Let $A$ be $mtimes n$ and $B$ be $ntimes m$ matrix over real number with $m<n$.
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1
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Let $A$ be $mtimes n$ and $B$ be $ntimes m$ matrix over real number
with $m<n$. Then
$AB$ is always nonsingular.
$AB$ is always singular.
$BA$ is always nonsingular
$BA$ is always singular.
False, Since $A=0$, $B=0$ $implies$ $AB=0$.
Need not be true. Let $x$ be $1times n$ matrix. Since, $x x^Tge 0$
Using the same reasoning 3 can be false.
What is the guarantee that $BA$ is always singular? If $A=xneq 0, B=x^t.$ Then $BA$ is singular. How do I prove In general?
linear-algebra matrices
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
add a comment |
up vote
1
down vote
favorite
Let $A$ be $mtimes n$ and $B$ be $ntimes m$ matrix over real number
with $m<n$. Then
$AB$ is always nonsingular.
$AB$ is always singular.
$BA$ is always nonsingular
$BA$ is always singular.
False, Since $A=0$, $B=0$ $implies$ $AB=0$.
Need not be true. Let $x$ be $1times n$ matrix. Since, $x x^Tge 0$
Using the same reasoning 3 can be false.
What is the guarantee that $BA$ is always singular? If $A=xneq 0, B=x^t.$ Then $BA$ is singular. How do I prove In general?
linear-algebra matrices
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be $mtimes n$ and $B$ be $ntimes m$ matrix over real number
with $m<n$. Then
$AB$ is always nonsingular.
$AB$ is always singular.
$BA$ is always nonsingular
$BA$ is always singular.
False, Since $A=0$, $B=0$ $implies$ $AB=0$.
Need not be true. Let $x$ be $1times n$ matrix. Since, $x x^Tge 0$
Using the same reasoning 3 can be false.
What is the guarantee that $BA$ is always singular? If $A=xneq 0, B=x^t.$ Then $BA$ is singular. How do I prove In general?
linear-algebra matrices
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
Let $A$ be $mtimes n$ and $B$ be $ntimes m$ matrix over real number
with $m<n$. Then
$AB$ is always nonsingular.
$AB$ is always singular.
$BA$ is always nonsingular
$BA$ is always singular.
False, Since $A=0$, $B=0$ $implies$ $AB=0$.
Need not be true. Let $x$ be $1times n$ matrix. Since, $x x^Tge 0$
Using the same reasoning 3 can be false.
What is the guarantee that $BA$ is always singular? If $A=xneq 0, B=x^t.$ Then $BA$ is singular. How do I prove In general?
linear-algebra matrices
linear-algebra matrices
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
edited Nov 29 at 6:35
Chinnapparaj R
5,0901825
5,0901825
asked Nov 29 at 5:33
Unknown x
2,50211025
asked Nov 29 at 5:33
Unknown x
2,50211025
asked Nov 29 at 5:33
Unknown x
2,50211025
2,50211025
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
$BA in mathbb{R}^{n times n}$,
but $rank(BA)le min{rank(A) ,rank(B) } le m<n$
Hence it must be singular.
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
add a comment |
up vote
1
down vote
Alternatively, $A$ is $m times n$ and $m<n$ implies $Ax=0$ has a non zero solution, say $x_0$. The same non zero $x_o$ is the solution of $(BA)x=0$ as well! So $BA$ must be singular
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
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
up vote
1
down vote
accepted
$BA in mathbb{R}^{n times n}$,
but $rank(BA)le min{rank(A) ,rank(B) } le m<n$
Hence it must be singular.
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
add a comment |
up vote
1
down vote
accepted
$BA in mathbb{R}^{n times n}$,
but $rank(BA)le min{rank(A) ,rank(B) } le m<n$
Hence it must be singular.
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$BA in mathbb{R}^{n times n}$,
but $rank(BA)le min{rank(A) ,rank(B) } le m<n$
Hence it must be singular.
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
$BA in mathbb{R}^{n times n}$,
but $rank(BA)le min{rank(A) ,rank(B) } le m<n$
Hence it must be singular.
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
answered Nov 29 at 5:36
Siong Thye Goh
98.3k1463116
98.3k1463116
add a comment |
add a comment |
up vote
1
down vote
Alternatively, $A$ is $m times n$ and $m<n$ implies $Ax=0$ has a non zero solution, say $x_0$. The same non zero $x_o$ is the solution of $(BA)x=0$ as well! So $BA$ must be singular
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
add a comment |
up vote
1
down vote
Alternatively, $A$ is $m times n$ and $m<n$ implies $Ax=0$ has a non zero solution, say $x_0$. The same non zero $x_o$ is the solution of $(BA)x=0$ as well! So $BA$ must be singular
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
add a comment |
up vote
1
down vote
up vote
1
down vote
Alternatively, $A$ is $m times n$ and $m<n$ implies $Ax=0$ has a non zero solution, say $x_0$. The same non zero $x_o$ is the solution of $(BA)x=0$ as well! So $BA$ must be singular
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
Alternatively, $A$ is $m times n$ and $m<n$ implies $Ax=0$ has a non zero solution, say $x_0$. The same non zero $x_o$ is the solution of $(BA)x=0$ as well! So $BA$ must be singular
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
answered Nov 29 at 6:32
Chinnapparaj R
5,0901825
5,0901825
add a comment |
add a comment |
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