
gray dust
if you have a row 0, 0, 0, 4 that implies 0 = 4, which is bullshit, so no solution
gray dust
look at the RREF matrix on the right
that's what you want to get your augmented matrix to
gray dust
those aren't other methods
that's just describing what other matrices look like
a square matrix has the same number of rows as columns
gray dust
no
north sierra
oh
gray dust
and RREF is NOT a method of solving
north sierra
oh
gray dust
you're referring to GAUSSIAN ELIMINATION
gray dust
if you can get the matrix into rref, that will directly give you the solution to the system
gray dust
the method you just used to do your hw
gray dust
$\begin{bmatrix}1 & 0 & 0 & 0 & 4\ 0 & 1 & 0 & 0 & 2 \ 0 & 0 & 1 & 0 & 1 \ 0 & 0 & 0 & 1 & 7 \end{bmatrix}$
stoic pythonBOT
RokettoJanpu:
gray dust
this matrix is in rref
gray dust
when you have a diagonal of all 1's... and every other element is a 0 (besides the right column)
stoic pythonBOT
RokettoJanpu:
gray dust
plug your solution into the original system
north sierra
like i heard you can add the values back into the original equation?
then what answer should i look for
gray dust
listen
it's like you solve 2x = 6
you solve it, x = 3
how to check it?
plug the solution into the original equation
e
2*3 = 6
6 = 6
makes sense, your solution is good
north sierra
i see
okay thank you so much
i have to sleep now
but i will continue learning tomorrow morning
gray dust
no problem. do your best to review how to solve matrices with elimination
have a good night
undone garnet
Prob:\
$A = \begin{bmatrix}1999&b&1\a&b&0\b&a&1\end{bmatrix}, a, b \in \mathbb{R}$\
A has an eigenvalue $\lambda = 1$\
Find $f(a,b) = 0$\
Proof:\
Let $x = 1999$\
$p(\lambda) = \det(A-\lambda E) = -\lambda^3 + (x+1+b)\lambda^2 - (xb-ab+x)\lambda +xb-ab+a^2-b^2$\
Because $\det(A-E)=0$, so\
$p(\lambda) = (1-\lambda)(\lambda^2-(x+b)\lambda+xb-ab-b)$\
and we need\
$xb-ab-b=xb-ab+a^2-b^2$\
$\Leftrightarrow a^2-b^2+b=0$\
$\Leftrightarrow f(a, b) = a^2-\left(b-\frac{1}{2}\right)^2+\frac{1}{4}$
that's my proof
but I find it not right when I choose a = b = 0
stoic pythonBOT
Nguyễn Thành Trung:
dusky epoch
"Find f(a,b) = 0"
:?
what does this mean
does this mean find a function f(a,b) that is zero iff your matrix has 1 as an eigenvalue
noble swallow
So the problem was "find a necessary and sufficient condition for a and b such that 1 is an eigenvalue of A"?
noble swallow
Oh ok, your solution seems alright
undone garnet
but when I choose a = b= 0, A doesn't have 1 as an eigenvalue
that makes me so confused
noble swallow
Why doesn't it?
stoic pythonBOT
Nguyễn Thành Trung:
dusky epoch
what's x
dusky epoch
let's see
A-E = [1998, b, 1; a, b-1, 0; b, a, 0]
we need det of that to be 0
det(A-E) = a^2 - b(b-1)
noble swallow
Ker(A-E) if a=b=0 is not {0}, so 1 is an eigenvalue
noble swallow
You're also forgetting a -λx term
heavy glacier
why is the span of the empty set the set containing the zero vector?
undone garnet
let a = b = 0
$A = \begin{bmatrix}1999&0&1\0&0&0\0&0&1\end{bmatrix}$
stoic pythonBOT
Nguyễn Thành Trung:
noble swallow
Aut(1)=span{(-1,0,1998)}
noble swallow
Has only 1 among real eigenvalues or among all eigenvalues?
undone garnet
find a, b if it's possible
thank you @dusky epoch
I'll try
oh well
it's not possible
noble swallow
@heavy glacier according to the definition of span(C) as the smallest subspace containing C then span(emptyset) would be indeed {0}
@undone garnet are you sure? I got some values of them that seem to work
undone garnet
$(1-\lambda)^3 = 1-3\lambda + 3\lambda^2-\lambda^3$\
$p_A(\lambda)=xb-ab+a^2-b^2-(xb-ab+x)\lambda+(x+1+b)\lambda^2-\lambda^3$
stoic pythonBOT
Nguyễn Thành Trung:
undone garnet
so we have 3 equations
$\begin{cases}xb-ab+a^2-b^2=1\xb-ab+x=3\x+1+b=3\end{cases}$
stoic pythonBOT
Nguyễn Thành Trung:
undone garnet
$b=3-x-1=3-1999-1=-1997$
stoic pythonBOT
Nguyễn Thành Trung:
undone garnet
$xb-ab+x=3 \Leftrightarrow a=\frac{xb+x-3}{b}$
stoic pythonBOT
Nguyễn Thành Trung:
stoic pythonBOT
Nguyễn Thành Trung:
noble swallow
Ah you're right sorry, I made a mistake
undone garnet
$A=\begin{bmatrix}2&1\2&3\end{bmatrix},B=\begin{bmatrix}-2&1\-1&4\end{bmatrix}$\
$X \in M_2$ satisfies $AX-mX=B$\
Which condition of $m$ to let $X$ has 1 as an eigenvalue
stoic pythonBOT
Nguyễn Thành Trung:
undone garnet
😐 well, I have a tedious equations and don't know how to solve
dusky epoch
(A - mI)X = B
noble swallow
Mm, B is also invertible
B^(-1)(A-mI)X=I
We could impose B^(-1)(A-mI) to be invertible and its inverse to have 1 as an eigenvalue
pallid swallow
Hmm, but if Xv=v, and note we have B^(-1)(A-mI)Xv=v, so (B^(-1)(A-mI)-I)Xv=0, so (B^(-1)(A-mI)-I)X is singular, but B^(-1)(A-mI)X=I, so X is invertible, so (B^(-1)(A-mI)-I) is singular
and then you can probably attack that @undone garnet
for 2x2 matrices, singular means it is $uv^T$ for column vectors $u, v$.
stoic pythonBOT
Element118:
undone garnet
well
i solved it
here my proof
if $X$ has 1 as an eigenvalue so exists vector $v \neq 0$ such that $Xv=v$\
$AX-mX=B$ \Leftrightarrow (A-mE)X=B \Leftrightarrow B^{-1}(A-mE)X=E$\
$\Rightarrow B^{-1}(A-mE)Xv=v = Xv \Leftrightarrow \left(B^{-1}(A-mE)-E\right)Xv=0 \Leftrightarrow \left(B^{-1}(A-mE)-E\right)v=0 \Rightarrow \det\left(\left(B^{-1}(A-mE)-E\right)\right)=0$\
after solving equation, I got $m=1, 2$
pallid swallow
... you use E for identity matrix?
hmm, next I think you might need to check that m=1, 2 work
can't accidentally introduce extra solutions
noble swallow
Oh very nice, essentially using that if M is invertible, λ is an eigenvalue of M iff 1/λ is an eigenvalue of M^(-1)
Then yes, to conclude it will suffice to show whether B^(-1)(A-mI) is invertible for m=1,2
thin bloom
Will the linear combination of any vectors in S still be in the span of S
I'm having a brain fart rn
dusky epoch
idk will it
thin bloom
oh got it
pallid swallow
hehe, that's direct from definition of span
thin bloom
"Now let W denote any subspace of V that contains S" Does the word contain mean span(S) is a subset of span(W) ?
pallid swallow
Well, S here is a set of vectors. Here it just says that the subspace W contains the vectors in S, well, which IMPLIES that the subspace W contains span(S)
thin bloom
or $S \subseteq W$
stoic pythonBOT
Nicoskiy:
stoic pythonBOT
Nicoskiy:
pallid swallow
W is the subspace, so if the set of vectors in W can be called W, then W = span(W)
wintry steppe
@noble swallow , sorry to bother you. When you get the chance could you chat with me about the problem I showed you yesterday
wintry steppe
wintry steppe
Yeah
pallid swallow
Okay, so how would you define a midpoint of AC?
pallid swallow
What is the position vector for that point?
Position vectors make sense to be used here
pallid swallow
Well, AC/2 works if A is the origin
So, now you need to find the midpoint of BD and show it's the same point
wintry steppe
The professor said to use a separate point as the origin
If my origin is the midpoint, then how do I go about expressing that?
pallid swallow
15 minutes
@final wing
What have you tried?
(And this is probably discrete math)
maybe abstract algebra if you try hard enough
quaint heart
the second one can be vaguely considered abstract algebra
also Map(X, X) is the worst notation I've ever seen
half ice
I like it. There's no mistaking what it is
dusky epoch
X^X tho
final wing
@pallid swallow tried considering x in X how many choices for f(x) there would be
Then doing that for some y in X
half ice
Every element in the domain (X) has between 0 or |X| arrows on it. How many arrows? Does that help?
half ice
Oh true, we also need the arrangements of those arrows
dusky epoch
ok you won't be missed
wintry steppe
Shut up
half ice
dusky epoch
you said you were leaving
wintry steppe
💩
dusky epoch
why don't you do that instead of throwing around insults only to delete them seconds later thinking nobody saw them
wintry steppe
Mods ban him
dusky epoch
oh, i see you're misgendering me too? how nice on your part.
wintry steppe
half ice
What's the undercomplication? I'm not sure I know the best way to calculate this
dusky epoch
for each element of X, you have n options for where to map it
half ice
Oh duh, yeah that's easier
dusky epoch
@wintry steppe are you gonna apologize for calling me a "he" when i'm so clearly not, or are you gonna keep being an asshole
wintry steppe
Sorry
half ice
So every element has 2^n maps out of it
dusky epoch
no
wintry steppe
Sorry
wintry steppe
half ice
Let's say X = {a, b, c}
Then a can map to:
a, b, c,
a and b, a and c, b and c,
all three,
None of them
wintry steppe
I’m sorry
winter reef
and n! bijctions
dusky epoch
@half ice but no
wintry steppe
Sorry
dusky epoch
we aren't talking about multivalued maps
@wintry steppe no need to apologize more than once
wintry steppe
half ice
I thought we were? It just says "map"
If we allow any general relation then there's 2^n ways to map each element, and n elements, meaning there's n(2^n) maps
But I'm open to other mapping types
dusky epoch
@half ice map = function
half ice
Oh, well in that case it's a bit easier
quaint heart
even number of relations is very easy
2^(2n)
you were just over complicating things
worn rock
how do they go from step 1 to step 2 in only 1 step
like i get that you could expand it all
stone drum
It's not obvious to me either.
north sierra
i’m having trouble solving this linear equation using Gauss-Jordan elimination
(matrix row reduction)
slow scroll
Isn’t gauss Jordan for computing matrix inverses tho?
north sierra
i dont knwo
green garden
@slow scroll you can use it to compute an inverse, but you can use it to just get a matrix into RREF as well
@north sierra same deal as last night
use the same operations you already have
green garden
you understand what RREF is now at least?
north sierra
Yeah
green garden
and you understand why we want to put a matrix into RREF?
north sierra
no
green garden
that might help w/ motivation at least..
north sierra
But like i dont know why this is not intuitive
green garden
remember how a matrix is a representation of a system of equations?
green garden
what happens if all the coefficients in an equation are zero, except for 1
0x + 0y + 3z = 9
north sierra
z=3?
green garden
right
better still,
0x + 0y + 1z = 3
right?
so if we can get every equation into that form, at the same time, the solution is literally sitting there on the page
north sierra
oh
green garden
that's literally all RREF is
north sierra
i see
lol
am i going in the right direction?
ive been stuck on this for like 40 minutes
green garden
so there is a pretty standard sequence you can follow that works reliably
wintry steppe
wow there you need to solve with the QR decomposition or the LU decomposition or the LUP decomposition or the SVD or the polar decomposition
green garden
instead of just eyeballing and guesing
north sierra
yeah right now im just eyeballing and guessing
i would like to know a good way to do this
green garden
let me see if i remember it, hah
wintry steppe
Hello
wintry steppe
I'm trying to solve this matrix
green garden
@north sierra here's a guide https://people.richland.edu/james/lecture/m116/matrices/pivot.html
green garden
that too
wintry steppe
or you can just back-substitute, which is the same thing
north sierra
lol i’m on that website salt
wintry steppe
north sierra
is there a video on youtube?
green garden
basically "put the biggest stuff at the top" and "work up from the bottom left"
wintry steppe
its saying add -4 times row 3 to row 1
$\left[\begin{matrix}- \frac{37}{254} & \frac{19}{254} & - \frac{5}{254}\\frac{101}{508} & - \frac{45}{508} & - \frac{55}{508}\\frac{33}{508} & - \frac{65}{508} & - \frac{23}{508}\end{matrix}\right]$
here's the inverse of your matrix
it looks uugggly
Uh what?
Perform the following 3 elementary row operations, one after the other, and give the resulting matrix at each step:
oh
green garden
hah i was wondering
wintry steppe
yikes
green garden
@wintry steppe so what are you stuck on exactly?
wintry steppe
well
green garden
i dont understand what those calculations are under the matrix
wintry steppe
you see how x1 is 10, x2 is -19, and x3 is 19
green garden
you're multiplying by 4? ok
wintry steppe
Im unsure what values that r1 has
green garden
what does the question state?
wintry steppe
Add -4 times r1 to r3
green garden
it sounds like they want you to modify r3 and leave r1 as-is
wintry steppe
Okay
Got it correct, thank you :)
How would I do this next problem
0 5 0 5 -25
1 -1 5 0 6
I have to solve it in RREF
@green garden
gray dust
did you try any row ops yet?
gray dust
you can multiply a row by a nonzero constant and add the result to another row
wintry steppe
do an interchange first
now it's in rref 🙂
oh wait no it's not
your work is correct
it's in rref
I think it’s correct
then what's the question?
I forgt to solve the lattice
Had to look at my 📝
matrix*
thin bloom
Is this even possible? I can't think of any examples
thin bloom
W1 = span([1, 0, 0]) and W2 = span([0, 0, 1]) works with the first condition
thin bloom
@pallid swallow Seems impossible
I got to a contradiction where m+n < dim(W1 + W2) AND m + n > dim(W1 + W2)
@pallid swallow ?
<@&286206848099549185>
Can any helper look the question and see if my answer is valid?
raw drum
@thin bloom, it's not impossible
thin bloom
wat
raw drum
Try L[(1,1,0)] and L[(0,1,1)]
You just need that their intersection isn't empty
And that W_2 is not contained in W_1
raw drum
No, it would be 3
wait
Wait, facepalm
Sorry
I meant L[(1,0,0);(0,1,0)] and L[(0,1,0);(0,0,1)]
thin bloom
o my ga
wintry steppe
what does this mean
like is this just every matrix with 1s and 0s as entries
\
wintry steppe
like is E just every matrix m x n with 0s and 1s
sonic osprey
They tell you what E^{ij} means
wintry steppe
i dont understand it though
sonic osprey
what don't you understand
green garden
1 0 0
0 0 0
0 0 0
^ that's E11
0 0 0
0 0 1
0 0 0
^ that's E23
wintry steppe
yeah I got it thank you so much
would
$$S=\left{\begin{pmatrix}1\ 0\end{pmatrix},:\begin{pmatrix}0\ 1\end{pmatrix}\right}$$
be such a set
stoic pythonBOT
apples:
slow scroll
well, those aren't diagonal matrices, but they are isomorphic to that representation so basically
sonic osprey
No this isn't right
wintry steppe
ohhh I see
sonic osprey
The elements of S are not in your subspace
wintry steppe
$$S=\left{\begin{pmatrix}1&0\ 0&0\end{pmatrix},:\begin{pmatrix}0&0\ 0&1\end{pmatrix}\right}$$
stoic pythonBOT
apples:
wintry steppe
how about now?
slow scroll
yes
wintry steppe
awesome thanks!!
north sierra
Pivots have to be exactly diagonal in the matrix?
raw drum
north sierra
111|3
010|0
002|2
000|-10
can i divide row three by just 1/2 to make the 2 into a 1?
earnest zinc
@north sierra yes
but you know the system is inconsistent regardless (last row, you cant have 0=-10) so its not necessary go on with row operations
north sierra
ok but i think i made a mistake
so something here is wrong but i can’t figure out what i did wrong
could someone point out my mistake please?
earnest zinc
@north sierra yeah one sec
uhm
i think you messed up step 2
try row 2 = row 2 - row 1
earnest zinc
step 1*
to put a matrix in rref
you try to get one pivot positions
sorry im in a movie kek
north sierra
LOL
earnest zinc
#❓how-to-get-help is always open
north sierra
but why wouldn't what i did work?
north sierra
lol
earnest zinc
yeah idk how you got that
north sierra
ur at the movies?
earnest zinc
ye ye
north sierra
enjoy the movie m8 lol
gray dust
you watching matrix reloaded?
north sierra
LOL
earnest zinc
😄
earnest zinc
@north sierra
north sierra
yeah\
earnest zinc
ok so
our goal is to try to get a pivot position in each row
so first, take the first column
the first row should be a 1, and all of the rows should be 0 in the first column below it
to do that:
row 2 = row 2- row 1
row 3 = row 3 - 2(row 1)
row 4 = row 4 - 5(row1)
north sierra
ok
earnest zinc
after doing that, the only nonzero in the first column is in the first row
north sierra
can you explain why my first step was wrong?
because by step 4 i have what you want
north sierra
how
earnest zinc
i just looked at it very shallowly, didnt realize it was right
usually you would take the top row and use multiples of that to get 0 in each row for the first column
ok ok ok ok
north sierra
ok
north sierra
LOL
earnest zinc
anyways, you have to make a deduction from the final matrix you wrote
north sierra
whats a deduction?
earnest zinc
basically, does it have a unique solution, infinite solutions, or no solution
do you know which one of the three ^ it is?
earnest zinc
yes
north sierra
cause 0 0 0 != whatever number
earnest zinc
BAM
north sierra
but the thing is
earnest zinc
o
north sierra
i think i did this wrong
earnest zinc
o
north sierra
yeah
earnest zinc
why
north sierra
the answer that my teacher posted is diffferent
earnest zinc
wait a min
what did you do in step 3 on your paper
i didnt look at the algebra lol
north sierra
the algebra is right
earnest zinc
but what did you do in step 3
to get from the 2nd to the third matrix on your paper
earnest zinc
basically how did row 3 go from 2 1 0 2 --> 0 1 2 4
north sierra
i took (2 * Row 1) - row 3
north sierra
lol
earnest zinc
when doing row operations, you have to add or subtract from row thats changing
so you cant do row x - row 3
you have to do row 3 - row x
north sierra
oh
earnest zinc
so you would do row 3 - 2(row 1)
north sierra
always the row you want changed - or + x row?
earnest zinc
yes
earnest zinc
😄
if there are any other mistakes just ping me ill be on discord
im not watching this movie i have noise cancelling headphones running
too scary
earnest zinc
my friends went
north sierra
oh
earnest zinc
bad movie
north sierra
lol
earnest zinc
: D
north sierra
you there @earnest zinc
north sierra
hi
earnest zinc
hi
north sierra
i need help
earnest zinc
im help
north sierra
yaya
earnest zinc
send send
earnest zinc
yay
north sierra
is 3rd step good?
earnest zinc
lets see
what was the 2nd step??
let me do it and then show what i did
then you can ask ?s 😄
north sierra
hi
earnest zinc
oh shoot i missed somethinng hold on
north sierra
when do we use swap rows?
earnest zinc
okok
north sierra
when should we swap rows*
north sierra
yeah its goodo
north sierra
damn i feel sad
ive been trying this for hours
and you finished it in 5 minutes
lol
earnest zinc
nah dude, this is hard to grasp
north sierra
im soooooooooo bad
earnest zinc
takes practice, but you'll do it in quicker
just do the assigned homework 😄
anyway, anything look iffy?
north sierra
what's the 1 called again?
north sierra
our goal is to try to get a pivot position in each row
so first, take the first column
the first row should be a 1, and all of the rows should be 0 in the first column below it
pivot
whats pivot
north sierra
i understand all that stuff you explained
earnest zinc
do you know what echeon form is
north sierra
just wanna know what pivot is
earnest zinc
reducing == doing all of the steps to get that last matrix i wrote
north sierra
so i can only call a non zero a pivot once i have reduced the matrix fully?
earnest zinc
yea, then its the first non zero number in each row
not important rn though
did you justify each step i did?
north sierra
yeah i understand everything
just thinking what comes next is the hard part for me
also whats the point of a switch
switching rows
and when should it be used
earnest zinc
you do a switch when you wanna make a matrix look better
if i didnt use a switch in the 5th matrix i drew
then the answer would look like this
compare this matrix to the one above it
the one above it just looks nicer
let me see if i can find a easy set of steps that tell you what to do
earnest zinc
in the top matrix
the "pivots" go left to right each column down
but in the bottom matrix it doesnt do that
north sierra
also, in the second matrix, you made the 1 in row 2 a 0, isn't that bad ?
shouldn't you keep all the diagnol 1's
north sierra
diagonal*
earnest zinc
switching rows makes that diagonal
north sierra
ohh
earnest zinc
if i didnt switch the rows earlier, i wouldve gotten that matrix ^^
north sierra
ok
makes sense
okay
but you know for mine
i did 2R2 - R3 to not get rid of that
1
in 1st matrix
earnest zinc
thats a very complicated way to do it
north sierra
okay
north sierra
so the MAIN goal is to just make the the first (row2,column1) a zero and not worry about weather i am going to change the 1 in the diagonal line?
north sierra
can you answer my Q above lol sorry
earnest zinc
yes yes
north sierra
what text book is this?
north sierra
wow
earnest zinc
no well
i couldnt go to lectures because i was busy
so i just took the tests and did the homework
actually ill DM it to you
earnest zinc
the structure might not match that of your college, so DM me any problems and ill find the corresponding notes in the book
anyway, the finding the solution to any matrix
you find the left most non zero column
north sierra
DUDE
earnest zinc
then you switch the rows so you can have a non zero in the first row and first columnn
north sierra
i just bought this book
earnest zinc
LOL
north sierra
for $135
undone garnet
how can we prove that A doesn't have an eigenvalue not equal 0 iff A is nilpotent matrix?
dusky epoch
"A doesn't have an eigenvalue not equal to 0" is a roundabout way to say "all the eigenvalues of A are 0"
undone garnet
but how to prove there's no another matrix such that all eigenvalues of it are 0 but it's not a nilpotent matrix
remote fable
Suppose A has a non-zero eigenvalue k. Then let v be an eigenvector corresponding to that eigenvalue. Then we have that: Av=kv. Since A is nilpotent, we can find n such that A^n=0 matrix. So we have A^n v=0. But this is not possible, because we already have A^n= k^n * v, and since k is non-zero, this can not be 0, a contradiction.
A^n v = k^n * v I meant
undone garnet
thank you
I can prove this state
but the reverse
A has all zero eigenvalues
A has to be a nilpotent
remote fable
oh sorry I misread the "iff" as "if"
scarlet hamlet
could someone explain this theorem in another way?
like what im getting from it
noble swallow
@undone garnet there are different ways of seeing that
scarlet hamlet
is: u have 2 vector spaces v & w, then the lin. combinations of the set of all linrar mappings that go from v to w is a vector space?
noble swallow
Have you seen the generalized eigenspaces decomposition of the space (should be dealt with in Jordan normal form theory I think)?
Or Cayley-Hamilton theorem?
noble swallow
Right, that gives quite a direct proof of it, given that if 0 is the only eigenvalue of A, then p_A(λ)=λ^n (with the appropriate sign)
So A^n=0
north basalt
Heyo folks! Im doing Linear n DiffeQ at uni rn, but when looking at the homework.
I think its more of a DiffeQ question, but not sure where to put it
gray dust
north basalt
Ight, thankya
slow scroll
I would just test u v and w for 3. For part a of 4, use properties of linear transformations and for part b you have
Ap=b
Aq=b
So think about how you can conclude that p-q is in the null space. Think: what does it mean for p-q to be in the null space?
@wintry steppe
gaunt mulch
is there any way that i can calulate rank using power itration or power method
meager tinsel
Should I post exercises for which I have the solution but which I just generally find interesting to solve
(I esp like my solution for this one)
And if yes then is there a specific place maybe.)
north sierra
@gray dust
you were saying to check if i got the matrix right plug the values back into the equation and check if they equal each other
?
gray dust
yes, similarly to if you solve 2x=6, you get x=3, then plug x=3 back into the original equation to check the solution
gray dust
like i never get the same number
i don't know what this means
north sierra
sending a pic
one sec
so those are the equations and thats my matrix in RREF
so how do i check if im right
i dont get it
1 + 0 + 0 = 3 is what i get but that's not right
and 0 + 1 + 0 = -2 is wrong too
and 0 + 0 + 1 = 2 is wrong too
@gray dust
nvm
i got it
gray dust
@north sierra yeah i saw what happened on the other server. plugging in the solution means plug in whatever x/y/z values you got into the original set of equations
north sierra
oh
yeah i thought the row and columns were x y z too lol
but they're just the coefficient right?
half ice
You've actually found x = 2, y = -2, z = 3
Which is the solution to a system, if that's what you've started with
north sierra
yeah
aright
but in the REFF
what are the other values mean?
100
010
001
000
before the |
half ice
You can read this as
1x + 0y + 0z =
0x + 1y + 0z =
0x + 0y + 1z =
0x + 0y + 0z =
Note your last equation reads 0 = 0
north sierra
so if i plug x = 2,y = -2 , z = 3 in the variables, i should get -2, -2 and 3?
half ice
What do you mean "get"?
north sierra
=
half ice
Your matrix just reads
x = 2, y = -2, z = 3
I don't know where you'd plug them in. This is a solution to a system
north sierra
yeah but if i plug those numbers in the variables i will get the same answer?
half ice
If you plug those numbers into the variables, you'll find that all four equations are true
north sierra
1(2) + 0(-2) + 0(3) = 2
0(2) + 1(-2) + 0(3) = -2
0(2) + 0(-2) + 1(3) = 3
0x + 0y + 0z = 0
this is what im trying to ask lol
is that True/Correct?
oh yeah it is this is cool
half ice
That's how we found x, y, z
I don't see what info your getting by then putting them back in
north sierra
yeah
half ice
north sierra
i was just wondering if it still works
half ice
That's the one you want to put them into
half ice
Because you found the x,y,z that solves that
north sierra
yeah
airhgt
thanks for the help
how come i can't get the right answer for the 2nd equation
half ice
There is no unique solution. That is something you'd expect; there's four unknowns but only three equations
However, you've found lots about the solutions that exist
north sierra
so there are infinite many solutions?
half ice
w + 3y = 1
x - 2y = 2
z = 3
w = 1 - 3y
x = 2 + 2y
z = 3
So you can select any y you want, and get a solution
half ice
So the reduced matrix can be written as
w + 3y = 1
x - 2y = 2
z = 3
Get comfortable with going between a matrix form and an equation form
w + 3y = 1
x - 2y = 2
z = 3
north sierra
oh i see now
im using the matrix coefficients
instead of the original one
so i use the matrix coefficients for when i have at least one free variable?
half ice
Ya ya.
The matrix could be seen as the same as the equation, except you're just not writing out the variables
Yeah. I just rearranged the equations to get
w = 1 - 3y
x = 2 + 2y
z = 3
north sierra
i see
half ice
So you can put in any y you want, and get a new solution to the system
Since there's one free parameter, the solutions form a line
Or, that the solutions form a one dimensional vector space
cobalt beacon
Hi i've been stuck on a simplex problem (Big-M/two-phase) for a few hours now and im not sure where im going wrong. Could you spare some time to help me please? Thanks very much!