Given an invertible matrix $A \in \mathbb{R}^n$, and given that $A \leadsto I_n$ is obtained such that $I_n = (E_k \cdot\ldots\cdot E_3\cdot E_2 \cdot E_1)A$, where $E_1,\ldots,E_k$ are a set of elementary matrices, then $A = E_1^{-1} \cdot \ldots \cdot E_k^{-1}$.

Here is my proof:

Let $P(k)$ be the proposition that given any invertible matrix $A$, then for any $k\in\mathbb{N}$, for $E_1,\ldots,E_k$ being the set of elementary operations such that $I = (E_k \cdot\ldots\cdot E_3\cdot E_2 \cdot E_1)A$ we have that $A = E_1^{-1} \cdot \ldots \cdot E_k^{-1}$.

For the base case, $k=1$, we have that $I = E_1\cdot A$. Multiplying by $E_1^{-1}$ on both sides, we have $$E_1^{-1}\cdot I = E_1^{-1} \cdot E_1\cdot A$$ $$E_1^{-1} = A$$ hence the base case is proven. Fix $k\in\mathbb{N}$, assume $P(k_0)$ is true for every $k_0 \in \{1,\ldots,k\}$. Let there exist a matrix $A^\prime$ such that $$E_{k+1}E_k\ldots E_1 \cdot A^\prime =I$$ Multiplying by $(E_{k+1}\ldots E_1)^{-1}$ on both sides, we have $$(E_{k+1}E_k\ldots E_1)^{-1}\cdot (E_{k+1}E_k\ldots E_1) \cdot A^\prime = (E_{k+1}\ldots E_1)^{-1} \cdot I$$ By successively applying the inductive hypothesis, we have $$A^\prime = E_1^{-1} \cdot \ldots \cdot E_{k+1}^{-1}$$ Hence $P(k_0)$ implies $P(k+1)$ for every $k_0 \in \{1,\ldots,k\}$. By the Principle of Strong Induction, $P(k)$ is true for every $k\in\mathbb{N}$.

Is my proof correct? Let me know. Thanks.


1 Answer 1


You claim that you applied the induction hypothesis. But where did you do that? What you use in fact is that$$(E_{k+1}\ldots E_1)^{-1}={E_1}^{-1}\ldots{E_{k+1}}^{-1}.$$And this, yes, can be proved by induction.

Note that the fact that your matrices $E_j$ are elementary is never used.


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