- #1
IKonquer
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I have very little experience with proofs, and I am trying to learn algebra on my own. The following problem is found within the text of Michael Artin's Algebra.
The first problem encountered was the following: Let A, B be n x n matrices. If both are invertible, then so is their product AB, and [tex] (AB)^{-1} = B^{-1}A^{-1} [/tex]
To show this is true for the right inverse I said,
Let X be an invertible matrix.
[tex]
\begin{flalign*}
XX^{-1} = I\\
(AB)(AB)^{-1} = I\\
A(BB^{-1})A^{-1} = I \\
AIA^{-1} = I\\
AA^{-1} = I\\
I = I\\
\end{flalign*}
[/tex]
Also the same can be said for the left inverse, which makes sense to me.
I'm having a lot of trouble understanding the following: If [tex]A_{1}, ... , A_{m} [/tex] are invertible, then so is the product [tex]A_{1} ... A_{m} [/tex] and its inverse is [tex]A_{m}^{-1} ... A_{1}^{-1} [/tex]
Let m = 1
[tex]A_{1}A^{-1} = I [/tex]
This shows that [tex]A_{1} [/tex] is invertible, and its product, which is itself, is invertible.
For m + 1
Let [tex]A_{1}, ... , A_{m}, A_{m+1} [/tex] be invertible matrices and let P be the product of [tex]A_{1}, ... , A_{m} [/tex].
"By the induction hypothesis, P is invertible." - I don't quite understand why you can say that. Could someone explain why this is true?
The first problem encountered was the following: Let A, B be n x n matrices. If both are invertible, then so is their product AB, and [tex] (AB)^{-1} = B^{-1}A^{-1} [/tex]
To show this is true for the right inverse I said,
Let X be an invertible matrix.
[tex]
\begin{flalign*}
XX^{-1} = I\\
(AB)(AB)^{-1} = I\\
A(BB^{-1})A^{-1} = I \\
AIA^{-1} = I\\
AA^{-1} = I\\
I = I\\
\end{flalign*}
[/tex]
Also the same can be said for the left inverse, which makes sense to me.
I'm having a lot of trouble understanding the following: If [tex]A_{1}, ... , A_{m} [/tex] are invertible, then so is the product [tex]A_{1} ... A_{m} [/tex] and its inverse is [tex]A_{m}^{-1} ... A_{1}^{-1} [/tex]
Let m = 1
[tex]A_{1}A^{-1} = I [/tex]
This shows that [tex]A_{1} [/tex] is invertible, and its product, which is itself, is invertible.
For m + 1
Let [tex]A_{1}, ... , A_{m}, A_{m+1} [/tex] be invertible matrices and let P be the product of [tex]A_{1}, ... , A_{m} [/tex].
"By the induction hypothesis, P is invertible." - I don't quite understand why you can say that. Could someone explain why this is true?