# Help Understanding a Proof

• Dec 23rd 2009, 07:37 PM
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

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

$\displaystyle Ix_0=I0$

$\displaystyle x_0=0$

(Thinking)

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

Quote:

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)$$\displaystyle \left(\begin{array}{c}x_1\\x_2\\.\\x_n\end{array}\ right)=\displaystyle \left(\begin{array}{c}b_1\\b_2\\.\\b_n\end{array}\ right) To get the right side to be equal to zero, they must have taken \displaystyle x_0=\left(\begin{array}{c}0\\0\\.\\0\end{array}\ri ght), but this isn't ANY solution. Can anyone help clarify this? • Dec 23rd 2009, 07:53 PM dedust Quote: Originally Posted by adkinsjr I'm having trouble understanding the proof of Theorem 1 on this page: Pauls Online Notes : Linear Algebra - Finding Inverse Matrices I don't understand how they proved the first conditional \displaystyle (a)\Rightarrow (b) If A in invertible, then this means that \displaystyle A^{-1} exists and that \displaystyle AA^{-1}=I=A^{-1}A. It seems that they assumed that \displaystyle AX=0, selected a solution \displaystyle x_0 to write \displaystyle Ax_0=0, and then mulitplied both sides by \displaystyle A^{-1} so that \displaystyle A^{-1}Ax_0=A^{-1}0 \displaystyle Ix_0=I0 \displaystyle x_0=0 (Thinking) I don't understand the logic here. It just looks like circular reasoning to me. Here is a quote from the proof: So it seems that the equation that they are multiplying by \displaystyle A^{-1} is: \displaystyle Ax_0=0 But where does that come from? How is it that substituting\displaystyle x_0 into \displaystyle AX gives \displaystyle Ax_0=0. If I look at it this way: \displaystyle \left(\begin{array}{ccc}a_{11}&a_{12}&...a_{1n}\\a _{21}&a_{22}&...a_{2n}\\.&.&.\\a_{n1}&a_{n2}&...a_ {nn}\end{array}\right)$$\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$\displaystyle x = 0$
• Dec 23rd 2009, 08:37 PM