When is the product of two non-square matrices invertible?
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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
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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
33.2k12261
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
33.2k12261
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
33.2k12261
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
33.2k12261
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
33.2k12261
edited Nov 18 at 2:54
answered Nov 18 at 2:09
Empy2
33.2k12261
answered Nov 18 at 2:09
Empy2
33.2k12261
answered Nov 18 at 2:09
Empy2
33.2k12261
33.2k12261
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 |
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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