Linked Questions

20
votes
3answers
28k views

Matrices: left inverse is also right inverse? [duplicate]

If $A$ and $B$ are square matrices, and $AB=I$, then I think it is also true that $BA=I$. In fact, this Wikipedia page says that this "follows from the theory of matrices". I assume there's a nice ...
6
votes
4answers
6k views

Assuming $AB=I$ prove $BA=I$ [duplicate]

Possible Duplicate: If $AB = I$ then $BA = I$ Most introductory linear algebra texts define the inverse of a square matrix $A$ as such: Inverse of $A$, if it exists, is a matrix $B$ such that $...
4
votes
3answers
6k views

Can we prove that matrix multiplication by its inverse is commutative? [duplicate]

We know that $AA^{-1} = I$ and $A^{-1}A = I$, but is there a proof for the commutative property here? Or is this just the definition of invertibility?
5
votes
2answers
8k views

Prove that a square matrix commutes with its inverse [duplicate]

The Question: This is a very fundamental and commonly used result in linear algebra, but I haven't been able to find a proof or prove it myself. The statement is as follows: let $A$ be an $n\...
2
votes
5answers
818 views

Why is the left inverse of a matrix equal to the right inverse? [duplicate]

Given a square matrix $A$ that has full row rank we know that the matrix is invertible. So there is a matrix $B$ such that $$ AB=1 $$ writing this in component notation, $$ A_{ij}B_{jk}=\delta_{ik} ...
1
vote
4answers
751 views

Are there matrices such that $AB=I$ and $BA \neq I$ [duplicate]

Are there matrices such that $AB=I$ and $BA \neq I$ ? $A$ and $B$ are square matrices
2
votes
2answers
2k views

What 's the short proof that for square matrices $AB = I$ implies $BA = I$? [duplicate]

Possible Duplicate: If $AB = I$ then $BA = I$ I'm trying to remember the one line proof that for square matrices $AB = I$ implies $BA = I$. I think it uses only elementary matrix properties and ...
1
vote
3answers
374 views

Can we prove $BA=E$ from $AB=E$? [duplicate]

I was wondering if $AB=E$ ($E$ is identity) is enough to claim $A^{-1} = B$ or if we also need $BA=E$. All my textbooks define the inverse $B$ of $A$ such that $AB=BA=E$. But I can't see why $AB=E$ ...
1
vote
1answer
2k views

For square matrix, right or left inverse is equivalent to inverse. [duplicate]

Definitions: Let $A$ be an $n\times n$ matrix. The $n\times n$ matrix $B(=A^{-1})$ is an inverse for $A$ if $AB=BA=I$. Let $A$ be a $k\times n$ matrix.The $n\times k$ matrix $B$ is a right ...
2
votes
1answer
1k views

Why does $A$ times its inverse equal to the identity matrix? [duplicate]

I was trying to come up with a proof of why: $AA^{-1} = I$. If we know that: $A^{-1}A = I$, then $A(A^{-1}A) = A \implies (AA^{-1})A = A$. However I don't like setting $AA^{-1} = I$ for fear that it ...
0
votes
1answer
1k views

Proof that $(AA^{-1}=I) \Rightarrow (AA^{-1} = A^{-1}A)$ [duplicate]

I'm trying to prove a pretty simple problem - commutativity of multiplication of matrix and its inverse. But I'm not sure, if my proof is correct, because I'm not very experienced. Could you, please, ...
3
votes
3answers
329 views

How to prove that a matrix is invertible $\iff$ invertible at right $\iff$ invertible at left? [duplicate]

Let $A$ a $n\times n$ invertible at left. In fact, I just want to prove that it's invertible at right (the rest is obvious). All what I can say is that there is a $B$ s.t. $BA=I.$ To prova $AB=I$, I ...
2
votes
1answer
268 views

For square matrices $A$, $B$, is $AB=I$ sufficient that $A$ and $B$ are inverse of each other? [duplicate]

Possible Duplicate: If $AB = I$ then $BA = I$ If $A$ and $B$ are two square matrices, and we know $AB=I$ where $I$ is the identity matrix. Is it sufficient that $BA=I$ as well so that $A$ and $B$ ...
0
votes
2answers
388 views

Proving $AB=BA=I$ [duplicate]

I have seem to forgot this important fact, and I am trying to prove it to myself by looking at $A$ as matrix of the elementary row operations. Where I seem to get stuck is that if $A$ are elementary ...
2
votes
1answer
446 views

Does $AA^T=I$ imply that $A^TA=I$? [duplicate]

Does $AA^T=I$ imply that $A^TA=I$? The wiki article defines the orthogonal group as: $$o(n,\Bbb C) = \{ A\in M_n(\Bbb C): AA^T=A^TA=I \}$$ My book writes: $$o(n,\Bbb C) = \{ A\in M_n(\Bbb C): AA^T=...

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