Need help, URGENT maths question: (6 Viewers)

1008 · May 24, 2016

InteGrand said:
$\noindent The second line there is invalid (that'd be saying the sum of product of terms is equal to the product of the sums, which is not true, e.g. $1\times 5 + 2\times 6 \neq (1+2)\times (5+6)$). It's rather simple in fact though, here's how to do it.$

$\noindent Let $A$ have entries denoted $a_{ij}$, and similarly $b_{ij}$ for $B$ ($A$ an $m \times n$ and $B$ an $n \times p$ matrix over some field, e.g. the reals or complex numbers). Let $\lambda$ be a scalar from this field. Then we have for any $i$ from $1$ to $m$ and $j$ from $1$ to $p$ that the $i-j$ entry of $AB$ is$

$\begin{align*}\left[A\left(\lambda B\right)\right] _{ij} &= \sum _{k=1}^{n} a_{ik}\left(\lambda b_{kj}\right) \quad (\text{since the }i-j \text{ entry of }\lambda B\text{ is }\lambda b_{jk} \text{ by definition of scalar multiplication of a matrix}) \\ &= \sum _{k=1}^{n} \left(\lambda a_{ik}\right) b_{kj}= \left[\left(\lambda A\right)B\right]_{ij} \quad (\text{due to associativity and commutative of multiplication in a field}) \\ &= \lambda \sum _{k=1}^{n} a_{ik} b_{kj}= \left[\left(\lambda A\right)B\right]_{ij} \quad (\text{due to the distributive law and associative law of multiplication in a field (the latter lets us firstly write }\left(\lambda a_{ik}\right) b_{kj} = \lambda a_{ik} b_{kj} \text{ and the former then lets us take the }\lambda \text{ outside the sum.} ) . \end{align*}$

In other words, the first part of the question follows due to the familiar properties of arithmetic in fields like the real/complex numbers.

(Also, your sums have been referring to the i-j components of matrices rather than the matrices themselves, so it wouldn't be right to say A(lambda B) equals those, for instance.)

Thanks! Yeah, just realised that I violated order of operations lol. x I'll get back to figuring out the second part of (1). Would this be right for (3) though?

EDIT: the first step is meant to have brackets around the whole thing, i.e.

InteGrand · May 24, 2016

1008 said:
Thanks! Yeah, just realised that I violated order of operations lol. x I'll get back to figuring out the second part of (1). Would this be right for (3) though?

EDIT: the first step is meant to have brackets around the whole thing, i.e.

Yeah that's correct, except make sure to put []_ij around the matrices, since those sums aren't equal to matrices, they're equal to entries of matrices. So what you've shown is that the ij entry of A(B+C) is equal to the ij entry of (AB + AC), so the matrices are equal since all their entries are equal.

(Actually, you showed that the ij entry of A(B+C) equals [AB]_ij + [AC]_ij, but from the definition of matrix addition, this is just [AB+AC]_ij.)

1008 · May 24, 2016

InteGrand said:
Yeah that's correct, except make sure to put []_ij around the matrices, since those sums aren't equal to matrices, they're equal to entries of matrices. So what you've shown is that the ij entry of A(B+C) is equal to the ij entry of (AB + AC), so the matrices are equal since all their entries are equal.

(Actually, you showed that the ij entry of A(B+C) equals [AB]_ij + [AC]_ij, but from the definition of matrix addition, this is just [AB+AC]_ij.)

Right, is this better?

Also, do I need to give reasons at each step like you do?

InteGrand · May 24, 2016

1008 said:
Right, is this better?

Also, do I need to give reasons at each step like you do?

I'm not sure about the necessity of giving reasons for each step (I mainly gave them for your own benefit). But it would probably be good to mention why the last line at least is true (reason being that the second last line is [AB]_ij + [AC]_ij, which is [AB + AC]_ij by definition of matrix addition).

1008 · May 24, 2016

InteGrand said:
I'm not sure about the necessity of giving reasons for each step (I mainly gave them for your own benefit). But it would probably be good to mention why the last line at least is true (reason being that the second last line is [AB]_ij + [AC]_ij, which is [AB + AC]_ij by definition of matrix addition).

Do you mean
"[AC]_ij + [BC]_ij, which is [AC + BC]_ij" or am I just hallucinating?

InteGrand · May 24, 2016

1008 said:
Do you mean
"[AC]_ij + [BC]_ij, which is [AC + BC]_ij" or am I just hallucinating?

Oh yeah you're right, sorry, forgot the Q. was right distributivity rather than left distributivity.

1008 · May 24, 2016

InteGrand said:
Oh yeah you're right, sorry, forgot the Q. was right distributivity rather than left distributivity.

Wait, I thought this was left distributivity lol

InteGrand · May 24, 2016

1008 said:
Wait, I thought this was left distributivity lol

It's considered right distributivity (because the thing we 'distribute' over the brackets is on the right, I guess).

You can see the definitions here for reference: https://en.wikipedia.org/wiki/Distributive_property#Definition .

1008 · May 24, 2016

InteGrand said:
It's considered right distributivity (because the thing we 'distribute' over the brackets is on the right, I guess).

You can see the definitions here for reference: https://en.wikipedia.org/wiki/Distributive_property#Definition .

Wow this contradicts with the UNSW notes, which calls this the left distributive law, and the other one the Right distributive law???

InteGrand · May 24, 2016

1008 said:
Wow this contradicts with the UNSW notes, which calls this the left distributive law, and the other one the Right distributive law???

I don't know what those notes say, but if they call them the other way round, then yes, there is a contradiction. You can search up online for other references to left- and right- distributivity. For example, here is a page on Proof Wiki where they prove a left distributivity thing, and you can see it matches the Wikipedia definition: https://proofwiki.org/wiki/Left_Distributive_Law_for_Natural_Numbers .

Wolfram says the same thing: http://mathworld.wolfram.com/Distributive.html .

1008 · May 24, 2016

InteGrand said:
I don't know what those notes say, but if they call them the other way round, then yes, there is a contradiction. You can search up online for other references to left- and right- distributivity. For example, here is a page on Proof Wiki where they prove a left distributivity thing, and you can see it matches the Wikipedia definition: https://proofwiki.org/wiki/Left_Distributive_Law_for_Natural_Numbers .

Wolfram says the same thing: http://mathworld.wolfram.com/Distributive.html .

Right thanks, I'll have to clarify this with my lecturer lol. Anyway, for (4), here's my working, if this is right?

Don't get how to do (3) tho...Can you still do it with this notation and how would you generalise that to A of any size and the square matrix I of any size?

Just for reference:
(3) prove AI = A and IA = A, where I represents identity matrices of the appropriate (possibly different) sizes

InteGrand · May 24, 2016

1008 said:
Right thanks, I'll have to clarify this with my lecturer lol. Anyway, for (4), here's my working, if this is right?

Don't get how to do (3) tho...Can you still do it with this notation and how would you generalise that to A of any size and the square matrix I of any size?

Just for reference:
(3) prove AI = A and IA = A, where I represents identity matrices of the appropriate (possibly different) sizes

Yeah, that working is essentially correct.

$\noindent And to show $AI = A = IA$, recall that $\left[I\right]_{ij}=\delta _{ij}$, where $\delta _{ij}$ is the \emph{Kronecker delta}, which is equal to $0$ if $i\neq j$ and $1$ if $i=j$ (if we think about it a bit, this is just an algebraic way of saying that the identity matrix has element 1 on the diagonal and 0 off the diagonal).$

$\noindent Now, let $A$ be $m\times n$. Then for any $i$-$j$ element of $AI_{n}$, we have$

$\left[AI\right]_{ij} &= \sum _{k=1}^{n} a_{ik}\delta _{kj} \\ &= a_{ij}\times 1 \quad (\text{because the Kronecker delta is 0 for all terms except the term where the summation index hits the value }j, \text{ when it is equal to 1}) \\ &= a_{ij} \\ \Rightarrow AI &= A.$

$\noindent The proof that $I_m A = A$ is essentially the same. Just note that in that case, the summation will range from $k=1$ to $m$, rather than $n$. I.e. the sum would be $\sum _{k=1}^{m}\delta _{ik} a_{kj} $. (And in this case, it's the $k=i$ term that remains, with all others being 0.)$

(And thanks for reposting the Q. here – saved me from having to go back and find it.

)

InteGrand · May 24, 2016

You can read up about the Kronecker delta here: https://en.wikipedia.org/wiki/Kronecker_delta .

1008 · May 24, 2016

InteGrand said:
You can read up about the Kronecker delta here: https://en.wikipedia.org/wiki/Kronecker_delta .

Yeah, never heard that before haha. But thanks, it makes sense.

And just for greater conceptual understanding, is there another way to prove it? Any hints?

If I knew about the Kronecker delta, I'd be able to solve so many problems more quickly omg

InteGrand · May 25, 2016

1008 said:
Yeah, never heard that before haha. But thanks, it makes sense.

And just for greater conceptual understanding, is there another way to prove it? Any hints?

If I knew about the Kronecker delta, I'd be able to solve so many problems more quickly omg

Yeah, there are other ways of proving it. It's easiest to just think about it intuitively I guess. The i-j entry of AI is the i-th row of A dotted with the j-th column of I. The j-th column of I is just the j-th standard basis vector e_j for Rⁿ (or Cⁿ in complex case) and dotting any vector with the j-th standard basis vector just returns the j-th entry of that vector. Hence

ij entry of AI = i-th row of A dotted with j-th column of I
= j-th entry of (i-th row vector of A)
= a_ij (essentially by definition of a_ij),

so the ij entry of AI = ij entry of A, so AI = A.

1008 · May 25, 2016

InteGrand said:
Yeah, there are other ways of proving it. It's easiest to just think about it intuitively I guess. The i-j entry of AI is the i-th row of A dotted with the j-th column of I. The j-th column of I is just the j-th standard basis vector e_j for Rⁿ (or Cⁿ in complex case) and dotting any vector with the j-th standard basis vector just returns the j-th entry of that vector. Hence

ij entry of AI = i-th row of A dotted with j-th column of I
= j-th entry of (i-th row vector of A)
= a_ij (essentially by definition of a_ij),

so the ij entry of AI = ij entry of A, so AI = A.

Yeah, that makes sense too. Just a question, how do you go about studying all these topics? Do you have a textbook you prefer, or is it more like you go online to clarify concepts? Or is it a website you go to?

InteGrand · May 25, 2016

1008 said:
Yeah, that makes sense too. Just a question, how do you go about studying all these topics? Do you have a textbook you prefer, or is it more like you go online to clarify concepts? Or is it a website you go to?

I guess textbooks, online resources / YouTube videos, and practice are good ways to study these types of things (as well as spending time thinking about them, mentally asking Q's about the topics and trying to answer them or investigating them, etc.).

1008 · May 25, 2016

InteGrand said:
I guess textbooks, online resources / YouTube videos, and practice are good ways to study these types of things (as well as spending time thinking about them, mentally asking Q's about the topics and trying to answer them or investigating them, etc.).

Do you have any textbook/resources in mind for vectors and matrices?

And how would you do this?

leehuan · May 25, 2016

1008 said:
Do you have any textbook/resources in mind for vectors and matrices?

And how would you do this?

For the first part just evaluate the determinant down the third column by brute force. The determinants of the minors (2x2 matrices) are actually really tidy.

I think InteGrand provided a proof to part b) somewhere.

(Sorry about last night... I ended up going to sleep!!!)

1008 · May 25, 2016

leehuan said:
For the first part just evaluate the determinant down the third column by brute force. The determinants of the minors (2x2 matrices) are actually really tidy.

I think InteGrand provided a proof to part b) somewhere.

(Sorry about last night... I ended up going to sleep!!!)

That's alright

and thanks. I guess the main trick is knowing where to get your determinant from in this question. Btw, do you still have your working for that particular question (proving all those properties of matrices)? Just wanna compare my working to yours...

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