Matrices

1)Is AB (λ µ)-nice?#

2)Describe the set of 1-nice matrix.

For part 1 would I use proof by induction?

(edited 7 years ago)

(1) Should be direct proofs and/or counterexamples.

(2) I've had a quick play. If you start by putting a, b, c, d in the top 2 x 2 corner and then fill in what the other values must be, you end up with a condition on d. At that point, you have every 1-nice matrix described by just a, b, and c. Then see if there's a nice way to describe this.

You might find it easier to look at 0-nice matrices and then use your previous result to relate 0-nice matrices to 1-nice ones, but I haven't played with that.

Reply 2

ghostwalker

Original post by DFranklin

(2) I've had a quick play. If you start by putting a, b, c, d in the top 2 x 2 corner and then fill in what the other values must be, you end up with a condition on d.

I didn't get any restriction on d. However, I'm about to go out until tomorrow, so don't have time to pursue this at present. I'll find out what I missed them.

Reply 3

DFranklin

Original post by ghostwalker

I didn't get any restriction on d. However, I'm about to go out until tomorrow, so don't have time to pursue this at present. I'll find out what I missed them.

I didn't spend much time on it either - might be my mistake. With six conditions on the coefficients of the matrix I think you'd expect only 3 free variables at the end though.

Reply 4

DFranklin

My inclination would be to rewrite what it means for A to be

\lambda

nice in terms of summations. That is, write down a formula for the sum of the kth row of A, and the kth column (and then use that these sums must = lambda).

Once you've done this, the actual proof will be trivial.

Reply 5

DFranklin

Original post by mollyjordansmith

Thank you, is the AB is (lambda x mu)- nice part false? I'm trying to prove it the same way as the addition one and seem to be failing?

My feeling is that it's false but I haven't actually tried at all and I could be wrong. Muck about with some examples and see what you get.

Reply 6

ghostwalker

Original post by mollyjordansmith

Is AB (λ µ)-nice?#

Surprisingly, it appears to be true, so I'd try for a proof.
If C=AB, then you need to show the sum of a row/column in C is equal to

\lambda\mu

. The trick is in rearranging the summation where you express the entries in C in terms of those in A and B.

Original post by DFranklin

With six conditions on the coefficients of the matrix I think you'd expect only 3 free variables at the end though.

Agreed, though in this case two of the constraints work out to be identical.

(edited 7 years ago)

Reply 7

Sam00

Are you trying to find the eigenvectors and eigenvalues?

Reply 8

DFranklin

Original post by Sam00

Are you trying to find the eigenvectors and eigenvalues?

No.

Reply 9

ghostwalker

Original post by mollyjordansmith

Am I along the right lines?

I don't follow your reasoning for the line I've highlighted in red.

The most direct approach I can see is to just flip the order of summation to start.

Edit: It only requires a couple of lines after that.

(edited 7 years ago)

Reply 10

ghostwalker

Original post by mollyjordansmith

Like this

Yes. Make sure you're happy with why you can do that.

Then notice that

b_{mk}

does not depend on i, so can be pulled out of the inner summation.

Reply 11

ghostwalker

Original post by mollyjordansmith

Thank you, I've never done summation stuff before hence why I'm struggling

No problem. Nested summations can throw people, even if you're used to a single summation. Can you finish from there?

With nested summations with fixed limits, you can think of the entries in a rectangular grid or array. With the summation one way round you're summing along the row, and then adding those sums. The other way round you're summing down the columns and then adding those sums. In both cases you just adding all the entries in the table, hence they're equal.

Reply 12

ghostwalker

Original post by mollyjordansmith

I'm not sure:
does the summation of bmk now equal mu?
and how can does aim become lambda as there is no k?

I think I'm going to have to do this in full detail, as I'm not sure of where you're coming from with the first question, and don't want to just give a Yes/No answer, in case your understanding is different to what I'm assuming.

So, from the rearrangement:

\displaystyle\sum_{m=1}^n\sum_{i=1}^na_{im}b_{mk}

If we put in parentheses to clarify, this actually means:

\displaystyle\sum_{m=1}^n\left( \sum_{i=1}^na_{im}b_{mk}\right)

Now

b_{mk}

does not depend on i, so we can pull it out of the inner summation, which is a summation over i, thus:

\displaystyle\sum_{m=1}^n\left(b_{mk}\sum_{i=1}^na_{im}\right)

Now we can work on the inner summation.

a_{im}

is being summed for a fixed m, and with i varying form 1 to n, so we are summing column m of the matrix A. That comes to

\lambda

So, we now have:

\displaystyle\sum_{m=1}^n\left(b_{mk}\lambda\right)

Now lambda is a constant. It doesn't depend on m, so we can pull it out in front of the summation, thus:

\displaystyle=\lambda \sum_{m=1}^nb_{mk}

Here we are summing the entries in column k of the matrix B, which comes to

\mu

And so this equals

\lambda\mu

.

Now, you need to do something similar for the rows of C.

Reply 13

ghostwalker

Original post by mollyjordansmith

I'm struggling to see what you mean by what the other values must be.

So, a11=a, a12=b, a21=c, a22=d.

It would help to draw the matrix out.

Since your matrix is 1-nice, what must a31 equal for the row to sum to 1?

And so on.

Reply 14

ghostwalker

Original post by mollyjordansmith

So would it be

a b 1-a-b
c d 1-c-d
1-a-c 1-b-d a+b+c+d-1

Yep.

Reply 15

ghostwalker

Original post by mollyjordansmith

I'm really sorry I'm struggling to see what to do next.

What are you currently working on in your course that's brought up this question on "nice" matrices?

Reply 16

ghostwalker

Original post by mollyjordansmith

We have been doing linear systems and matrices

I was hoping for something a bit more specific.

However, if it's just general, then I'd just leave the description with describing the set in terms of the elements, in standard set notation.

Reply 17

ghostwalker

Original post by mollyjordansmith

I've been told to use these equstions to create a linear system to solve but I dont
see how that works.

Based on what you've put in your first post, the only way you could create a linear system, that I can see, is in 2),where you've done the work already.
Set up the row and column constraints as linear equations, but it's rather making a mountain out of a molehill.

E.g.

a_{11}+a_{12}+a_{13}=1

Etc. and solve.

If you were meant to do it that way, it would have helped to have known from the first post.

Reply 18

ghostwalker

Original post by mollyjordansmith

How do you solve that system?

I'd put it into an augmented matrix, and use Gaussian elimination. But, I imagine, this is what your course is all about and something you should have covered. I am somewhat surprised at your question.

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