Matrix ring with zero divisors
$begingroup$
Given a matrix ring
$K = left{ begin{bmatrix}0 & 0\ a & bend{bmatrix} ,Biggm|, a, b in mathbb{R}right}$ and element $x = begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} $.
In the book it says that $x$ is the right divisor of zero but I think that this is a mistake.
I need someone to confirm with better knowledge.
For example, right divisor is defined as when we take $A in K$ and $A neq 0$ then $A times x = 0$ which makes sense but then it follows: $begin{bmatrix}0 & 0\ A_1 & A_2end{bmatrix} times begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} = begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0$.
Therefore, $x$ is not the right divisor of the ring.
Can someone confirm this?
Thanks.
linear-algebra abstract-algebra
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
$endgroup$
add a comment |
$begingroup$
Given a matrix ring
$K = left{ begin{bmatrix}0 & 0\ a & bend{bmatrix} ,Biggm|, a, b in mathbb{R}right}$ and element $x = begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} $.
In the book it says that $x$ is the right divisor of zero but I think that this is a mistake.
I need someone to confirm with better knowledge.
For example, right divisor is defined as when we take $A in K$ and $A neq 0$ then $A times x = 0$ which makes sense but then it follows: $begin{bmatrix}0 & 0\ A_1 & A_2end{bmatrix} times begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} = begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0$.
Therefore, $x$ is not the right divisor of the ring.
Can someone confirm this?
Thanks.
linear-algebra abstract-algebra
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
$endgroup$
1
$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23
add a comment |
$begingroup$
Given a matrix ring
$K = left{ begin{bmatrix}0 & 0\ a & bend{bmatrix} ,Biggm|, a, b in mathbb{R}right}$ and element $x = begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} $.
In the book it says that $x$ is the right divisor of zero but I think that this is a mistake.
I need someone to confirm with better knowledge.
For example, right divisor is defined as when we take $A in K$ and $A neq 0$ then $A times x = 0$ which makes sense but then it follows: $begin{bmatrix}0 & 0\ A_1 & A_2end{bmatrix} times begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} = begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0$.
Therefore, $x$ is not the right divisor of the ring.
Can someone confirm this?
Thanks.
linear-algebra abstract-algebra
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
$endgroup$
Given a matrix ring
$K = left{ begin{bmatrix}0 & 0\ a & bend{bmatrix} ,Biggm|, a, b in mathbb{R}right}$ and element $x = begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} $.
In the book it says that $x$ is the right divisor of zero but I think that this is a mistake.
I need someone to confirm with better knowledge.
For example, right divisor is defined as when we take $A in K$ and $A neq 0$ then $A times x = 0$ which makes sense but then it follows: $begin{bmatrix}0 & 0\ A_1 & A_2end{bmatrix} times begin{bmatrix}0 & 0\ 1 & 1end{bmatrix} = begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0$.
Therefore, $x$ is not the right divisor of the ring.
Can someone confirm this?
Thanks.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
edited Jan 14 at 23:21
Arturo Magidin
267k34591922
267k34591922
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
asked Jan 14 at 23:18
mcdonald reversedmcdonald reversed
212
212
1
$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23
add a comment |
1
$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23
1
1
$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23
$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23
add a comment |
1 Answer
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$begingroup$
No: how do you know that $$begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $Aneq 0$ such that $Acdot x=0$, so this is not required to hold for arbitrary $A$.
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
add a comment |
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$begingroup$
No: how do you know that $$begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $Aneq 0$ such that $Acdot x=0$, so this is not required to hold for arbitrary $A$.
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
add a comment |
$begingroup$
No: how do you know that $$begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $Aneq 0$ such that $Acdot x=0$, so this is not required to hold for arbitrary $A$.
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
$endgroup$
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
add a comment |
$begingroup$
No: how do you know that $$begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $Aneq 0$ such that $Acdot x=0$, so this is not required to hold for arbitrary $A$.
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
$endgroup$
No: how do you know that $$begin{bmatrix}0 & 0\ A_2 & A_2end{bmatrix} neq 0?$$ If $A_2=0$, then this matrix is $0$. (And as long as $A_1neq 0$, $A$ will still be nonzero.) Note that the definition of $x$ being a right zero divisor is just that there exists $Aneq 0$ such that $Acdot x=0$, so this is not required to hold for arbitrary $A$.
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
answered Jan 14 at 23:24
Eric WofseyEric Wofsey
193k14222353
193k14222353
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
add a comment |
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
$begingroup$
Thanks for clarifying, now makes sense.
$endgroup$
– mcdonald reversed
Jan 14 at 23:35
add a comment |
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$begingroup$
I doubt the book says that $x$ is "the" (singular definite article) right divisor of zero. It is a (singular indefinite article) right divisor of zero, but it isn't the only one, so it can't be "the" right divisor of zero. Your calculations are correct, but you seem to have overlooked the possibility that the result of the operation is equal to zero, even if the first factor is not. For example, what happens when $A_1=1$ and $A_2=0$?
$endgroup$
– Arturo Magidin
Jan 14 at 23:23