Is {x, y ER^n such that ||x+y||> 0} a subspace?
How to go about answering part C? Believe there is a typo and it is supposed to say col(B)
I know vector b must be 4x1.
This is exercise 2.B.3.c from Axler's "Linear algebra done right": https://imgur.com/a/0w87ZSi
The basis I found for U at a) was {(3,1,0,0,0), (0,0,7,1,0), (0,0,0,0,1)}. I extended it to a basis of R^(5) by adding (3,2,7,1,1) and (3,1,6,1,1).
I need to find a subspace so that the direct sum of the span of those first 3 vectors and this subspace equals R^(5). My intuition said that the span of the last two vectors I added could be that subspace.
So I tried to solve this in two ways and failed:
First method: Let V be the span of those last 2 vectors and U the span of those 3. We know that if U⊕V = R^5 then their intersection is 0. Assume that they are not a direct sum => there is a non-zero vector v that is part of the intersection between U and V, v = (x1, x2, x3, x4, x5), x1=3x2 , x3=7x4, x4=x5 , x1=3x4. =>
=> v = (3x2, x2, 7x2, x2, x2) and this respects all the 4 rules => v = x2(3,1,7,2,2). => Any scaling of (3,1,7,2,2) is part of the intersection between U and V. So they are not a direct sum???
Second method: We know that if U⊕V = R^5 then the only solution to U+W is 0.
U = (3x2, x2, 7x4, x4, x5) by replacing x1 and x3 with the rules in the imgur photo.
W = a(3,2,7,1,1) + b(3,1,6,1,1) = (3a+3b, 2a+b, 7a+6b, a+b, a+b) = (3a1, a2, a3, a1, a1)
U+W = {(3x2 + 3a1, x2 + a2, 7x4 + a3, x4 + a1, x5 + a1)}. U+W = 0 so we have the system of equations
3x2 + 3a1 = 0
x2 + a2 = 0
7x4 + a3 = 0
x4 + a1 = 0
x5 + a1 = 0
We know from the last two that a1 = -x4 = -x5. From the second one: a2 = -x2. From the third one: a3 = -7x4
What now? How do we get to show that 3x2 = x2 = 7x4 = x4 = x5 = 3a1 = a2 = a3 = a1 = a1 = 0 ?
EDIT: In the solutions the author said that the subspace W is the span of the second vectors they added at b) so I followed the same approach as them.
https://preview.redd.it/glmkjl5loj281.png?width=1167&format=png&auto=webp&s=9a7ef944c913d27b73decf50eaec99dbe849d805
I have a question with this example: Why can´t dim U=4? A subspace U of V can have the same dimension as V, right? In that case, U=V. Why can´t that be the case in this example? More specifically why can´t we extend the basis of U trivially (i.e. only adding 0-vectors) to a basis of V?
What am I missing?
So I am currently working through Linear Algebra Done Right by Sheldon Axler and although I have had some exposure to college linear algebra, it wasn't proof based.
I have a question regarding this problem: https://i.imgur.com/CNqQcQq.png
The solution says: https://i.imgur.com/Tnu7wng.png
My work:
https://i.imgur.com/xCShJYy.jpg
I understand the condition for being a subspace but I am having some trouble understanding the definition of both linear transformations. Is is it necessary to define the linear maps in terms of m and j and if so ,why? If anyone could tell me why my proof is wrong it would be great. Thanks!
If we take the projection P of a vector x onto the orthonormal vector set {v1,v2..vn}, then are P and x - P orthogonal? If yes, why?
Im not looking for a formal proof, just some intuition. I know that x - P is orthogonal to v1,v2...vn, but why is this the case with P as well?
In the example, the two vectors are linearly independent
How do I modify H such that H is no longer a subspace of R4? If H is now the set of vectors of the form {(,a,b,c)}, then H = span{<1,0,0>, <0,1,0>, <0,0,1>}. Since there are only 3 entries for each vector, H spans R3. However, R3 is not a subset of R4 and therefore the basis vectors for H are not in R4. Thus, {a,b,c} is not a subspace of R4 by theorem 1.
If H is now the set of vectors of the form {(a,b,c,d,e)}, then H = span {<1,0,0,0,0>, <0,1,0,0,0>, <0,0,1,0,0>, <0,0,0,1,0>, <0,0,0,0,1>} Since there are 5 entries for each vector, H spans R5 and R5 is bigger than R4. The basis vectors in H are not in R4 and therefore {(a,b,c,d,e)} is not a subspace of R4.
Is there any other way (or am I completely off the mark)? Thanks for any help!
So I have this question, and I'm clueless on how to approach it, I'm getting that B is linearly dependent hence it can't be a basis anyway.... If I could be given some direction, I'd be grateful
V is a subspace that is spanned by the functions f , g :R->R which are defined by:
g(x) = CosX and f(x) = SinX.
We'll also define the functions h, k:R->R by:
h(x) = 2sinX+cosX and k(x) = 3cosX
Prove that the groups B = {f,g} and C = {h,k} are basis for V.
So when I tried solving this I thought to myself, well ok let's look at B first, it's told that V is spanned by f,g so the Span{B}=V, I just need to prove that B is Linearly independent and it's a basis!
So B is linearly independent if for the two scalars a,b: a*f(x)+b*g(y)= 0 only and only if a=b=0
But f(0) = sin0 = 0 and g(pi/2) = cos(pi/2) = 0
So a*f(0)+b*g(pi/2)=0 for any a,b.....
I'm clearly doing something wrong here, I'm not sure how to approach this.
EDIT: Even for the same x for g(x) and f(x) there's a solution in which a,b are not 0 scalars
V = [{(2, 3, −6), (- 1, −2, 4), (0, −1, 2)}], W = [{(7, −7, 14), (- 1) , 1, −2)}].
Determine the base and dimension for spaces V, W, V + W, and V ∩ W.
Edit: Solved!
Problem: https://imgur.com/a/425TSWI
Question: Is this set a subspace of R^3 ? I know the set is closed under addition, but what I cannot figure out is if it is closed under scalar multiplication if the scalar is a real number.
Cause I'm dyslexic and you lookin fineaf
I know and underatand that (0,0), W={(x,y): ax+by=0} and R^2 itself are linear subspaces of R^2, but why isn't V={(x,y): ax^3 +by=0} or any curve that has (0,0) as an internal point a subspace of R^2? I mean, all these curves should respect the axioms of a vectorial space
Suppose I have a linear transformation T: V ->V and want to prove that kerT + ker(T-I) is in V. Is this proof trivial? because T is a linear transformation from V to V, so V must be the biggest space in the problem, so kerT + ker(T-I) must be in V correct? Or do I have to prove it?
Hi guys, I have this exercise in which I am given two vector subspaces of R4 U1=Span((0,1,-1,1),(1,1,-2,3),(1,-1,-2,1)) U2=Span((1,-2,-3,0),(2,0,1,-1),(1,2,2,-1). Is it correct to put the six vector in a matrix, then do a gauss elimination, choose the parameter and write the base with the parameter/parameters I have?
The repetition of questions at the end are to focus where I need help. If you can answer any or all of these questions, please help and thank you so much. The 3 topics are separated by -'s and pair with the title.
Vector Space:
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A vector space is a collection of vectors closed under addition and multiplication. More specifically with an example: The collection of vectors with exactly two positive real valued components is not a vector space. The sum of any two vectors in that collection is again in the collection, but multiplying any vector by, say, −5, gives a vector that’s not in the collection. We say that this collection of positive vectors is closed under addition but not under multiplication.
This doesn't make sense to me.
I'd like to use a 2x2 matrix to understand this, with vectors A = (1, 1) and B = (1, 0) (a lower triangular matrix of 1's).
First, the sum, as explained above, A + B = (2, 1) is in the collection.
What is the collection? My brain just thinks (2, 1) = A + B. This is not A nor B, nor a scalar of A or B. It is A + B. *So how is this in the collection of vectors (it's not A nor B)?
Second, why is multiplying by -5 giving a vector that's not in the collection?
-5A = (-5, -5). Again my brain just thinks (-5, -5) = -5A. This is not A nor B, this is a scalar of A. *So how is this not in the collection of vectors (it is cA)?
"Finally, if a collection of vectors that is closed under addition and multiplication, then it is a vector space."
Subspaces:
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A subspace is, in simple terms, c1*A + c2*B and the zero vector must be included in this addition and multiplication of A and B.
With the addition and multiplication of A and B in mind (c1*A + c2*B), I don't understand why an unsymmetric matrix does not form a subspace. I'd like to use a 2x2 matrix to understand this, with vectors A = (1, 1) and B = (1, 0) (a lower triangular matrix of 1's).
A and B are subspaces (lines). c1*A + c2*B give some b (c1*A + c2*B = b). I can reach any point in R^2 with c1*A + c2*B, in other words c1*A + c2*B span all of R^2. How come this A and B aren't a subspace?
Looking at matrices in general:
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... keep reading on reddit ➡If we have a linear map T : V -> W, and V and W are finite dimensional vector spaces and T is surjective, This means that Rank(T) = Dim(W). So by rank nullity theorem we have that Rank(T) + Nullity(T) = Dim(V). We can replace Rank(T) with Dim(W) so that Dim(W) + Nullity(T) = Dim(V). SO
Dim(V) >= Dim(W).
So the problem here is, can I replace v in V into complementary subspaces such that v = x + y (direct sum), where x and y are subspaces of v. So T(v) = w => T(x+y) => T(x) + T(y). If we make Dim(x) = Dim(W). Does that mean i can create a new linear map T : x -> W, which can be written as T: V -> W such that this new T: x-> W is surjective?
If not what steps am i missing?
Hi,
I'm looking for help finding bases for two separate subspaces:
If you can't see the picture:
Question A: find a basis for P: p element of Pol3, domain [0,2], Real, with requirement: p(0)=p(1)=p(2)
Question B: Find a basis for Q: p element of Pol2, domain [0,2], Complex, with requirement: xp'(x)=p(x)
What I've tried
Question A
I've tried finding the first one by giving the inputs: 0, 1 and 2 into the standard-basis:
p(0) = (1, 0, 0, 0)
p(1) = (1, 1, 1, 1)
p(2) = (1, 2, 4, 8)
With the vectors relative to the standard basis (1, x, x^2, x^3)
Solving this did give me one eigenvector; (0, 2, -3, 1), giving me one basis function: 2x - 3x^2 + x^3, but I can't seem to find the second basis function for this question (which should be the function: 1)
Question B
I actually don't know how I should approach this question.
Cause I'm dyslexic and you lookin fineaf
