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Thread: 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 $\displaystyle (a)\Rightarrow (b)$

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

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

    $\displaystyle A^{-1}Ax_0=A^{-1}0$

    $\displaystyle Ix_0=I0$

    $\displaystyle x_0=0$



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

    $\displaystyle Ax_0=0$

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

    $\displaystyle \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)$$\displaystyle \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\ right)$=$\displaystyle \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 $\displaystyle x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri ght)$, 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 $\displaystyle (a)\Rightarrow (b)$

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

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

    $\displaystyle A^{-1}Ax_0=A^{-1}0$

    $\displaystyle Ix_0=I0$

    $\displaystyle x_0=0$



    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 $\displaystyle A^{-1}$ is:

    $\displaystyle Ax_0=0$

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

    $\displaystyle \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)$$\displaystyle \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\ right)$=$\displaystyle \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 $\displaystyle x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri ght)$, but this isn't ANY solution. Can anyone help clarify this?
    we need to check all of the solution to the system $\displaystyle Ax = 0$
    at the end, we find that the only solution is$\displaystyle x = 0$
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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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