Help with this matrix problem
up vote
-1
down vote
favorite
Let $A in mathbb{Q}^{mtimes n}$. Proof that if A has at least two distinct right-inverses, then A has infinitely many right-inverses.
I began this proof as follows:
Let suppose that $B,C in K^{ntimes m}$ are two distinct right-inverses for $A$. Then, $AB=I_{n}$ and $AC=I_{n}$. Then we have $AB+AC=2I_{n} iff A(dfrac{1}{2}(B+C))=I_{n}$.
Then $dfrac{1}{2}(B+C) in K^{m times n}$ is also a right-inverse for $A$. I think repeating this process I could prove the proposition, but I'm not really sure. Is there another way? (without Gauss-Jordan algorithm)
linear-algebra matrices
asked Nov 17 at 5:07
Abraham Hernández
31
add a comment |
up vote
-1
down vote
favorite
Let $A in mathbb{Q}^{mtimes n}$. Proof that if A has at least two distinct right-inverses, then A has infinitely many right-inverses.
I began this proof as follows:
Let suppose that $B,C in K^{ntimes m}$ are two distinct right-inverses for $A$. Then, $AB=I_{n}$ and $AC=I_{n}$. Then we have $AB+AC=2I_{n} iff A(dfrac{1}{2}(B+C))=I_{n}$.
Then $dfrac{1}{2}(B+C) in K^{m times n}$ is also a right-inverse for $A$. I think repeating this process I could prove the proposition, but I'm not really sure. Is there another way? (without Gauss-Jordan algorithm)
linear-algebra matrices
asked Nov 17 at 5:07
Abraham Hernández
31
1
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $A in mathbb{Q}^{mtimes n}$. Proof that if A has at least two distinct right-inverses, then A has infinitely many right-inverses.
I began this proof as follows:
Let suppose that $B,C in K^{ntimes m}$ are two distinct right-inverses for $A$. Then, $AB=I_{n}$ and $AC=I_{n}$. Then we have $AB+AC=2I_{n} iff A(dfrac{1}{2}(B+C))=I_{n}$.
Then $dfrac{1}{2}(B+C) in K^{m times n}$ is also a right-inverse for $A$. I think repeating this process I could prove the proposition, but I'm not really sure. Is there another way? (without Gauss-Jordan algorithm)
linear-algebra matrices
asked Nov 17 at 5:07
Abraham Hernández
31
Let $A in mathbb{Q}^{mtimes n}$. Proof that if A has at least two distinct right-inverses, then A has infinitely many right-inverses.
I began this proof as follows:
Let suppose that $B,C in K^{ntimes m}$ are two distinct right-inverses for $A$. Then, $AB=I_{n}$ and $AC=I_{n}$. Then we have $AB+AC=2I_{n} iff A(dfrac{1}{2}(B+C))=I_{n}$.
Then $dfrac{1}{2}(B+C) in K^{m times n}$ is also a right-inverse for $A$. I think repeating this process I could prove the proposition, but I'm not really sure. Is there another way? (without Gauss-Jordan algorithm)
linear-algebra matrices
linear-algebra matrices
asked Nov 17 at 5:07
Abraham Hernández
31
asked Nov 17 at 5:07
Abraham Hernández
31
asked Nov 17 at 5:07
Abraham Hernández
31
asked Nov 17 at 5:07
Abraham Hernández
31
asked Nov 17 at 5:07
Abraham Hernández
31
31
1
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18
add a comment |
1
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18
1
1
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001998%2fhelp-with-this-matrix-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
If $AB=AC=I$ then $A(tB+(1-t)C) = I$ for any scalar $t$.
– copper.hat
Nov 17 at 5:10
Thanks a lot !
– Abraham Hernández
Nov 17 at 5:18