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    How would I find

    1. A real 2x2 matrix A that satisfies A^2=-I where I is the identity matrix?

    2. 2x2 matrices A, B that satisfy AB=-BA ?
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    1-Write A as (a b)
    (c d).
    Solve the simultaneous equations that arise.
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    (Original post by TheEd)
    A 2x2 matrix A that satisfies A^2=-I where I is the identity matrix?
    \begin{pmatrix} a & b \\c & d \end{pmatrix}\begin{pmatrix} a & b \\c & d \end{pmatrix} = \begin{pmatrix} -1 & 0 \\0 & -1 \end{pmatrix}
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    (Original post by Pheylan)
    \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.
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    (Original post by 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.
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    If we are told

    \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:

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

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

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

    A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
    and
    A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

    We also have two complex solutions A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a = b = ±i
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    (Original post by tgodkin)
    If we are told

    \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:

    a^2+bc=-1
    ab+bd=0
    ac+cd=0
    bc+d^2=-1
    OK cheers I'll have a go at solving them.
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    For your second problem, to get you on the right track:

    AB=-BA is saying…

    \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}
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    (Original post by tgodkin)
    If we are told

    \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:

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

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

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

    A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
    and
    A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

    We also have two complex solutions A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a = b = ±i
    Can you please explain how you arrived at those real solutions from solving the simultaneous equations?
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    (Original post by tgodkin)
    If we are told

    \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:

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

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

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

    A=\begin{pmatrix} 0 & -1 \\1 & 0 \end{pmatrix}
    and
    A=\begin{pmatrix} 0 & 1 \\-1 & 0 \end{pmatrix}

    We also have two complex solutions A=\begin{pmatrix} ±i & 0 \\0 & ±i \end{pmatrix} where a = 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.
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    (Original post by 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 = 0 is a solution, then bc=-1.

    Two answers satisfy this equation which are:
    a=±1
    b=±1 where b = -a
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    (Original post by 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 a^2=b^2 could also imply a=-b
    This however doesn't change the values of our real solutions.

    But yes I can see there would be many complex solutions, thanks.
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    (Original post by tgodkin)
    True I overlooked that. If we only consider real solutions then a^2=b^2 could also imply a=-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.
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    (Original post by 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.
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    (Original post by tgodkin)
    For your second problem, to get you on the right track:

    AB=-BA is saying…

    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=\begin{pmatrix} 0 & -1 \\1 & 0\end{pmatrix}, B=\begin{pmatrix} 1 & 0 \\0 & -1\end{pmatrix}

    and

    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.
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    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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