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$A$ is a square matrix, and there is a matrix $D$ such that $AD=I$.

I need help to prove either (a) or (b) below:

(a) Show that for each $\vec b$, $A\vec x=\vec b$ has a unique solution.


(b) Show that $A$ is invertible.

For (a), $AD\vec b=I\vec b = \vec b$, so obviously the given equation has at least one solution $\vec x =D\vec b$. But how to show that the solution is unique?

For (b), I'm guessing we should show either that $DA=I$ or that $DA=AD$ (and thus $D=A^{-1}$), but I'm not sure how to do this without assuming that A is invertible, which is what we are needing to show.

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Note that by the definition of an Inverse Matrix it is required to be the unique matrix such that: $${\bf A}{\bf A}^{-1}={\bf A}^{-1}{\bf A}={\bf I}$$ – Shaktal Jul 19 '12 at 15:20
What is your definition of "invertible matrix"? – DonAntonio Jul 19 '12 at 15:20
@DonAntonio By "invertible matrix", I mean it must be left- and right- invertible. – Ryan Jul 19 '12 at 15:36
probably duplicate of – chaohuang Jul 19 '12 at 15:40
up vote 3 down vote accepted

Here is a proof which does not use group theory or vector spaces:

I will establish equivalence of the following. The details are left for you to be filled in. Once you have established (b) by the sketch below, establishing (a) is trivial.

The following are equivalent for any square matrix $A$:

  1. $A$ has a left inverse.
  2. $Ax = 0$ has only the trivial solution.
  3. The reduced row echelon form of $A$ is the identity matrix.
  4. $A$ is a product of elementary matrices.
  5. $A$ is invertible.
  6. A has a right inverse.

$1\implies 2$: If $A$ has a left inverse $L$ then $Ax=0 \implies LAx=L0\implies Ix=x=0$.

$2\implies 3$: The augmented matrix for $Ax=0$ must have been reduced to $Ix=0$ by Gauss Jordan elimination.

$3\implies 4$: Since $E_1\cdots E_kA=I$ and each elementary matrix is invertible (and the inverse is also an elementary matrix) so $A=E_k^{-1}\cdots E_1^{-1}$.

$4\implies 5$: Each elementary matrix is invertible.

$5\implies 6$: Trivial.

$6\implies 1$: If $R$ is the right inverse of $A$, then $A$ is the left inverse of $R$ and by $1\implies 5$ $R$ is invertible with inverse $A$, following which $R$ is the left inverse of $A$.

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It is such cleverness to invoke the priorly-established <$1 \implies 5$> within the crucial last step <$6 \implies 1$>! And yes, that your proof involves only beginning linear algebra concepts instead of vector spaces or group theory makes it immensely useful and worthy, IMO. This is precisely the type of first-principles proof I was after, although I really like the simplicity of Ed's determinant argument too. – Ryan Jul 19 '12 at 18:25
Does this still work if we omit the initial assumption that A has a left inverse? – Emily Jul 19 '12 at 18:49
This proof shows that all the statements are equivalent. But your problem statement does not make any of those statements. In other words, there is nothing here that guarantees that your matrix A must have a left inverse, or that D is the right inverse of A. All it says is that if D is the right inverse of A, then A is invertible, so we still need to prove that that is the case. – Emily Jul 19 '12 at 19:10
@Ed For the purpose of answering my original question, <$5 \implies 6$> can be omitted, but 1 is important because <$1 \implies 5$> is used to prove <$6 \implies 1$>. What Shahab has done is to prove part of the "Invertible Matrix Theorem", which contains a set of equivalent statements about the invertibility of A. He has done a better job than my textbook author David Lay, who made a small but critical mistake. This is actually the motivation for my original question. :) – Ryan Jul 19 '12 at 19:16
@Ed The supposition is found in my original question: "A is a square matrix, and there is a matrix D such that AD=I." I.e. I was asking for a proof that <$6 \implies 1$>. Mathstackexchange is one of the best things about the internet. – Ryan Jul 19 '12 at 19:19

Let's use a determinant argument.

$AD=I \Longrightarrow \det(AD)=\det(I) \Longrightarrow \det(A)\det(D)=1.$

If $A$ or $D$ are square and non-invertible, then $\det(A)=0$ or $\det(D)=0$. By above, this cannot be the case.

Therefore, both $A$ and $D$ are invertible and belong to $GL_n(\mathbb{F})$. The inverse is unique, so $D = A^{-1}$.

Showing part (b) implies part (a): if the inverse $A^{-1}$ exists, it is unique, so $A^{-1}Ax = A^{-1}b \Longrightarrow x = A^{-1}b$.

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$A$ is not invertible, it only has by assumption an inverse to the right. If you assume that invertible matrices are a group with law being the product of matrices, then you implicitely say that $A$ is invertible, which is not clear. – Louis La Brocante Jul 19 '12 at 15:37
This is nice @Ed, yet I can't understand why you can't deduce that both $\,A,D\,$ are invertible after you showed $\,\det A\cdot \det D=1\neq 0\,$...? – DonAntonio Jul 19 '12 at 17:02
Yeah, actually I just noticed that, too. I have updated the response to use this stronger argument. Group theoretic considerations are still used to show uniqueness -- probably not necessary, but illuminating. – Emily Jul 19 '12 at 17:02
@DonAntonio if either $\det(A)$ or $\det(D)$ would be equal to zero, then $\det(A)\cdot\det(D)$ should not be $1$ :). – Louis La Brocante Jul 19 '12 at 18:10
@Ed This is a great proof, and by far the most straightforward, thus the most elegant, IMO. Thank you! – Ryan Jul 19 '12 at 18:17

Claim: A square matrix $\,A\,$ is invertible iff $\,AD=I\,$ for some (square) matrix $\,D\,$

The proof of the above depends on the following

Proposition: in $\,x\in G\,$ is a group with unit $\,1\,$ and $\,xx=x\,$ , then $\,x=1\,$

So using now associativity: $$(DA)(DA)=D(AD)A=DIA=DA\Longrightarrow DA=I=AD$$

Putting $\,D=A^{-1}\,$ gives us the usual notation for inverse, and the above solves positively both questions: $$(a) \,\,Ax=b\Longrightarrow A^{-1}Ax=A^{-1}b\Longrightarrow A^{-1}b$$ $(b)\,$It's done above

Added: Following the remarks in the comments, and since we cannot assume any of $\,D,A,DA, AD\,$ is invertible (and thus talking of a group is out of order), I shall try an approach proposed by Rolando (assume the matrices are $\,n\times n\,$):

We're given $\,AD=I,$ . Either in terms of matrices or of linear operators, this means that $\,\dim Im(AD)=n\,$ (is full, i.e. $\,AD\,$ is onto). Now we have a general claim whose proof is elementary:

Claim: If we have functions $\,f:A\to B\,\,,\,g:B\to C\,$ , then $\,g\circ f\,$ onto $\,\Longrightarrow\,g\,$ onto.

From this it follows that $\,A\,$ is onto $\,\Longleftrightarrow \dim Im(A)=n\Longleftrightarrow \dim\ker A=0\,$ ,which means $\,A\,$ is a bijection and, thus, invertible.

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Proposition: in x∈G is a group with unit 1 and xx=1 , then x=1 ???? – Louis La Brocante Jul 19 '12 at 15:30
@Rolando, thanks: that was a stupid typo. I already fixed it. – DonAntonio Jul 19 '12 at 15:31
I think you want to say "if $G$ is a group with unit $1$ and $x\in G$ is such that $xx=x$, then $x=1$.", or something like that. – Matthew Pressland Jul 19 '12 at 15:32
But to use that you must be aware that the matrix product is a group law for invertible matrices, so you assume that $DA$ is invertible, which is not clear imo. – Louis La Brocante Jul 19 '12 at 15:35
Well, I do not need $\,DA\,$ being invertible but each of them being, and you're right: the claim, which is painfully easy to prove, cannot be used unless we already know we're in, at least, a monoid (no need of group for this). Good point. – DonAntonio Jul 19 '12 at 15:37

For any $b$ in your vector space, you have $$AD(b)=I(b)=b$$ In particular $A$ is surjective, and $D$ is injective. Since an endomorphism of a vector space is surjective if and only if it is bijective by the rank nullity theorem, $A$ is invertible.

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Since $AD=I$, the range of $A$ is the full vector space: For any vector $\vec v$ you can find a vector $\vec w$ so that $A\vec w=\vec v$: Just use $\vec w=D\vec v$. Denote the dimension of the vector space with $d$ (I assume we are in a finite-dimensional vector space, because otherwise I would not know how to define a square matrix). Thus the range of $A$ has dimension $d$.

Now assume there are two vectors $\vec v_1\neq\vec v_2$ so that $A\vec v_1=A\vec v_2$. Then by linearity, $A(\vec v_1-\vec v_2)=\vec 0$. Now you can take the vector $\vec b_1 := \vec v_1-\vec v_2$ (which by the assumption is nonzero) and extend that with $d-1$ other vectors to a basis $\vec b_1,\dots,\vec b_n$ of the vector space. Now look at the image of an arbitrary vector $w=\sum w_i\vec b_i$: Since $\vec b_1$ is, by construction, always mapped to $\vec 0$, every vector is mapped to the $\sum_{k=\color{red}2}^d w_i A\vec b_i$. But the space spanned by those vectors can only have the maximal dimension $d-1$, which contradicts the previously shown fact that the range of $A$ is the whole vector space. Thus the assumption $\vec v_1\neq\vec v2$ but $A\vec v_1=A\vec v_2$ cannot be true.

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Fisrt, we need to prove that $A$ is invertible. Given $ AD=I \Rightarrow |AD|=1 \Rightarrow |A||D|=1$ $\Rightarrow$ $|A| \neq 0$ and $|D|\neq 0\,.$ Now, since $|A|\neq 0$ then $A$ is invertible.

The other step, you need to prove that $D=A^{-1}$. One way is the following or the other ways as in the other answers

$ AD=I \Rightarrow AD=AA^{-1} \Rightarrow AD-AA^{-1}=0 \Rightarrow A(D-A^{-1})=0 \Rightarrow (D-A^{-1})=0\,, $ if not then $AD \neq I\,,$ $\Rightarrow D = A^{-1}\,,$

since A does not equal the zero matrix.

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and what is $A^{-1}$? – Louis La Brocante Jul 19 '12 at 15:42
@Mhenni In Step 1, we're not allowed to assume that $A^{-1}$ exists. And in Step 4, it is quite possible that the product of two non-zero matrices is zero. – Ryan Jul 19 '12 at 15:50
See the last edit. – Mhenni Benghorbal Jul 21 '12 at 13:31

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