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Why is the product of elementary matrices necessarily invertible? I understand that each elementary matrix is invertible, but why is their product also invertible? Is it the indirect result of this theorem?

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    $\begingroup$ I would say "direct result" of the theorem. You could also make an argument from determinants. $\endgroup$ – B. Goddard Sep 5 '18 at 13:59
  • $\begingroup$ Try induction on the number of elementary matrices that appear as factors. The theorem you showed gives the induction step (as well as the base case if you start from two factors). $\endgroup$ – Jyrki Lahtonen Sep 6 '18 at 7:52
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Each elementary matrix $E_i$ is invertible so $\det(E_i)\ne0$.

As $$\det(E_iE_j)=\det(E_i)\det(E_j)\ne0$$ then $E_iE_j$ is invertible.

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  • $\begingroup$ I'd be interested to understand the down vote. $\endgroup$ – PM. Sep 6 '18 at 7:26
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It has nothing to do with linear algebra. It is a general fact that in any unital ring, a product of invertible elements: $a_1a_2\dots a_n$, is invertible, and its invertible, and its inverse is the product of the inverses of the factors, in reverse order. In other words: $$(a_1a_2\dots a_n)^{-1}=a_n^{-1}\dots a_2^{-1} a_1^{-1}. $$

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That $A$ is invertible means precisely that there is another matrix $A^{-1}$ such that $$AA^{-1}=I=A^{-1}A.$$

It is easy to show that the inverse, if it exists, must be unique.

Now suppose $A,B$ to be invertible, and denote $C:=AB$. $C$ will be invertible if we can find $C^{-1}$ such that $CC^{-1}=I=C^{-1}C$. Observe that $$C(B^{-1}A^{-1})=(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=AIA^{-1}=AA^{-1}=I.$$

Similarly we get $(B^{-1}A^{-1})C=I$. Therefore, since inverses are unique, by definition we can conclude that $C^{-1}=B^{-1}A^{-1}$.

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  • $\begingroup$ What does := mean? $\endgroup$ – Jwan622 Sep 5 '18 at 14:08
  • $\begingroup$ @Jwan622 It means "we define the left hand side as the right hand side". I have introduced a symbol not used before, $C$, and I have defined it in terms of symbols previously defined $\endgroup$ – Jose Brox Sep 5 '18 at 14:09
  • $\begingroup$ Why not just use the = sign? $\endgroup$ – Jwan622 Sep 5 '18 at 14:11
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    $\begingroup$ @Jwan622 Because the identity sign is used to state that two already meaningful different expressions actually represent the same mathematical object (as in $2-2=0$). This is totally different to defining a value for a new symbol. $\endgroup$ – Jose Brox Sep 5 '18 at 14:15

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