Check proof: Invertible matrix is equal to product of inverses of elementary matrices

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

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.