Help Understanding a Proof

**adkinsjr** · December 23rd 2009, 07:37 PM

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?

**dedust** · December 23rd 2009, 07:53 PM

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 $(a)\Rightarrow (b)$

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 $x = 0$

**adkinsjr** · December 23rd 2009, 08:37 PM

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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