Sum of two invertible matrices [duplicate]
$begingroup$
This question already has an answer here:
Product or sum of invertible matrix give an invertible matrix?
3 answers
If A and B are two n x n invertible matrices, would the matrix result from A+B be invertible?
I think it would because for a matrix to be invertible its determinant would have to be greater than 0, and if you add the determinants of two matrices greater than 0 you would have to get a non zero answer. But is there any way to prove this?
linear-algebra matrices
edited Dec 19 '18 at 19:51
Bernard
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
$endgroup$
marked as duplicate by Dietrich Burde, Community♦ Dec 19 '18 at 19:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Product or sum of invertible matrix give an invertible matrix?
3 answers
If A and B are two n x n invertible matrices, would the matrix result from A+B be invertible?
I think it would because for a matrix to be invertible its determinant would have to be greater than 0, and if you add the determinants of two matrices greater than 0 you would have to get a non zero answer. But is there any way to prove this?
linear-algebra matrices
edited Dec 19 '18 at 19:51
Bernard
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
$endgroup$
marked as duplicate by Dietrich Burde, Community♦ Dec 19 '18 at 19:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49
add a comment |
$begingroup$
This question already has an answer here:
Product or sum of invertible matrix give an invertible matrix?
3 answers
If A and B are two n x n invertible matrices, would the matrix result from A+B be invertible?
I think it would because for a matrix to be invertible its determinant would have to be greater than 0, and if you add the determinants of two matrices greater than 0 you would have to get a non zero answer. But is there any way to prove this?
linear-algebra matrices
edited Dec 19 '18 at 19:51
Bernard
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
$endgroup$
This question already has an answer here:
Product or sum of invertible matrix give an invertible matrix?
3 answers
If A and B are two n x n invertible matrices, would the matrix result from A+B be invertible?
I think it would because for a matrix to be invertible its determinant would have to be greater than 0, and if you add the determinants of two matrices greater than 0 you would have to get a non zero answer. But is there any way to prove this?
This question already has an answer here:
Product or sum of invertible matrix give an invertible matrix?
3 answers
linear-algebra matrices
linear-algebra matrices
edited Dec 19 '18 at 19:51
Bernard
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
edited Dec 19 '18 at 19:51
Bernard
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
edited Dec 19 '18 at 19:51
Bernard
123k741117
edited Dec 19 '18 at 19:51
Bernard
123k741117
edited Dec 19 '18 at 19:51
Bernard
123k741117
123k741117
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
asked Dec 19 '18 at 19:47
Finalsock23Finalsock23
11
11
marked as duplicate by Dietrich Burde, Community♦ Dec 19 '18 at 19:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Dietrich Burde, Community♦ Dec 19 '18 at 19:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49
add a comment |
1
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49
1
1
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer is generally no. For instance, consider
$$
A = pmatrix{1&0\0&1}, quad B = pmatrix{-1&0\0&-1}
$$
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is generally no. For instance, consider
$$
A = pmatrix{1&0\0&1}, quad B = pmatrix{-1&0\0&-1}
$$
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
add a comment |
$begingroup$
The answer is generally no. For instance, consider
$$
A = pmatrix{1&0\0&1}, quad B = pmatrix{-1&0\0&-1}
$$
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
add a comment |
$begingroup$
The answer is generally no. For instance, consider
$$
A = pmatrix{1&0\0&1}, quad B = pmatrix{-1&0\0&-1}
$$
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
The answer is generally no. For instance, consider
$$
A = pmatrix{1&0\0&1}, quad B = pmatrix{-1&0\0&-1}
$$
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
answered Dec 19 '18 at 19:49
OmnomnomnomOmnomnomnom
129k792185
129k792185
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
add a comment |
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
$begingroup$
I see, thank you for your explanation!
$endgroup$
– Finalsock23
Dec 19 '18 at 19:53
add a comment |
1
$begingroup$
1+(-1)=0$phantom{}$.
$endgroup$
– user1551
Dec 19 '18 at 19:49