Thread: First Year Mathematics B (Integration, Series, Discrete Maths & Modelling)

Originally Posted by InteGrand
What other method do you have in mind? (Induction helps takes care of it nicely if you want to prove it straight from the axioms.)

$\noindent (The statement being asked to prove is not just true by definition like say the statement nm = \underbrace{m+ m + \cdots m}_{n \text{ times}} for positive integers m. The statement being asked to prove is saying that adding the vector \mathbf{v} n times is the same as multiplying the vector by the scalar n. This is proved with the help of the integer version of this, which is true by definition.)$
For some reason I kept thinking what was the problem with doing it component wise...

In that case, how could you prove this then?
$\lambda \textbf{0}=\textbf{0}$

As a reference, they proved $0\textbf{v}=\textbf{0}$ like this
$\textbf{v}+\textbf{0}=\textbf{v}=1\textbf{v}=(1+0) \textbf{v}=1\textbf{v}+0\textbf{v}=\textbf{v}+0 \textbf{v}\\ \text{Then by cancellation property }0\textbf{v}=\textbf{0}$

Also gonna take a stab at the axioms they used...
1. Existence of 0
2. Scaling factor of 1
3. arithmetic (or just existence of 0 in field of scalars)
4. Scalar distributive law
5. Scaling factor of 1

Originally Posted by leehuan
For some reason I kept thinking what was the problem with doing it component wise...

In that case, how could you prove this then?
$\lambda \textbf{0}=\textbf{0}$

As a reference, they proved $0\textbf{v}=\textbf{0}$ like this
$\textbf{v}+\textbf{0}=\textbf{v}=1\textbf{v}=(1+0) \textbf{v}=1\textbf{v}+0\textbf{v}=\textbf{v}+0 \textbf{v}\\ \text{Then by cancellation property }0\textbf{v}=\textbf{0}$

Also gonna take a stab at the axioms they used...
1. Existence of 0
2. Scaling factor of 1
3. arithmetic (or just existence of 0 in field of scalars)
4. Scalar distributive law
5. Scaling factor of 1
$\noindent Note$

\begin{align*}\lambda \bold{0} + \lambda\bold{0} &= \lambda \left(\bold{0} +\bold{0}\right) \quad (\text{distributivity axiom}) \\&= \lambda \bold{0} \quad (\text{as }\bold{0}+\bold{0} =\bold{0} \text{ due to definition of zero vector}) \\ \Rightarrow \lambda \bold{0} &= \bold{0} \quad (\text{uniqueness of zero vector in a vector space}).\end{align*}

We can't just do these component-wise because they are results about arbitrary vector spaces (not necessarily ones where the vectors are n-tuples).

Originally Posted by leehuan
For some reason I kept thinking what was the problem with doing it component wise...

In that case, how could you prove this then?
$\lambda \textbf{0}=\textbf{0}$

As a reference, they proved $0\textbf{v}=\textbf{0}$ like this
$\textbf{v}+\textbf{0}=\textbf{v}=1\textbf{v}=(1+0) \textbf{v}=1\textbf{v}+0\textbf{v}=\textbf{v}+0 \textbf{v}\\ \text{Then by cancellation property }0\textbf{v}=\textbf{0}$

Also gonna take a stab at the axioms they used...
1. Existence of 0
2. Scaling factor of 1
3. arithmetic (or just existence of 0 in field of scalars)
4. Scalar distributive law
5. Scaling factor of 1
man i feel sorry if they expect u to properly remember the axioms, for us we just had to use 2 or 3.

They really aren't hard to remember.

Originally Posted by seanieg89
They really aren't hard to remember.
annoying as fuck tho

The axioms are just annoying.

Actually knowing when and how to use them like there though is ...

S = {x is an element of R^4 | (x1)-5(x3) = 2(x4)}

Find one non-zero element in S.

Do you just sub in random values for x1,x3,x4 (also where's x2)?

Originally Posted by Flop21
S = {x is an element of R^4 | (x1)-5(x3) = 2(x4)}

Find one non-zero element in S.

Do you just sub in random values for x1,x3,x4 (also where's x2)?
x=(2,0,0,1) x2 can be whatever u want doesnt affect the eqn

Don't mind if this question isn't finished off for me but can someone please set it up? Finding it hard to set everything out neatly here.

$Show that U=\{A\textbf{x}:\textbf{x}\in \mathbb{R}^5\} is a subspace of \mathbb{R}^3\\ where A is a fixed 3\times 5 matrix.$

Originally Posted by leehuan
Don't mind if this question isn't finished off for me but can someone please set it up? Finding it hard to set everything out neatly here.

$Show that U=\{A\textbf{x}:\textbf{x}\in \mathbb{R}^5\} is a subspace of \mathbb{R}^3\\ where A is a fixed 3\times 5 matrix.$
Is there missing detail like Ax = 0 or Ax is an element of ___

Originally Posted by HeroicPandas
Is there missing detail like Ax = 0 or Ax is an element of ___
Nope, and I know it can't be Ax=0 because that's a system of linear equations.

Pretty sure by assumption Ax is an element of R3 if the whole set is a subspace of R3

Originally Posted by leehuan
Nope, and I know it can't be Ax=0 because that's a system of linear equations.

Pretty sure by assumption Ax is an element of R3 if the whole set is a subspace of R3
Woops, I did a mistake in dimension and thought that Ax is still 3x5 matrix

Originally Posted by leehuan
Don't mind if this question isn't finished off for me but can someone please set it up? Finding it hard to set everything out neatly here.

$Show that U=\{A\textbf{x}:\textbf{x}\in \mathbb{R}^5\} is a subspace of \mathbb{R}^3\\ where A is a fixed 3\times 5 matrix.$
i forgot the proper proof for these

U is subset of r3 which is known v.s <- not sure if u need to include this for first year, but im used to writing it anyway..

0 vector is obvious
Ax+Ay=A(x+y) y element of U etc i.e A(0)+A(0)=A(0+0) = 0
A*(Landa)(x)=(Landa) Ax landa is real, can be complex if u want A*Landa(0)-Landa A(0) =0

Originally Posted by leehuan
Don't mind if this question isn't finished off for me but can someone please set it up? Finding it hard to set everything out neatly here.

$Show that U=\{A\textbf{x}:\textbf{x}\in \mathbb{R}^5\} is a subspace of \mathbb{R}^3\\ where A is a fixed 3\times 5 matrix.$
$\noindent In fact for \emph{any} m-by-n matrix A, that set (with \mathbf{x} coming from \mathbb{F}^{n}, where \mathbb{F} is the field of scalars) is a subspace of \mathbb{F}^{m}. This subspace is the \textsl{image} of A (alternatively called \textsl{range} or \textsl{column space}).$

$\noindent Basically to show it's a subspace, you have to show it's non-empty (or show it contains the zero vector), closed under addition, and closed under scalar multiplication. Use matrix arithmetic properties outlined in RenegadeMx's post to do this.$

$\noindent E.g. To show it's closed under addition, let \bold{y}_1, \bold{y}_2 \in U. Then by definition \bold{y}_i = A \bold{x}_i for some \bold{x}_i \in \mathbb{F}^{n} (i=1,2). Then \bold{y}_1 + \bold{y}_2 = A\bold{x}_1 + A\bold{x}_2 = A\left(\bold{x}_1 + \bold{x}_2\right), which is in U since \bold{x}_1 + \bold{x}_2 \in \mathbb{F}^{n}. Hence U is closed under addition.$

In other words, to show a set S is closed under addition, it means we have to show that if a and b are arbitrary elements of S, then so is a+b.

To show the set U is closed under scalar multiplication, it means we have to show that if c is an arbitrary scalar and v an arbitrary vector from U, then cv is also in U.

heres another one where u need to think more carefully compared to the easy stuff like 3x1+2x3=0 (if u want the practice)

For this polynomial question, I get the same matrix but R1 and R3 swapped. Is this still okay, or is my method wrong?

Originally Posted by Flop21
For this polynomial question, I get the same matrix but R1 and R3 swapped. Is this still okay, or is my method wrong?

As long as you have the same rows, you'll end up at the same answer (because then you're starting off with the same set of simultaneous equations, just written in a different order, which clearly won't affect the solutions).

heres another one where u need to think more carefully compared to the easy stuff like 3x1+2x3=0 (if u want the practice)

I'll go for it if I get the time

Anyway

$Show that the set S=\{p\in \mathbb{P}_2: p(0)=1\} is not a subspace of \mathbb{P}_2$

Is this just because the zero polynomial can't be in S?

Originally Posted by leehuan

Anyway

$Show that the set S=\{p\in \mathbb{P}_2: p(0)=1\} is not a subspace of \mathbb{P}_2$

Is this just because the zero polynomial can't be in S?
yes

to prove its not a v.s in general u either show 0 isnt there (easiest) or that it doesnt satisfy addition/scalar multiplication

only need to show 1 of those conditions as well

yes
Lol sweet.
_______________________

Even if there IS a typo in the notation, this question is actually confusing me to the maximum...

This was the answer I got to part (ii)

$z^2-2\left(\cos \frac{2\pi}{5}+\cos \frac{4\pi}{5}\right)z+2\left(\cos \frac{2\pi}{5}+\cos \frac{6\pi}{5}\right)$

to prove its not a v.s in general u either show 0 isnt there (easiest) or that it doesnt satisfy addition/scalar multiplication

only need to show 1 of those conditions as well
Yeah sweet, that reaffirms what I know

Originally Posted by leehuan
Lol sweet.
_______________________

Even if there IS a typo in the notation, this question is actually confusing me to the maximum...

This was the answer I got to part (ii)

$z^2-2\left(\cos \frac{2\pi}{5}+\cos \frac{4\pi}{5}\right)z+2\left(\cos \frac{2\pi}{5}+\cos \frac{6\pi}{5}\right)$
$z^2 -(\alpha + \alpha^4 + \alpha^2 + \alpha^3)z + (\alpha^3 + \alpha^2 + \alpha + \alpha^4)$

You forgot that α⁵=1

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