When does $AB$ have linearly independent columns, if $A$ and $B$ are non-square matrices?
$begingroup$
If
$A$ is $m times n$ ($m<n$), and its rows are independent
$B$ is $n times p$ ($p<n$), and its columns are independent
We also know $mge n$.
does $AB$ have linearly independent columns?
Or what additional requirements are needed for $AB$ to have linearly independent columns?
linear-algebra determinant matrix-rank
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
$endgroup$
add a comment |
$begingroup$
If
$A$ is $m times n$ ($m<n$), and its rows are independent
$B$ is $n times p$ ($p<n$), and its columns are independent
We also know $mge n$.
does $AB$ have linearly independent columns?
Or what additional requirements are needed for $AB$ to have linearly independent columns?
linear-algebra determinant matrix-rank
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
$endgroup$
add a comment |
$begingroup$
If
$A$ is $m times n$ ($m<n$), and its rows are independent
$B$ is $n times p$ ($p<n$), and its columns are independent
We also know $mge n$.
does $AB$ have linearly independent columns?
Or what additional requirements are needed for $AB$ to have linearly independent columns?
linear-algebra determinant matrix-rank
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
$endgroup$
If
$A$ is $m times n$ ($m<n$), and its rows are independent
$B$ is $n times p$ ($p<n$), and its columns are independent
We also know $mge n$.
does $AB$ have linearly independent columns?
Or what additional requirements are needed for $AB$ to have linearly independent columns?
linear-algebra determinant matrix-rank
linear-algebra determinant matrix-rank
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
edited Dec 7 '18 at 4:26
Oliver Chang
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
asked Dec 7 '18 at 4:06
Oliver ChangOliver Chang
83
83
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $A$ has rank $m$ and $B$ has rank $p$, $AB$ has rank at most $min(m,p)$.
$AB$ is $mtimes p$, so it could have linearly independent columns if $m ge p$,
but not if $m < p$.
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
$endgroup$
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
add a comment |
$begingroup$
For this we need $AB $ to have rank $p $ (since $p$ columns). All we are garanteed however is that the rank of $AB$ is less orequal to the minimum of ${m,n,p}$. So this will fail for example when $m <p$.
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
$endgroup$
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $A$ has rank $m$ and $B$ has rank $p$, $AB$ has rank at most $min(m,p)$.
$AB$ is $mtimes p$, so it could have linearly independent columns if $m ge p$,
but not if $m < p$.
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
$endgroup$
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
add a comment |
$begingroup$
Since $A$ has rank $m$ and $B$ has rank $p$, $AB$ has rank at most $min(m,p)$.
$AB$ is $mtimes p$, so it could have linearly independent columns if $m ge p$,
but not if $m < p$.
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
$endgroup$
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
add a comment |
$begingroup$
Since $A$ has rank $m$ and $B$ has rank $p$, $AB$ has rank at most $min(m,p)$.
$AB$ is $mtimes p$, so it could have linearly independent columns if $m ge p$,
but not if $m < p$.
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
$endgroup$
Since $A$ has rank $m$ and $B$ has rank $p$, $AB$ has rank at most $min(m,p)$.
$AB$ is $mtimes p$, so it could have linearly independent columns if $m ge p$,
but not if $m < p$.
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
answered Dec 7 '18 at 4:20
Robert IsraelRobert Israel
321k23210462
321k23210462
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
add a comment |
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
1
1
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
So it is possible that AB has linearly independent columns if $mge p$, but not always?
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:22
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
$begingroup$
For an example with $m=p=2$ and $n=3$, try $A = pmatrix{1 & 0 & 0cr 0 & 1 & 0cr}$, $B = pmatrix{0 & 0cr 1 & 0cr 0 & 1cr}$.
$endgroup$
– Robert Israel
Dec 7 '18 at 13:46
add a comment |
$begingroup$
For this we need $AB $ to have rank $p $ (since $p$ columns). All we are garanteed however is that the rank of $AB$ is less orequal to the minimum of ${m,n,p}$. So this will fail for example when $m <p$.
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
$endgroup$
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
add a comment |
$begingroup$
For this we need $AB $ to have rank $p $ (since $p$ columns). All we are garanteed however is that the rank of $AB$ is less orequal to the minimum of ${m,n,p}$. So this will fail for example when $m <p$.
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
$endgroup$
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
add a comment |
$begingroup$
For this we need $AB $ to have rank $p $ (since $p$ columns). All we are garanteed however is that the rank of $AB$ is less orequal to the minimum of ${m,n,p}$. So this will fail for example when $m <p$.
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
$endgroup$
For this we need $AB $ to have rank $p $ (since $p$ columns). All we are garanteed however is that the rank of $AB$ is less orequal to the minimum of ${m,n,p}$. So this will fail for example when $m <p$.
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
edited Dec 7 '18 at 4:26
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
answered Dec 7 '18 at 4:24
AnyADAnyAD
2,098812
2,098812
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
add a comment |
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
what if we also know $m ge p$
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:26
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
$begingroup$
@Oliver Chang The columns should not necessarily be independent. I'd try to find an example.
$endgroup$
– AnyAD
Dec 7 '18 at 4:32
1
1
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
$begingroup$
I guess it's actually quite easy to make the rows of A orthogonal to columns of B, if n is large.
$endgroup$
– Oliver Chang
Dec 7 '18 at 4:57
add a comment |
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