# Introductory linear algebra questions

• Aug 13th 2013, 04:38 AM
Nemesis10192
Introductory linear algebra questions
Hi guys, having some issues with some basic introductory linear algebra questions, any help much appreciated! :)- 1) Find the dimension and a basis of the following vector spaces V over the given field K.

(i) V is the set of all vectors (a,b,y) in R^3 with a+2b-2y=0 with K=R.

My answer: Here I'm completely unsure, rearranging the equation for each variable we get the set of vectors (2y-2b,(2y-a)/2, (a+2b)/2). So does this have dimension 3? Clearly any vector can be constructed by (2y-2b)(1,0,0)+((2y-a)/2)(0,1,0)+((a+2b)/2)(0,0,1) and of course (1,0,0),(0,1,0),(0,0,1) are linearly independant. Is this correct?

(ii) V set of vectors (a,b,y) in R^3 with a-b=-y and a+2b=y with K=R.

My answer: Again unsure what to do with the supplied equations. Setting both equations equal to a, then combining them and solving for y gives y=(3/2)b so we have set of vectors (a,b,(3/2)b). Now here I get really confused. Again any vector can be constructed by a(1,0,0)+b(0,1,0)+(3/2)b(0,0,1) and this is the standard basis whose components we know to be linearly independant (or it wouldn't be a basis) which would imply that the vector space has dimension 3. However the any vector can also be constructed by a(1,0,0)+b(0,1,(3/2)) (which are clearly LI) which would imply a dimension of two. Here I'm really confused as the answers are contradictory. I believe the correct dimension to be 2 since the vector space on visualisation is I believe a plane, and so one would expect a dimension of 2 but why do I get two different answers? What am I doing wrong?

This can be generalised to me being unsure of how to approach a situation where I have a set of vectors (a,b,b). This can be spanned in two ways: (1) a(1,0,0)+b(0,1,0)+b(0,0,1) implying a dimension of 3 or (2) a(1,0,0)+b(0,1,1) implying a dimension of 2. In both cases the set of spanning vectors are linearly independant. What gives?

(iii) V=C^2, K=R.

My answer: dimension 4, basis (1,0), (0,i), (0,1), (0,i). Pretty basic textbook example here so no problems (i hope). If K=C then the space would have dimension 2 and a basis (1,0), (0,1), correct?

(iv) V is set of all polynomials over R with degree at most n, in which the sum of the coefficients is 0, K=R.

My answer: So we have a_0+a_1x+a_2x^2+...+a_nx^n with a_0+a_1+a_2+...+a_n=0. Not really sure how the condition of the coefficients restricts things. Normally the set of all polynomials over R with degree at most n would have dimension n+1 and (standard) basis 1,x,x^2,...,x^n. Can someone give me a hint on how to start this one?

(v) K=R and and V is the set of functions from R to R which are solutions of the differential equation d^2f/dx^2-5f=0.

My answer: Solving the (second order) differential equation via standard methods gives a general solution f(x)=c_1e^(root(5)x)+c_2e^(-root(5)x) so does this just have dimension 2 and basis e^(root(5)x), e^(-root(5)x)?
Thanks in advance for any help!
• Aug 13th 2013, 03:59 PM
ANDS!
Re: Introductory linear algebra questions
Quote:

So does this have dimension 3?
No. y = (a+2b)/2. There is no need to write the other elements of the vector, because that doesn't give you anything. Instead what you have is (a,b,[a+2b]/2), where a and b are real numbers. It's not too hard to show that the basis of this space is (1,0,1/2), (0,1,1).

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Now here I get really confused.
You defined a to be one of b-y or y-2b. As you have defined it, a is not a "free variable".

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What am I doing wrong?
You have y=b-a and y=a+2b. This is only possible if b=-2a. Then y = -3a. Then the set of vectors is in fact those of the form (a,-2a,-3a), where a is a real number. The dimension should be obvious here.

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dimension 4, basis (1,0), (0,i), (0,1), (i,0).
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Not really sure how the condition of the coefficients restricts things.
If you know all but one of the coefficients, then you know ALL of the coefficients. Suppose we are in p^2, with the condition on the coefficients. Then a vector is ax^2+ bx+c. But a+b+c= 0 implies that c = -a-b. Then (in vector form) what you really have is (a,0,-a) + (0,b,-b). It should be clear that the basis vectors are (1,0,-1) and (0,1,-1) - with dimension 2.
• Aug 13th 2013, 04:21 PM
HallsofIvy
Re: Introductory linear algebra questions
Quote:

Originally Posted by Nemesis10192
Hi guys, having some issues with some basic introductory linear algebra questions, any help much appreciated! :)- 1) Find the dimension and a basis of the following vector spaces V over the given field K.

(i) V is the set of all vectors (a,b,y) in R^3 with a+2b-2y=0 with K=R.

In addition to ANDS! "y= (a+ 2b)/2" you could also write a= 2y- 2b so that (a, b, y)= (2y- 2b, b, y)= (2y, 0, y)+ (-2b, b, 0)= y(2, 0, 1)+ b(-2, 1, 0) which should make it clear what a basis and the dimension are.

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(ii) V set of vectors (a,b,y) in R^3 with a-b=-y and a+2b=y with K=R.
a- b= -y is the same as y= b- a so (a, b, y)= (a, b, b- a)= (a, 0, -a)+ (0, b, b)= a(1, 0, -1)+ b(0, 1, 1).

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This can be generalised to me being unsure of how to approach a situation where I have a set of vectors (a,b,b). This can be spanned in two ways: (1) a(1,0,0)+b(0,1,0)+b(0,0,1) implying a dimension of 3
NO. Those are not two different "b"s so you cannot write them separately.

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or (2) a(1,0,0)+b(0,1,1) implying a dimension of 2. In both cases the set of spanning vectors are linearly independant. What gives?
This is correct because it has just the one value of "b".

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(iii) V=C^2, K=R.

My answer: dimension 4, basis (1,0), (0,i), (0,1), (0,i). Pretty basic textbook example here so no problems (i hope). If K=C then the space would have dimension 2 and a basis (1,0), (0,1), correct?
Yes, $C^2$, as a vector space over the real numbers has dimension 4, as a vector space over the complex numbers, dimension 2..

Quote:

(iv) V is set of all polynomials over R with degree at most n, in which the sum of the coefficients is 0, K=R.

My answer: So we have a_0+a_1x+a_2x^2+...+a_nx^n with a_0+a_1+a_2+...+a_n=0. Not really sure how the condition of the coefficients restricts things. Normally the set of all polynomials over R with degree at most n would have dimension n+1 and (standard) basis 1,x,x^2,...,x^n. Can someone give me a hint on how to start this one?
The "set of all polynomials over R with degree at most n" is a vector space of dimension n, but the condition that $a_0+a_1+a_2+...+a_n=0$ allows you to write one as a linear combination of the others, $a_0= -(a_1+a_2+...+a_n)$, so that you can replace that one with the others. That means you are left with n-1 "independent" numbers so any such polynomial can be written as a linear combination of those n-1 terms.

Quote:

(v) K=R and and V is the set of functions from R to R which are solutions of the differential equation d^2f/dx^2-5f=0.

My answer: Solving the (second order) differential equation via standard methods gives a general solution f(x)=c_1e^(root(5)x)+c_2e^(-root(5)x) so does this just have dimension 2 and basis e^(root(5)x), e^(-root(5)x)?
Thanks in advance for any help!
Yes, that's correct. More generally, the set of all solutions to any nth order linear, homogeneous, differential equation is a vector space of dimension n.
• Aug 20th 2013, 10:28 AM
Nemesis10192
Re: Introductory linear algebra questions
Many thanks for your replies guys, they cleared up things immensely for me. That said, I would like to check my answers to some new questions since and have decided its probably better just to continue within this thread. Hopefully you guys will have the time to help me again :). Read up on how to use basic latex+ got an image of the problem sheet so hopefully my post will be a bit more readable this time! Here is the problem sheet(insert picture wasn't working-I press upload and nothing happens):
Nemesis10192's Library | Photobucket

(1) (i) yes (ii) yes (iii) yes (iv) no- Rewriting using the supplied equation we have $(\alpha,\beta,\frac{\alpha+\beta-1}{7})$ thus testing that scalar multiplication and vector addition returns another vector in the subspace we have $a(\alpha_{1}, \beta_{1}, \frac{\alpha_{1}+\beta_{1}-1}{7}) + b(\alpha_{2}, \beta_{2}, \frac{\alpha_{2}+\beta_{2}-1}{7})=(a\alpha_{1}+b\alpha_{2}, a\beta_{1}+b\beta_{2},\frac{(a\alpha_{1}+b\alpha_{ 2})+(a\beta_{1}+(b\beta_{2})-(a+b)}{7})$ which is of the form $(\alpha, \beta, \frac{\alpha+\beta-(a+b)}{7})$ and so by the (a+b) term at the end instead of just 1 we see that the answer to (iv) is no. Are these correct?

(2) Not too bothered about these, fairly routine- dimension= number of vectors in the basis, can find basis by sifting and we may stop once we have 4 vectors anyhow since vectors with 4 components cannot possibly span anything with more than dim=4, correct? About (ii) though- I'm unsure what to make of the condition about the finite field, does it literally just restrict the components of each vector to being 0 or 1, or does it affect the answer in some other way?

(3) Would really like an assessment of how good my proofs are for the following question. I'm new to proof writing so don't really know what's acceptable/how nitpicky I need to be etc. If for any part there is a more succinct way of proving it, please do share :).

(i) Well this is very obvious, so much so that I'm not really sure how to prove it. My best attempt is something along the lines of:
Since $w_{1},...,w_{m}$ do not span W we can pick a $w_{m+1}$ s.t. it is not a linear combination of $w_{1},...,w_{m}$ but since $w_{1},...,w_{m}$ are also linearly independent we have $a_{1}w_{1}+a_{2}w_{2}+...+a_{m}w_{m}+a_{m+1}w_{m+1 }=0 \Rightarrow a_{1}=a_{2}=a_{m}=a_{m+1}=0$ which is the definition of linearly independence hence $w_{1},...,w_{m+1}$ are linearly independent.

(ii) Proof of finiteness of dimension of W: Since V is finite dimensional lets call its dimension n. Assume W is infinite dimensional, then we can find vectors $w_{1},w_{2},...,w_{m} m>n$ that are linearly independent since an infinite dimensional space has infinitely many linearly independent vectors in it. But since W is a subgroup of V these m linearly independent vectors must also be in V which contradicts dim(v)=n since dimension of a vector space is defined to be the number of linearly independent vectors in any basis for it. Therefore W is finite-dimensional.

Proof that $dim(W)\leq dim(V):$ $Let w_{1},...,w_{m}$ be a basis for W. Since these vectors form a basis for W, they are linearly independent, and since W is a subgroup of V then $w_{1},...,w_{n} n\geq m$ is a linearly independent set of vectors in V. Therefore we can extend the basis for W into a basis for V; $w_{1},...,w_{m},v_{1}...,v_{k} k\geq 0$. Hence $dim(V)=m+k \geq m=dim(W).$

(iii) Since V is finite-dimensional, we know by (ii) that $W_{1}$ and $W_{2}$ are finite-dimensional. Let $w_{1},...,w_{m}$ be a basis for $W_{1}.$Then since $W_{1}$ is a subgroup of $W_{2}, w_{1},...,w_{n}$ is a set of vectors in $W_{2}.$By definition of basis $w_{1},...,w_{m}$ is a linearly independent set and so can be extended to a basis for $W_{2}; w_{1},...,w_{m},v_{1},...v_{k} k \geq 0$. Since $dim(W_{1})=dim(W_{2})$we have $m=dim(W_{1})=dim(W_{2})=m+k \Rightarrow m=m+k \Rightarrow k=0$ and so by the way we have set up our bases we see that $W_{1}$ and $W_{2}$ have the same basis $w_{1},...,w_{m}$and so have the same span $\Rightarrow W_{1}=W_{2}.$

(4) W is again finite dimensional by (ii). Let a basis for W be $w_{1},...,w_{m}$ then since W is a subgroup of V, $w_{1},...,w_{n}$ is a set of vectors in V. Since $w_{1},...,w_{m}$ is a basis it is a set of linearly independent vectors and so can be extended to a basis for V; $w_{1},...,w_{m},x_{1},...,x_{k} k \geq 0$. Since the aforementioned set is a basis, its members are linearly independent, therefore $x_{1},...,x_{k}$ are linearly independent. Define a new subspace X of V generated by the basis $x_{1},...,x_{k} k\geq 0.$ Then V=W+X and $W\bigcap X=0$ and so we see W has a complementary subgroup in V.

Critique of the proofs very much wanted!

(5) (i) basis of image (1,0,1,0),(0,0,1,1),(1,1,0,0),(0,1,0,1); Rank=4. By rank-nullity we therefore have nullity=0 and a basis of the kernel= $0_{V}$?
(ii) So in general we have the mapping of $a+bx+cx^2+dx^3+ex^4+fx^5$ to $24e+120fx$. Therefore a basis of the image is 1,x and rank=2? Therefore nullity=4 and a basis of the kernel is $1,x,x^2,x^3$? Bit confused about this one, as well as the concept of a kernel.

(6) (i) Stretch scale factor 2 and rotation 180 degrees about origin.
(ii) Projection onto the y axis.
(iii) reflection in y=x.
(iv) Don't know how to describe the geometrically- x coordinate maps to x+y and y coordinate maps to y-x then a scale factor of $\frac{1}{\sqrt{2}}$is applied to the whole thing but I have no idea how to describe this succinctly?

(7) No idea here, can someone give me some starting hints please?

Wow, that took depressingly long to type out! Hope I become faster with familiarity :). Thanks in advance for any help!