Mixing
If $,operatorname{trace}(C) = 0; $ then there exist matrices $A$ and $B$ such that $AB - BA =C$
$begingroup$
If $c_{11} + c_{22}=0$ then there exist matrices $A$ and $B$ such that $AB - BA =C$, where
$$C = begin{pmatrix} c_{11} & c_{12} \ c_{21} & c_{22} end{pmatrix}.$$
I cannot understand where to start. This question is from Linear Algebra by Hoffman & Kunje. I thought this problem would be on this site, but I could not find it.
Can anyone give me a hint?
I really could not understand the answer to this question here.
I have two questions:
1) Is the statement true for any Square Matrix?
2) Is there any easy way to prove it for the matrix of order $2 times 2$?
linear-algebra matrices vector-spaces
edited Jan 9 at 19:27
the_fox
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
$endgroup$
|
show 4 more comments
$begingroup$
If $c_{11} + c_{22}=0$ then there exist matrices $A$ and $B$ such that $AB - BA =C$, where
$$C = begin{pmatrix} c_{11} & c_{12} \ c_{21} & c_{22} end{pmatrix}.$$
I cannot understand where to start. This question is from Linear Algebra by Hoffman & Kunje. I thought this problem would be on this site, but I could not find it.
Can anyone give me a hint?
I really could not understand the answer to this question here.
I have two questions:
1) Is the statement true for any Square Matrix?
2) Is there any easy way to prove it for the matrix of order $2 times 2$?
linear-algebra matrices vector-spaces
edited Jan 9 at 19:27
the_fox
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
$endgroup$
2
$begingroup$
The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
$endgroup$
– Servaes
Jan 9 at 17:00
$begingroup$
Can you give any hint ?@Servaes
$endgroup$
– cmi
Jan 9 at 17:43
2
$begingroup$
In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
$endgroup$
– Daniel Schepler
Jan 9 at 18:12
2
$begingroup$
It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
$endgroup$
– hardmath
Jan 10 at 5:45
1
$begingroup$
@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
$endgroup$
– Jean Marie
Jan 10 at 8:53
|
show 4 more comments
$begingroup$
If $c_{11} + c_{22}=0$ then there exist matrices $A$ and $B$ such that $AB - BA =C$, where
$$C = begin{pmatrix} c_{11} & c_{12} \ c_{21} & c_{22} end{pmatrix}.$$
I cannot understand where to start. This question is from Linear Algebra by Hoffman & Kunje. I thought this problem would be on this site, but I could not find it.
Can anyone give me a hint?
I really could not understand the answer to this question here.
I have two questions:
1) Is the statement true for any Square Matrix?
2) Is there any easy way to prove it for the matrix of order $2 times 2$?
linear-algebra matrices vector-spaces
edited Jan 9 at 19:27
the_fox
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
$endgroup$
If $c_{11} + c_{22}=0$ then there exist matrices $A$ and $B$ such that $AB - BA =C$, where
$$C = begin{pmatrix} c_{11} & c_{12} \ c_{21} & c_{22} end{pmatrix}.$$
I cannot understand where to start. This question is from Linear Algebra by Hoffman & Kunje. I thought this problem would be on this site, but I could not find it.
Can anyone give me a hint?
I really could not understand the answer to this question here.
I have two questions:
1) Is the statement true for any Square Matrix?
2) Is there any easy way to prove it for the matrix of order $2 times 2$?
linear-algebra matrices vector-spaces
linear-algebra matrices vector-spaces
edited Jan 9 at 19:27
the_fox
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
edited Jan 9 at 19:27
the_fox
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
edited Jan 9 at 19:27
the_fox
2,69221535
edited Jan 9 at 19:27
the_fox
2,69221535
edited Jan 9 at 19:27
the_fox
2,69221535
2,69221535
asked Jan 9 at 16:46
cmicmi
1,141212
asked Jan 9 at 16:46
cmicmi
1,141212
asked Jan 9 at 16:46
cmicmi
1,141212
1,141212
2
$begingroup$
The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
$endgroup$
– Servaes
Jan 9 at 17:00
$begingroup$
Can you give any hint ?@Servaes
$endgroup$
– cmi
Jan 9 at 17:43
2
$begingroup$
In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
$endgroup$
– Daniel Schepler
Jan 9 at 18:12
2
$begingroup$
It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
$endgroup$
– hardmath
Jan 10 at 5:45
1
$begingroup$
@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
$endgroup$
– Jean Marie
Jan 10 at 8:53
|
show 4 more comments
2
$begingroup$
The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
$endgroup$
– Servaes
Jan 9 at 17:00
$begingroup$
Can you give any hint ?@Servaes
$endgroup$
– cmi
Jan 9 at 17:43
2
$begingroup$
In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
$endgroup$
– Daniel Schepler
Jan 9 at 18:12
2
$begingroup$
It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
$endgroup$
– hardmath
Jan 10 at 5:45
1
$begingroup$
@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
$endgroup$
– Jean Marie
Jan 10 at 8:53
2
2
$begingroup$
The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
$endgroup$
– Servaes
Jan 9 at 17:00
$begingroup$
The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
$endgroup$
– Servaes
Jan 9 at 17:00
$begingroup$
Can you give any hint ?@Servaes
$endgroup$
– cmi
Jan 9 at 17:43
$begingroup$
Can you give any hint ?@Servaes
$endgroup$
– cmi
Jan 9 at 17:43
2
2
$begingroup$
In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
$endgroup$
– Daniel Schepler
Jan 9 at 18:12
$begingroup$
In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
$endgroup$
– Daniel Schepler
Jan 9 at 18:12
2
2
$begingroup$
It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
$endgroup$
– hardmath
Jan 10 at 5:45
$begingroup$
It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
$endgroup$
– hardmath
Jan 10 at 5:45
1
1
$begingroup$
@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
$endgroup$
– Jean Marie
Jan 10 at 8:53
$begingroup$
@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
$endgroup$
– Jean Marie
Jan 10 at 8:53
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The issue is to give explicit solutions $(A,B)$ to relationship :
$$AB-BA=underbrace{begin{pmatrix}p & q\
r& -pend{pmatrix}}_Ctag{1}$$
We are going to consider 3 cases, from the most general to the most particular.
- Let us assume first $p neq 0$. Relationship (1) is verified by taking :
$$A:=begin{pmatrix}0& p\
0 & r\ end{pmatrix} text{and}
B:=begin{pmatrix}0&0\
1 & q/pend{pmatrix}.tag{2}$$
Let us now consider the case $p=0$ that we divide into 2 subcases :
Case $p=0$ and $r neq -q$ : it suffices to take :
$$A:=tfrac{1}{2(q+r)}begin{pmatrix}r& -q\
r& -qend{pmatrix} text{and}
B:=begin{pmatrix} q& q\
-r&-rend{pmatrix}tag{3}$$
to get $$AB-BA=begin{pmatrix}0 & q\
r& 0end{pmatrix}.$$
- Case $p=0$ and $r=-q$ : take :
$$A:=begin{pmatrix}1& 1\
1 & 0\ end{pmatrix} text{and}
B:=begin{pmatrix}0&q\
q & 0end{pmatrix}.tag{4}$$
giving
$$AB-BA=begin{pmatrix} 0& q\
-q & 0\ end{pmatrix} .$$
Remark about the degrees of freedom we have for (2), (3) and (4) :
Out of relationship (2) for example (the same is true for (3) and (4)), it is possible to generate a lot of other solutions to (1). For example by taking, for any $a$ and any $b$ :
$$A:=begin{pmatrix}a& p\
0 & a+r\ end{pmatrix} text{and}
B:=begin{pmatrix}b&0\
1 & b+q/pend{pmatrix}.tag{5}$$
The idea behind (5) is in fact very natural : if in (1), for a given $C$, we have a solution $(A,B)$, then, for any $k$, $(A,B+kI_2)$ is as well a solution because we still have
$$A(B+kI_2)-(B+kI_2)A = C.$$
(and $(A+k'I_2,B)$ as well) i.e., we can add a multiple of the identity matrix, to any of the solution matrices without changing the result. More generally, we can add to $B$ any polynomial $p(A)$ in $A$ :
$$A(B+p(A))-(B+p(A))A = C,$$
and, of course, to $A$ any polynomial in $B$.
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The issue is to give explicit solutions $(A,B)$ to relationship :
$$AB-BA=underbrace{begin{pmatrix}p & q\
r& -pend{pmatrix}}_Ctag{1}$$
We are going to consider 3 cases, from the most general to the most particular.
- Let us assume first $p neq 0$. Relationship (1) is verified by taking :
$$A:=begin{pmatrix}0& p\
0 & r\ end{pmatrix} text{and}
B:=begin{pmatrix}0&0\
1 & q/pend{pmatrix}.tag{2}$$
Let us now consider the case $p=0$ that we divide into 2 subcases :
Case $p=0$ and $r neq -q$ : it suffices to take :
$$A:=tfrac{1}{2(q+r)}begin{pmatrix}r& -q\
r& -qend{pmatrix} text{and}
B:=begin{pmatrix} q& q\
-r&-rend{pmatrix}tag{3}$$
to get $$AB-BA=begin{pmatrix}0 & q\
r& 0end{pmatrix}.$$
- Case $p=0$ and $r=-q$ : take :
$$A:=begin{pmatrix}1& 1\
1 & 0\ end{pmatrix} text{and}
B:=begin{pmatrix}0&q\
q & 0end{pmatrix}.tag{4}$$
giving
$$AB-BA=begin{pmatrix} 0& q\
-q & 0\ end{pmatrix} .$$
Remark about the degrees of freedom we have for (2), (3) and (4) :
Out of relationship (2) for example (the same is true for (3) and (4)), it is possible to generate a lot of other solutions to (1). For example by taking, for any $a$ and any $b$ :
$$A:=begin{pmatrix}a& p\
0 & a+r\ end{pmatrix} text{and}
B:=begin{pmatrix}b&0\
1 & b+q/pend{pmatrix}.tag{5}$$
The idea behind (5) is in fact very natural : if in (1), for a given $C$, we have a solution $(A,B)$, then, for any $k$, $(A,B+kI_2)$ is as well a solution because we still have
$$A(B+kI_2)-(B+kI_2)A = C.$$
(and $(A+k'I_2,B)$ as well) i.e., we can add a multiple of the identity matrix, to any of the solution matrices without changing the result. More generally, we can add to $B$ any polynomial $p(A)$ in $A$ :
$$A(B+p(A))-(B+p(A))A = C,$$
and, of course, to $A$ any polynomial in $B$.
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
$endgroup$
add a comment |
$begingroup$
The issue is to give explicit solutions $(A,B)$ to relationship :
$$AB-BA=underbrace{begin{pmatrix}p & q\
r& -pend{pmatrix}}_Ctag{1}$$
We are going to consider 3 cases, from the most general to the most particular.
- Let us assume first $p neq 0$. Relationship (1) is verified by taking :
$$A:=begin{pmatrix}0& p\
0 & r\ end{pmatrix} text{and}
B:=begin{pmatrix}0&0\
1 & q/pend{pmatrix}.tag{2}$$
Let us now consider the case $p=0$ that we divide into 2 subcases :
Case $p=0$ and $r neq -q$ : it suffices to take :
$$A:=tfrac{1}{2(q+r)}begin{pmatrix}r& -q\
r& -qend{pmatrix} text{and}
B:=begin{pmatrix} q& q\
-r&-rend{pmatrix}tag{3}$$
to get $$AB-BA=begin{pmatrix}0 & q\
r& 0end{pmatrix}.$$
- Case $p=0$ and $r=-q$ : take :
$$A:=begin{pmatrix}1& 1\
1 & 0\ end{pmatrix} text{and}
B:=begin{pmatrix}0&q\
q & 0end{pmatrix}.tag{4}$$
giving
$$AB-BA=begin{pmatrix} 0& q\
-q & 0\ end{pmatrix} .$$
Remark about the degrees of freedom we have for (2), (3) and (4) :
Out of relationship (2) for example (the same is true for (3) and (4)), it is possible to generate a lot of other solutions to (1). For example by taking, for any $a$ and any $b$ :
$$A:=begin{pmatrix}a& p\
0 & a+r\ end{pmatrix} text{and}
B:=begin{pmatrix}b&0\
1 & b+q/pend{pmatrix}.tag{5}$$
The idea behind (5) is in fact very natural : if in (1), for a given $C$, we have a solution $(A,B)$, then, for any $k$, $(A,B+kI_2)$ is as well a solution because we still have
$$A(B+kI_2)-(B+kI_2)A = C.$$
(and $(A+k'I_2,B)$ as well) i.e., we can add a multiple of the identity matrix, to any of the solution matrices without changing the result. More generally, we can add to $B$ any polynomial $p(A)$ in $A$ :
$$A(B+p(A))-(B+p(A))A = C,$$
and, of course, to $A$ any polynomial in $B$.
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
$endgroup$
add a comment |
$begingroup$
The issue is to give explicit solutions $(A,B)$ to relationship :
$$AB-BA=underbrace{begin{pmatrix}p & q\
r& -pend{pmatrix}}_Ctag{1}$$
We are going to consider 3 cases, from the most general to the most particular.
- Let us assume first $p neq 0$. Relationship (1) is verified by taking :
$$A:=begin{pmatrix}0& p\
0 & r\ end{pmatrix} text{and}
B:=begin{pmatrix}0&0\
1 & q/pend{pmatrix}.tag{2}$$
Let us now consider the case $p=0$ that we divide into 2 subcases :
Case $p=0$ and $r neq -q$ : it suffices to take :
$$A:=tfrac{1}{2(q+r)}begin{pmatrix}r& -q\
r& -qend{pmatrix} text{and}
B:=begin{pmatrix} q& q\
-r&-rend{pmatrix}tag{3}$$
to get $$AB-BA=begin{pmatrix}0 & q\
r& 0end{pmatrix}.$$
- Case $p=0$ and $r=-q$ : take :
$$A:=begin{pmatrix}1& 1\
1 & 0\ end{pmatrix} text{and}
B:=begin{pmatrix}0&q\
q & 0end{pmatrix}.tag{4}$$
giving
$$AB-BA=begin{pmatrix} 0& q\
-q & 0\ end{pmatrix} .$$
Remark about the degrees of freedom we have for (2), (3) and (4) :
Out of relationship (2) for example (the same is true for (3) and (4)), it is possible to generate a lot of other solutions to (1). For example by taking, for any $a$ and any $b$ :
$$A:=begin{pmatrix}a& p\
0 & a+r\ end{pmatrix} text{and}
B:=begin{pmatrix}b&0\
1 & b+q/pend{pmatrix}.tag{5}$$
The idea behind (5) is in fact very natural : if in (1), for a given $C$, we have a solution $(A,B)$, then, for any $k$, $(A,B+kI_2)$ is as well a solution because we still have
$$A(B+kI_2)-(B+kI_2)A = C.$$
(and $(A+k'I_2,B)$ as well) i.e., we can add a multiple of the identity matrix, to any of the solution matrices without changing the result. More generally, we can add to $B$ any polynomial $p(A)$ in $A$ :
$$A(B+p(A))-(B+p(A))A = C,$$
and, of course, to $A$ any polynomial in $B$.
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
$endgroup$
The issue is to give explicit solutions $(A,B)$ to relationship :
$$AB-BA=underbrace{begin{pmatrix}p & q\
r& -pend{pmatrix}}_Ctag{1}$$
We are going to consider 3 cases, from the most general to the most particular.
- Let us assume first $p neq 0$. Relationship (1) is verified by taking :
$$A:=begin{pmatrix}0& p\
0 & r\ end{pmatrix} text{and}
B:=begin{pmatrix}0&0\
1 & q/pend{pmatrix}.tag{2}$$
Let us now consider the case $p=0$ that we divide into 2 subcases :
Case $p=0$ and $r neq -q$ : it suffices to take :
$$A:=tfrac{1}{2(q+r)}begin{pmatrix}r& -q\
r& -qend{pmatrix} text{and}
B:=begin{pmatrix} q& q\
-r&-rend{pmatrix}tag{3}$$
to get $$AB-BA=begin{pmatrix}0 & q\
r& 0end{pmatrix}.$$
- Case $p=0$ and $r=-q$ : take :
$$A:=begin{pmatrix}1& 1\
1 & 0\ end{pmatrix} text{and}
B:=begin{pmatrix}0&q\
q & 0end{pmatrix}.tag{4}$$
giving
$$AB-BA=begin{pmatrix} 0& q\
-q & 0\ end{pmatrix} .$$
Remark about the degrees of freedom we have for (2), (3) and (4) :
Out of relationship (2) for example (the same is true for (3) and (4)), it is possible to generate a lot of other solutions to (1). For example by taking, for any $a$ and any $b$ :
$$A:=begin{pmatrix}a& p\
0 & a+r\ end{pmatrix} text{and}
B:=begin{pmatrix}b&0\
1 & b+q/pend{pmatrix}.tag{5}$$
The idea behind (5) is in fact very natural : if in (1), for a given $C$, we have a solution $(A,B)$, then, for any $k$, $(A,B+kI_2)$ is as well a solution because we still have
$$A(B+kI_2)-(B+kI_2)A = C.$$
(and $(A+k'I_2,B)$ as well) i.e., we can add a multiple of the identity matrix, to any of the solution matrices without changing the result. More generally, we can add to $B$ any polynomial $p(A)$ in $A$ :
$$A(B+p(A))-(B+p(A))A = C,$$
and, of course, to $A$ any polynomial in $B$.
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
edited Jan 10 at 9:38
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
answered Jan 9 at 20:06
Jean MarieJean Marie
29.5k42050
29.5k42050
add a comment |
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2
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The question you link states that this is true for any square matrix, if you take $c_{11}+c_{22}=0$ to mean $operatorname{Tr}(C)=0$.
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– Servaes
Jan 9 at 17:00
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Can you give any hint ?@Servaes
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– cmi
Jan 9 at 17:43
2
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In the specific case of $2times 2$ matrices, if $C$ has trace 0 then by Jordan normal form, $C$ must be similar either to $begin{bmatrix} lambda & 0 \ 0 & -lambda end{bmatrix}$ or to $begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix}$. Or, in case the scalar field has characteristic 2, then $begin{bmatrix} lambda & 1 \ 0 & lambda end{bmatrix}$ is also a possibility.
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– Daniel Schepler
Jan 9 at 18:12
2
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It might very well be that we can help you understand the previous Accepted Answer ("make" is perhaps too strong a word), but Readers will need to have you explain where you got stuck following that Answer. In other words, let's break down the "gap" in understanding so that any new explanation can be precisely directed to assist you.
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– hardmath
Jan 10 at 5:45
1
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@hardmath I agree with you. I think that the difficulty that the OP has to "understand the answer math.stackexchange.com/q/1045396 (that I will name Q hereafter) is well understandable because it was erroneous (see the comment I posted yesterday to the answer to A). This said, I didn't see a way to be didactic with the OP by exploiting A. I saw another direct explicit way that I wrote down ; in other cases, I try not to give a complete solution at once.
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– Jean Marie
Jan 10 at 8:53