When is the product of two non-square matrices invertible?
up vote
2
down vote
favorite
Suppose $A$ is a $k times n$ matrix and $B$ is an $n times k$ matrix. Suppose $n geq k$. Both $A$ and $B$ have rank $k$. Can we say $AB$ is invertible? Also, what happens if $n<k$?
I'm a beginner in linear algebra. I know this question has been answered many times but most answers are too technical for me (involving kernel etc.) and I could not understand them. I'm familiar with the terminology of rank, linear dependence and invertibility, but not much beyond that.
Thank you for your help.
linear-algebra matrices inverse
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
add a comment |
up vote
2
down vote
favorite
Suppose $A$ is a $k times n$ matrix and $B$ is an $n times k$ matrix. Suppose $n geq k$. Both $A$ and $B$ have rank $k$. Can we say $AB$ is invertible? Also, what happens if $n<k$?
I'm a beginner in linear algebra. I know this question has been answered many times but most answers are too technical for me (involving kernel etc.) and I could not understand them. I'm familiar with the terminology of rank, linear dependence and invertibility, but not much beyond that.
Thank you for your help.
linear-algebra matrices inverse
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $A$ is a $k times n$ matrix and $B$ is an $n times k$ matrix. Suppose $n geq k$. Both $A$ and $B$ have rank $k$. Can we say $AB$ is invertible? Also, what happens if $n<k$?
I'm a beginner in linear algebra. I know this question has been answered many times but most answers are too technical for me (involving kernel etc.) and I could not understand them. I'm familiar with the terminology of rank, linear dependence and invertibility, but not much beyond that.
Thank you for your help.
linear-algebra matrices inverse
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
Suppose $A$ is a $k times n$ matrix and $B$ is an $n times k$ matrix. Suppose $n geq k$. Both $A$ and $B$ have rank $k$. Can we say $AB$ is invertible? Also, what happens if $n<k$?
I'm a beginner in linear algebra. I know this question has been answered many times but most answers are too technical for me (involving kernel etc.) and I could not understand them. I'm familiar with the terminology of rank, linear dependence and invertibility, but not much beyond that.
Thank you for your help.
linear-algebra matrices inverse
linear-algebra matrices inverse
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
edited Nov 18 at 2:26
Theo Bendit
15.9k12147
15.9k12147
asked Nov 18 at 2:02
Canine360
229115
asked Nov 18 at 2:02
Canine360
229115
asked Nov 18 at 2:02
Canine360
229115
229115
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06
add a comment |
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $ntimes n$ matrix $C$ so its columns are a basis of $mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.
answered Nov 18 at 2:09
Empy2
33k12159
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $ntimes n$ matrix $C$ so its columns are a basis of $mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.
answered Nov 18 at 2:09
Empy2
33k12159
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
add a comment |
up vote
0
down vote
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $ntimes n$ matrix $C$ so its columns are a basis of $mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.
answered Nov 18 at 2:09
Empy2
33k12159
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
add a comment |
up vote
0
down vote
up vote
0
down vote
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $ntimes n$ matrix $C$ so its columns are a basis of $mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.
answered Nov 18 at 2:09
Empy2
33k12159
No, if one of the columns of $B$ happens to be orthogonal to all the rows of $A$ then that column of $AB$ will be all zeros.
I expect the rank has to be at least $2k-n$. Extend $B$ to an $ntimes n$ matrix $C$ so its columns are a basis of $mathbb{R}^n$. $C$ is invertible, so $AC$ has rank $k$. Now remove the $n-k$ extra columns of $AC$, and what's left has rank between $k$ and $k-(n-k)$.
answered Nov 18 at 2:09
Empy2
33k12159
edited Nov 18 at 2:54
answered Nov 18 at 2:09
Empy2
33k12159
answered Nov 18 at 2:09
Empy2
33k12159
answered Nov 18 at 2:09
Empy2
33k12159
33k12159
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
add a comment |
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
Is there any sufficient condition for $AB$ to be invertible?
– Canine360
Nov 18 at 2:11
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
I don't know that either, sorry.
– Empy2
Nov 18 at 2:15
add a comment |
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%2f3003059%2fwhen-is-the-product-of-two-non-square-matrices-invertible%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
When $n<k$ the matrices cannot have rank $k$. In any matrix the rank cannot exceed either the number of rows or the number of columns.
– Oscar Lanzi
Nov 18 at 2:09
May be relevant: math.stackexchange.com/questions/674310/… and math.stackexchange.com/questions/1136495/…
– NoChance
Nov 18 at 3:06