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Math Help - Help Understanding a Proof

  1. #1
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    Help Understanding a Proof

    I'm having trouble understanding the proof of Theorem 1 on this page:

    Pauls Online Notes : Linear Algebra - Finding Inverse Matrices

    I don't understand how they proved the first conditional (a)\Rightarrow (b)

    If A in invertible, then this means that A^{-1} exists and that AA^{-1}=I=A^{-1}A.

    It seems that they assumed that AX=0, selected a solution x_0 to write Ax_0=0, and then mulitplied both sides by A^{-1} so that

    A^{-1}Ax_0=A^{-1}0

    Ix_0=I0

    x_0=0



    I don't understand the logic here. It just looks like circular reasoning to me. Here is a quote from the proof:

    We start by assuming that x_0 is any solution to the system. Plug this into the system and then mulitply both sides by A^{-1}
    So it seems that the equation that they are multiplying by A^{-1} is:

    Ax_0=0

    But where does that come from? How is it that substituting x_0 into AX gives Ax_0=0. If I look at it this way:

    \left(\begin{array}{ccc}a_{11}&a_{12}&...a_{1n}\\a  _{21}&a_{22}&...a_{2n}\\.&.&.\\a_{n1}&a_{n2}&...a_  {nn}\end{array}\right) \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\  right)= \left(\begin{array}{c}b_1\\b_2\\.\\b_n\end{array}\  right)

    To get the right side to be equal to zero, they must have taken x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri  ght), but this isn't ANY solution. Can anyone help clarify this?
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  2. #2
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    Smile

    Quote Originally Posted by adkinsjr View Post
    I'm having trouble understanding the proof of Theorem 1 on this page:

    Pauls Online Notes : Linear Algebra - Finding Inverse Matrices

    I don't understand how they proved the first conditional (a)\Rightarrow (b)

    If A in invertible, then this means that A^{-1} exists and that AA^{-1}=I=A^{-1}A.

    It seems that they assumed that AX=0, selected a solution x_0 to write Ax_0=0, and then mulitplied both sides by A^{-1} so that

    A^{-1}Ax_0=A^{-1}0

    Ix_0=I0

    x_0=0



    I don't understand the logic here. It just looks like circular reasoning to me. Here is a quote from the proof:



    So it seems that the equation that they are multiplying by A^{-1} is:

    Ax_0=0

    But where does that come from? How is it that substituting x_0 into AX gives Ax_0=0. If I look at it this way:

    \left(\begin{array}{ccc}a_{11}&a_{12}&...a_{1n}\\a  _{21}&a_{22}&...a_{2n}\\.&.&.\\a_{n1}&a_{n2}&...a_  {nn}\end{array}\right) \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\  right)= \left(\begin{array}{c}b_1\\b_2\\.\\b_n\end{array}\  right)

    To get the right side to be equal to zero, they must have taken x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri  ght), but this isn't ANY solution. Can anyone help clarify this?
    we need to check all of the solution to the system Ax = 0
    at the end, we find that the only solution is  x = 0
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  3. #3
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    I think I understand now. We are just trying to prove that the solution to the system is 0. I was just reading the theorem wrong. I was thinking the we were trying to prove that if A is invertible then Ax=0, which didn't make any sense to me.
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