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Matrices

How would I find

1. A real 2x2 matrix AA that satisfies A2=−IA^2=-I where II is the identity matrix?

2. 2x2 matrices A,BA, B that satisfy AB=−BAAB=-BA ?
Reply 1
Avatar for Slumpy
Slumpy
16
1-Write A as (a b)
(c d).
Solve the simultaneous equations that arise.
Reply 2
Avatar for Pheylan
Pheylan
TheEd
A 2x2 matrix AA that satisfies A2=−IA^2=-I where II is the identity matrix?

(abcd)(abcd)=(−100−1)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix} = \begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}
Reply 3
Avatar for TheEd
TheEd
OP
Pheylan
(abcd)(abcd)=(−100−1)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix} = \begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}


I forgot to mention it has to be REAL, not imaginary.
Reply 4
Avatar for Slumpy
Slumpy
16
TheEd
I forgot to mention it has to be REAL, not imaginary.


You'll discover conditions on the matrix from doing this, which should lead you towards the answer.
Reply 5
Avatar for tgodkin
tgodkin
4
If we are told

(abcd)(abcd)=−(1001)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}=-\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

Then we have:

a2+bc=−1a^2+bc=-1 (1)
ab+bd=0ab+bd=0 (2)
ac+cd=0ac+cd=0 (3)
bc+d2=−1bc+d^2=-1 (4)

From (1) and (4) we have bc+d2=a2+bcbc+d^2 = a^2+bc
Or, a=da = d
Then, from (2), ab=0ab=0

So, for real solutions, b=±1b=±1 and c=±1c=±1 where c does not equal b. That is to say our solutions are:

A=(0−110)A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
and
A=(01−10)A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

We also have two complex solutions A=(±i00±i)A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a=b=±ia = b = ±i
Reply 6
Avatar for TheEd
TheEd
OP
tgodkin
If we are told

(abcd)(abcd)=−(1001)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}=-\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

Then we have:

a2+bc=−1a^2+bc=-1
ab+bd=0ab+bd=0
ac+cd=0ac+cd=0
bc+d2=−1bc+d^2=-1


OK cheers I'll have a go at solving them.
Reply 7
Avatar for tgodkin
tgodkin
4
For your second problem, to get you on the right track:

AB=-BA is saying…

(abcd)(efgh)=−(efgh)(abcd)\begin{pmatrix} a & b \\c & d\end{pmatrix}\begin{pmatrix} e & f \\g & h\end{pmatrix}=-\begin{pmatrix} e & f \\g & h\end{pmatrix}\begin{pmatrix} a & b \\c & d\end{pmatrix}
Reply 8
Avatar for TheEd
TheEd
OP
tgodkin
If we are told

(abcd)(abcd)=−(1001)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}=-\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

Then we have:

a2+bc=−1a^2+bc=-1 (1)
ab+bd=0ab+bd=0 (2)
ac+cd=0ac+cd=0 (3)
bc+d2=−1bc+d^2=-1 (4)

From (1) and (4) we have bc+d2=a2+bcbc+d^2 = a^2+bc
Or, a=da = d
Then, from (2), ab=0ab=0

So, for real solutions, b=±1b=±1 and c=±1c=±1 where c does not equal b. That is to say our solutions are:

A=(0−110)A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
and
A=(01−10)A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

We also have two complex solutions A=(±i00±i)A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a=b=±ia = b = ±i


Can you please explain how you arrived at those real solutions from solving the simultaneous equations?
Reply 9
Avatar for Slumpy
Slumpy
16
tgodkin
If we are told

(abcd)(abcd)=−(1001)\begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix}=-\begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}

Then we have:

a2+bc=−1a^2+bc=-1 (1)
ab+bd=0ab+bd=0 (2)
ac+cd=0ac+cd=0 (3)
bc+d2=−1bc+d^2=-1 (4)

From (1) and (4) we have bc+d2=a2+bcbc+d^2 = a^2+bc
Or, a=da = d
Then, from (2), ab=0ab=0

So, for real solutions, b=±1b=±1 and c=±1c=±1 where c does not equal b. That is to say our solutions are:

A=(0−110)A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
and
A=(01−10)A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

We also have two complex solutions A=(±i00±i)A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a=b=±ia = b = ±i


There's an infinite number of solutions with real coefficients, you've missed a fair few.

a^2=d^2 doesn't imply a=d.
Reply 10
Avatar for tgodkin
tgodkin
4
TheEd
Can you please explain how you arrived at those real solutions from solving the simultaneous equations?

If b or c were to be 0, it is clear from (1) than a is a complex number.
So, let us assume a=0a = 0 is a solution, then bc=−1bc=-1.

Two answers satisfy this equation which are:
a=±1a=±1
b=±1b=±1 where b=−ab = -a
Reply 11
Avatar for tgodkin
tgodkin
4
Slumpy
There's an infinite number of solutions with real coefficients, you've missed a fair few.

a^2=d^2 doesn't imply a=d.


True I overlooked that. If we only consider real solutions then a2=b2a^2=b^2 could also imply a=−ba=-b
This however doesn't change the values of our real solutions.

But yes I can see there would be many complex solutions, thanks.
Reply 12
Avatar for Slumpy
Slumpy
16
tgodkin
True I overlooked that. If we only consider real solutions then a2=b2a^2=b^2 could also imply a=−ba=-b
This however doesn't change the values of our real solutions.

But yes I can see there would be many complex solutions, thanks.


It really does, there are, as I say, an infinite number of them.
Pick a=-d.
Now pick bc such that b=-c, and |bc|=a^2+1.
Reply 13
Avatar for tgodkin
tgodkin
4
Slumpy
It really does, there are, as I say, an infinite number of them.
Pick a=-d.
Now pick bc such that b=-c, and |bc|=a^2+1.

Yeah I see that now, thanks.
Reply 14
Avatar for TheEd
TheEd
OP
tgodkin
For your second problem, to get you on the right track:

AB=-BA is saying…

A=(abcd)(efgh)=−(efgh)(abcd)A=\begin{pmatrix} a & b \\c & d\end{pmatrix}\begin{pmatrix} e & f \\g & h\end{pmatrix}=-\begin{pmatrix} e & f \\g & h\end{pmatrix}\begin{pmatrix} a & b \\c & d\end{pmatrix}


Well I've found 2 solutions for this question

A=(0−110),B=(100−1)A=\begin{pmatrix} 0 & -1 \\1 & 0\end{pmatrix}, B=\begin{pmatrix} 1 & 0 \\0 & -1\end{pmatrix}

and

A=(0−110),B=(−1001)A=\begin{pmatrix} 0 & -1 \\1 & 0\end{pmatrix}, B=\begin{pmatrix} -1 & 0 \\0 & 1\end{pmatrix}

but that was pure guess work! I tried it with simultaneous equations but how can you do it? They're all unknowns.

I got the simultaneous equations down to ae=dh but couldn't get any further.
For (2), A = B = zero matrix will work.

Also, it may help to think of (1) geometrically. Let A represent a rotation, for example.

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