RuneFreeze
Badges: 16
Rep:
?
#1
Report Thread starter 1 year ago
#1
Name:  Capture.JPG
Views: 46
Size:  22.4 KB

I can sort of do it by trial and error but I'm struggling to find a better and more systematic approach.

All I've done is write the Matrix A in terms of a, b, c, d and write out 4 expressions equal to zero.
0
reply
RuneFreeze
Badges: 16
Rep:
?
#2
Report Thread starter 1 year ago
#2
bump
0
reply
Louisb19
Badges: 14
Rep:
?
#3
Report 1 year ago
#3
If A^2 = 0 then the eigen-values of A must all be 0 (aka characteristic poly x^2 = 0). Maybe this will help you.
0
reply
Deranging
Badges: 9
Rep:
?
#4
Report 1 year ago
#4
(Original post by RuneFreeze)
Name:  Capture.JPG
Views: 46
Size:  22.4 KB

I can sort of do it by trial and error but I'm struggling to find a better and more systematic approach.

All I've done is write the Matrix A in terms of a, b, c, d and write out 4 expressions equal to zero.
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.

Anyway, I am not sure there is a particularly better way of doing it. Let your matrix be
 A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) .
Now because we have some choice of scale here and all the entries are non-zero, we can choose  a = 1 . Then if you calculate  A^2 you get some simultaneous equations.
0
reply
RuneFreeze
Badges: 16
Rep:
?
#5
Report Thread starter 1 year ago
#5
(Original post by Louisb19)
If A^2 = 0 then the eigen-values of A must all be 0 (aka characteristic poly x^2 = 0). Maybe this will help you.
Hmm I haven't (or won't) do eigenvalues. This is only the second chapter I've done on matrices (matrix multiplication) so it should just require very basic matrices stuff.
0
reply
Louisb19
Badges: 14
Rep:
?
#6
Report 1 year ago
#6
name the entries a,b,c,d then eigenvals satisfy

x^2 - x(a+d) + (ad - bc) = 0

so a = -d and then choose b,c such that (ad - bc) = 0. Then you are all good since then eigvals are both 0.
0
reply
Louisb19
Badges: 14
Rep:
?
#7
Report 1 year ago
#7
(Original post by RuneFreeze)
Hmm I haven't (or won't) do eigenvalues. This is only the second chapter I've done on matrices (matrix multiplication) so it should just require very basic matrices stuff.
oh nevermind then.
0
reply
RuneFreeze
Badges: 16
Rep:
?
#8
Report Thread starter 1 year ago
#8
(Original post by Deranging)
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.

Anyway, I am not sure there is a particularly better way of doing it. Let your matrix be
 A = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) .
Now because we have some choice of scale here and all the entries are non-zero, we can choose  a = 1 . Then if you calculate  A^2 you get some simultaneous equations.
Ok so I've just done four simultaneous equations by multiplying

\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix}\ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

which give the simultaneous equations

a^{2} + bc = 0
b(a+d) = 0
c(a+d) = 0
d^{2} + bc = 0

From the second and third equation, when  a+d = 0
\Rightarrow  b = c = 0 \mathrm{(inadmissible)}

So we know that the solution must follow from  a+d = 0
\Rightarrow  a = -d and
a^{2} = -bc

So I have reduced the four simultaneous equations to two simpler ones, at this point I guess the only method is trial and error which is simple enough.

e.g. \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}

Excuse the superfluousness of this btw, just wanted to try and learn latex

Thanks everyone for responses
1
reply
RDKGames
Badges: 20
Rep:
?
#9
Report 1 year ago
#9
(Original post by RuneFreeze)
From the second and third equation, when  a+d = 0
\Rightarrow  b = c = 0 \mathrm{(inadmissible)}
Your working is fine, but this is not a correct statement. Just because a+d=0 it does not imply that b=c=0.
You might just be misusing the \Rightarrow symbol here, but this is just a small pick on your working.
0
reply
ghostwalker
  • Study Helper
Badges: 16
#10
Report 1 year ago
#10
(Original post by RDKGames)
Your working is fine, but this is not a correct statement. Just because a+d=0 it does not imply that b=c=0.
You might just be misusing the \Rightarrow symbol here, but this is just a small pick on your working.
From the context, I suspect it was just a typo, and they meant a+d\not=0
0
reply
RDKGames
Badges: 20
Rep:
?
#11
Report 1 year ago
#11
(Original post by ghostwalker)
From the context, I suspect it was just a typo, and they meant a+d\not=0
That would make sense. Didn't notice that they might've meant this.
0
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 1 year ago
#12
(Original post by Deranging)
This is a bit of a silly question really isn't it? Why would you do part (a) if you can just do part (b) and then repeat your answer :P.
Well, it's fairly easy to do (a) without much calculation:

Spoiler:
Show

Use the matrix that maps (x, y) to (0, x).


It's a little harder to find a more general matrix, although I agree that since you're going to have to do it, there's some truth to your argument that you should just do that and let (a) come out for free.

However, the idea that lets you write down the matrix for (a) should also let you do (b) without relatively little calculation.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 18 Dec '19
  • The University of Law
    Open Day – GDL and LPC - Chester campus Postgraduate
    Sat, 4 Jan '20
  • University of East Anglia
    Mini Open Day Undergraduate
    Mon, 6 Jan '20

Did you vote in the 2019 general election?

Yes (360)
44.94%
No (84)
10.49%
I'm not old enough (357)
44.57%

Watched Threads

View All