So, Matrix A is [-1000 0 0 0;
0 -1000 0 0;
0 0 50 100;
0 0 -100 0]
and Matrix B is [1000;
1000;
0;
100]
The controllability matrix that I end up getting when using Matlab is:
(1*10^12) * [0 0 0.001 -1;
0 0 0.001 -1;
0 0 0 -0.0001;
0 0 0 -0.0001]
By doing rref, I end up getting this result:
[1 0 0 -10000000;
0 1 0 40000;
0 0 1 -950;
0 0 0 0]
So, we see that it is not controllable. From this rref, we see that the first three columns of our controllability matrix should be the basis of linearly independent vectors. However, it doesn't seem like they are linearly independent. Can someone help me figure out why this is the case and if this span of vectors can still be considered?
I hope the way that I wrote out the matrix makes sense.
My understanding is that a maximal linearly independent set and a basis essentially have the same definition. I'm wondering if there's a subtle difference that exists so that we would prefer to use one over the other in different situations. Is there a difference? Or is my textbook just being weird by using both (but not necessarily interchangeably)?
Thanks for any help!
Hello! I'm confused about the fact that C only has 2 elements... I don't know how to view them as vectors. Also the problem doesn't say if x ∈ ℝ^2 but I'm assuming that we're talking about R^2. So, what I'm thinking is:
Let's say
a(-x) + b(x²-2x)= 0
If 0 is the only value possible for a and b then the set is linearly independent right?
But what if we also consider a= b(x-2) ? Then we have that
b(x-2)(-x) + b(x²-2x)= -(b(x²-2x) + b(x²-2x) = 0
With that i think the set C wouldn't be linearly independent.
The problem says that i need to prove that C is linearly independent tho... is the problem a trap? Lol So I'm stuck i don't know how should I proceed or where my analysis is wrong .
Given the vectors v1 = (3,2,7,1,1) and v2 = (3,1,6,1,1) my intuition says they are linearly independent since one is not a scaled up version of the other. However if we solve the system
3a + 2b + 7c + d + e = 0
3a + b + 6c + d + e = 0
We subtract the 2nd from the 1st equation
b + c = 0 <=> b = -c
We rewrite the system
3a -2c + 7c + d + e = 0
3a -c + 6c + d + e = 0
Both equations are equal to each 3a+5c+d+e. That means 0=0 so any solutions are true. So are they dependent in the end? What to do from here?
I mean that didn't prove anything: non-trivial? non-zero?
For my intro to computer science class (that I have to take for my major), I have to write a program that takes a matrix, A, and creates another matrix, B, composed of only the linearly independent columns of A.
My problem is that I have yet to take a class on linear algebra, so my understanding basically derives from google searches and what I can gather from the lecture and slides of my intro to computer science class, which aren't as comprehensive as I'd really like. So I have no clue how I'd solve the problem, much less how to write a program that would solve it.
This is the example matrix I was given
This is the collection of linearly independent columns of the example matrix
So, I know which columns are linearly independent for the example matrix, since that was given to me. I also know that column 2 of A is linearly dependent on column 1 because it's just a multiple of it. Completely lost on the rest of it though. I basically just want to know how you'd go about finding which columns are and aren't linearly independent, since as long as I get the general process I should be able to figure out how to get that into code.
For a question I have to show that something is a linearly independent subset of R^(4). So I started finding the zero vector etc. but realised this is for a subspace right? So is there a way to show something is a subset or am I overthinking it and I just need to show it is linearly independent and it is already a subset. Thank you in advance.
What I need to show it from: {[1,0,1,2],[-2,-2,1,3],[1,3,0,0]}
After searching YT I have found that a neat way to prove linear independence of 3 Vectors in R3 is combining them into a matrix, reducing that matrix to echelon form and then reading the solution straight off that that the three vectors can all be scaled and added together to make 0.
The works well, but what if one had to determine whether 3 numbers in R were linearly independent.
Lets say a b and c belonged to R, how would we determine if these were linearly independent? Isn't it just natural that (0)a+(0)b+(0)c=0 and therefore they would be linearly independent??
Any help or advice would be greatly appreciated!!
If you plug in y = e^(-3x) to y‘’ - 6y’ + 9y = 0, it wouldn’t even satisfy the equation.
(e^(-3x))’’ -6 (e^(-3x))’ + 9 (e^(-3x))
= (-3 e^(-3x))’ - 6 (-3 e^(-3x)) + 9(e^(-3x))
= 9 e^(-3x) + 18e^(-3x) + 9 e^(-3x)
= 36 e^(-3x)
!= 0
https://preview.redd.it/m3w8uvtfzgs51.jpg?width=1248&format=pjpg&auto=webp&s=4747bf3c31acfd63c6b0c8c8d03411db409b2af4
My very basic understanding of QFT is that it describes all of physics in terms of fields, and particles are simply waves within these fields. So we not only have our basic fields like electric and magnetic, but also fields that describe the location and properties of electrons, quarks, etc. As such, the universe could, at least conceptually, be entirely described by the values of these fields at every point.
The standard model describes all known particles, but I can't find anything that discusses the fields from which each particle arises.
So my question is, how many of these fields are there, and of what types (scalar, vector (always 3 dimensional?), etc.)? In other words, how many distinct numbers would we need to fully describe all the properties of a single point in space.
Bonus points if you can list all of these fields for me.
Or if I have completely misunderstood QFT, please clarify.
How do i prove that for a finite dimensional linear vector space, if the basis is n-dimensional, no linearly independent set will have a dimension greater than n? Is it because the basis is a spanning set of the vector space and since it is also linearly independent, no other linearly independent set can be bigger?
In my physics intro, we're learning vectors from scratch. So two vectors V1 and v2 are linearly independent if c1v1+c2v2=0. So I put ∫f(x)= F(x)+c and then set a1f(x)+a2(F(x)+x)=0 but i don’t know what to do next
I am so sorry to be asking this on here but my professor hasn't been answering his emails since it is a holiday and this assignment is due Monday.
Anyway, So I am supposed to prove two vectors are linearly independent given that they are perpendicular.
Since they are perpendicular that means the dot product is zero. So I know I am supposed to use that somehow but I am just not sure how.
<x,y>=0 -> x1y1+x2y2+...+xnyn=0
I know if two vectors are linearly independent then only the trivial solution exists. I have tried manipulating this expression algebraically with no luck. I would love some hints.
In other words this: alpha*X+alpha2*Y =0 where alpha 1&2 =0
-
Are the vector valued functions [sin(t); cos(t)] and [sin(2t); cos(2t)] linearly dependent or independent for all t?
-
Consider x1= [cos(t); 0; 0], x2=[sin(t); cos(t); cos(t)], x3 = [cos(t); sin(t); cos(t)]. Are these 3 vector valued functions linearly independent for all t? Is there a 1st order homogeneous linear system for which they are solutions?
Determine if the functions : x; sin(x); cos(x); e^x; ln (x) are LINEARLY independent.
I'm currently reviewing for an exam and I don't remember how I solved this problem . We are given two 5x5 matrices and we apply reduced row echelon form to obtain their respective linearly independent vectors . Now I know that the union is just putting the linearly independent vectors into one matrix and reducing it to rref. My question is how do I get the intersection then ?
I also know that the if the dimension of (A) =3 and dim(B) = 3 and dim(A+B) = 5 then dim(A union B) = 1 , although I'm not sure how to apply it in this scenario if need be.
I understand why it matters for the rows to be linearly independent and what it signifies. But why do we care if the columns are linearly independent and why do we care to find a basis for them. The rank is the same for either.
For the row space each value in the row represents a different variable or dimension. For the column each value represents the same variable so it doesn't make sense to say they are linearly independent.
I need to prove it for a system of 3 equations with equations like:
ax + by +cz = d
How do you do this?
I tried eliminating cos^2 (x) by substituting x with 0. Then I was left with bsin^2 (x)+csin(2x)=0. Next I eliminated sin^2 (x) by dividing everything by sin^2 (x), then took the limit of both sides so I was left with c*(sin(2x))=0. I know the answer is that it is independent, but I don’t understand why sin(2x) can’t be zero? If I plugged in 0 into x, then it’s possible that c won’t be zero, right? Did I do something wrong?
Suppose that A is an mxn matrix and B is an nxp matrix. Suppose further that the columns of A are linearly independent, and that B does not have any free variables when row-reduced. Show that the columns of AB are linearly independent.
(Hint: You might want to first translate the problem into language that doesn't involve matrices.)
So, Matrix A is [-1000 0 0 0;
0 -1000 0 0;
0 0 50 100;
0 0 -100 0]
and Matrix B is [1000;
1000;
0;
100]
The controllability matrix that I end up getting when using Matlab is:
(1*10^12) * [0 0 0.001 -1;
0 0 0.001 -1;
0 0 0 -0.0001;
0 0 0 -0.0001]
By doing rref, I end up getting this result:
[1 0 0 -10000000;
0 1 0 40000;
0 0 1 -950;
0 0 0 0]
So, we see that it is not controllable. From this rref, we see that the first three columns of our controllability matrix should be the basis of linearly independent vectors. However, it doesn't seem like they are linearly independent. Can someone help me figure out why this is the case and if this span of vectors can still be considered?
I hope the way that I wrote out the matrix makes sense.
So, Matrix A is [-1000 0 0 0;
0 -1000 0 0;
0 0 50 100;
0 0 -100 0]
and Matrix B is [1000;
1000;
0;
100]
The controllability matrix that I end up getting when using Matlab is:
(1*10^12) * [0 0 0.001 -1;
0 0 0.001 -1;
0 0 0 -0.0001;
0 0 0 -0.0001]
By doing rref, I end up getting this result:
[1 0 0 -10000000;
0 1 0 40000;
0 0 1 -950;
0 0 0 0]
So, we see that it is not controllable. From this rref, we see that the first three columns of our controllability matrix should be the basis of linearly independent vectors. However, it doesn't seem like they are linearly independent. Can someone help me figure out why this is the case and if this span of vectors can still be considered?
I hope the way that I wrote out the matrix makes sense.
