For a $2 times 2$ matrix $C$, prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$.Solve the following systems of equations ( Matrices)If $A$ is an $mtimes n$ matrix, $B$ is an $ntimes m$ matrix and $n<m$, then $AB$ is not invertible.Linear Algebra Matrix and InverseDetermine if this matrix can be written as a linear combination of these matricesDo equal eigenvalues in a SS matrix imply algebraic equivalence?Prove that a square matrix can be expressed as a product of a diagonal and a permutation matrix.Bound on the dimension of sum of subspaces (Axler)Fundamental Matrix of LTV SystemCompare classical and modified algorithm of CholeskySolving a simple system of equations through matrix operations
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For a $2 times 2$ matrix $C$, prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$.
Solve the following systems of equations ( Matrices)If $A$ is an $mtimes n$ matrix, $B$ is an $ntimes m$ matrix and $n<m$, then $AB$ is not invertible.Linear Algebra Matrix and InverseDetermine if this matrix can be written as a linear combination of these matricesDo equal eigenvalues in a SS matrix imply algebraic equivalence?Prove that a square matrix can be expressed as a product of a diagonal and a permutation matrix.Bound on the dimension of sum of subspaces (Axler)Fundamental Matrix of LTV SystemCompare classical and modified algorithm of CholeskySolving a simple system of equations through matrix operations
$begingroup$
This question appears in Hoffman-Kunze Linear Algebra (exercise 1.5, question 8). The book has only introduced linear equations, elementary row operations, row-reduced echolon matrices, and matrix multiplication upto this point so the potential duplicate of this question using trace does not apply to this question.
The first part of the question asks "when it is possible to find $2 times 2$ matrices $A$ and $B$ such that $C = AB - BA$". I don't really understand the meaning of this query as there are no conditions placed on $C$ so is it okay to assume the answer is always as the product $AB$ and $BA$ always exists if $A$ and $B$ are $2 times 2$ matrices?
The second part states "prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$".
For this I did the following:
$$
textLet A =
beginbmatrix
a_11 & a_12 \
a_21 & a_22 \
endbmatrix;
B =
beginbmatrix
b_11 & b_12 \
b_21 & b_22 \
endbmatrix;
$$
$$
beginequationbeginaligned
AB =
beginbmatrix
a_11b_11+a_12b_21 & a_11b_12+a_12b_22 \
a_21b_11+a_22b_21 & a_21b_12+a_22b_22 \
endbmatrix;
endalignedendequationtag1labeleq1
$$
$$
beginequationbeginaligned
BA =
beginbmatrix
a_11b_11+a_21b_12 & a_12b_11+a_22b_12 \
a_11b_21+a_21b_22 & a_12b_21+a_22b_22 \
endbmatrix;
endalignedendequationtag2labeleq2
$$
So
$$
beginequationbeginaligned
C &=AB-BA\ &=
beginbmatrix
a_12b_21-a_21b_12 & a_11b_12+a_12b_22 - (a_12b_11+a_22b_12) \
a_21b_11+a_22b_21-(a_11b_21+a_21b_22) & a_21b_12-a_12b_21 \
endbmatrix;
endalignedendequationtag3labeleq3
$$
From this, it is obvious that $C_11 = -C_22 implies C_11 + C_22 = 0$.
But here (pg. 19) they go on to further solve $C=AB-BA$ by forming an augmented matrix of $AB-BA$ and coefficients of $C$ and solving that to show that $C_11 + C_22 = 0$. Can anyone please explain to me why this step was necessary when one can clearly see in $eqrefeq3$ that $C_11 + C_22 = 0$? To me it is obvious (or seems to be obvious in any case) that $eqrefeq3$ gives the general form of $C$ and all $2 times 2$ matrices $C$ such that $C=AB - BA$ will be of this form so why then go through the trouble of solving the augmented matrix to show that $C_11 +C_22 = 0$? What am I missing? Thanks.
linear-algebra matrices
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
$endgroup$
add a comment |
$begingroup$
This question appears in Hoffman-Kunze Linear Algebra (exercise 1.5, question 8). The book has only introduced linear equations, elementary row operations, row-reduced echolon matrices, and matrix multiplication upto this point so the potential duplicate of this question using trace does not apply to this question.
The first part of the question asks "when it is possible to find $2 times 2$ matrices $A$ and $B$ such that $C = AB - BA$". I don't really understand the meaning of this query as there are no conditions placed on $C$ so is it okay to assume the answer is always as the product $AB$ and $BA$ always exists if $A$ and $B$ are $2 times 2$ matrices?
The second part states "prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$".
For this I did the following:
$$
textLet A =
beginbmatrix
a_11 & a_12 \
a_21 & a_22 \
endbmatrix;
B =
beginbmatrix
b_11 & b_12 \
b_21 & b_22 \
endbmatrix;
$$
$$
beginequationbeginaligned
AB =
beginbmatrix
a_11b_11+a_12b_21 & a_11b_12+a_12b_22 \
a_21b_11+a_22b_21 & a_21b_12+a_22b_22 \
endbmatrix;
endalignedendequationtag1labeleq1
$$
$$
beginequationbeginaligned
BA =
beginbmatrix
a_11b_11+a_21b_12 & a_12b_11+a_22b_12 \
a_11b_21+a_21b_22 & a_12b_21+a_22b_22 \
endbmatrix;
endalignedendequationtag2labeleq2
$$
So
$$
beginequationbeginaligned
C &=AB-BA\ &=
beginbmatrix
a_12b_21-a_21b_12 & a_11b_12+a_12b_22 - (a_12b_11+a_22b_12) \
a_21b_11+a_22b_21-(a_11b_21+a_21b_22) & a_21b_12-a_12b_21 \
endbmatrix;
endalignedendequationtag3labeleq3
$$
From this, it is obvious that $C_11 = -C_22 implies C_11 + C_22 = 0$.
But here (pg. 19) they go on to further solve $C=AB-BA$ by forming an augmented matrix of $AB-BA$ and coefficients of $C$ and solving that to show that $C_11 + C_22 = 0$. Can anyone please explain to me why this step was necessary when one can clearly see in $eqrefeq3$ that $C_11 + C_22 = 0$? To me it is obvious (or seems to be obvious in any case) that $eqrefeq3$ gives the general form of $C$ and all $2 times 2$ matrices $C$ such that $C=AB - BA$ will be of this form so why then go through the trouble of solving the augmented matrix to show that $C_11 +C_22 = 0$? What am I missing? Thanks.
linear-algebra matrices
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
$endgroup$
$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59
add a comment |
$begingroup$
This question appears in Hoffman-Kunze Linear Algebra (exercise 1.5, question 8). The book has only introduced linear equations, elementary row operations, row-reduced echolon matrices, and matrix multiplication upto this point so the potential duplicate of this question using trace does not apply to this question.
The first part of the question asks "when it is possible to find $2 times 2$ matrices $A$ and $B$ such that $C = AB - BA$". I don't really understand the meaning of this query as there are no conditions placed on $C$ so is it okay to assume the answer is always as the product $AB$ and $BA$ always exists if $A$ and $B$ are $2 times 2$ matrices?
The second part states "prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$".
For this I did the following:
$$
textLet A =
beginbmatrix
a_11 & a_12 \
a_21 & a_22 \
endbmatrix;
B =
beginbmatrix
b_11 & b_12 \
b_21 & b_22 \
endbmatrix;
$$
$$
beginequationbeginaligned
AB =
beginbmatrix
a_11b_11+a_12b_21 & a_11b_12+a_12b_22 \
a_21b_11+a_22b_21 & a_21b_12+a_22b_22 \
endbmatrix;
endalignedendequationtag1labeleq1
$$
$$
beginequationbeginaligned
BA =
beginbmatrix
a_11b_11+a_21b_12 & a_12b_11+a_22b_12 \
a_11b_21+a_21b_22 & a_12b_21+a_22b_22 \
endbmatrix;
endalignedendequationtag2labeleq2
$$
So
$$
beginequationbeginaligned
C &=AB-BA\ &=
beginbmatrix
a_12b_21-a_21b_12 & a_11b_12+a_12b_22 - (a_12b_11+a_22b_12) \
a_21b_11+a_22b_21-(a_11b_21+a_21b_22) & a_21b_12-a_12b_21 \
endbmatrix;
endalignedendequationtag3labeleq3
$$
From this, it is obvious that $C_11 = -C_22 implies C_11 + C_22 = 0$.
But here (pg. 19) they go on to further solve $C=AB-BA$ by forming an augmented matrix of $AB-BA$ and coefficients of $C$ and solving that to show that $C_11 + C_22 = 0$. Can anyone please explain to me why this step was necessary when one can clearly see in $eqrefeq3$ that $C_11 + C_22 = 0$? To me it is obvious (or seems to be obvious in any case) that $eqrefeq3$ gives the general form of $C$ and all $2 times 2$ matrices $C$ such that $C=AB - BA$ will be of this form so why then go through the trouble of solving the augmented matrix to show that $C_11 +C_22 = 0$? What am I missing? Thanks.
linear-algebra matrices
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
$endgroup$
This question appears in Hoffman-Kunze Linear Algebra (exercise 1.5, question 8). The book has only introduced linear equations, elementary row operations, row-reduced echolon matrices, and matrix multiplication upto this point so the potential duplicate of this question using trace does not apply to this question.
The first part of the question asks "when it is possible to find $2 times 2$ matrices $A$ and $B$ such that $C = AB - BA$". I don't really understand the meaning of this query as there are no conditions placed on $C$ so is it okay to assume the answer is always as the product $AB$ and $BA$ always exists if $A$ and $B$ are $2 times 2$ matrices?
The second part states "prove that $C = AB - BA$ can only be found if and only if $C_11 + C_22 = 0$".
For this I did the following:
$$
textLet A =
beginbmatrix
a_11 & a_12 \
a_21 & a_22 \
endbmatrix;
B =
beginbmatrix
b_11 & b_12 \
b_21 & b_22 \
endbmatrix;
$$
$$
beginequationbeginaligned
AB =
beginbmatrix
a_11b_11+a_12b_21 & a_11b_12+a_12b_22 \
a_21b_11+a_22b_21 & a_21b_12+a_22b_22 \
endbmatrix;
endalignedendequationtag1labeleq1
$$
$$
beginequationbeginaligned
BA =
beginbmatrix
a_11b_11+a_21b_12 & a_12b_11+a_22b_12 \
a_11b_21+a_21b_22 & a_12b_21+a_22b_22 \
endbmatrix;
endalignedendequationtag2labeleq2
$$
So
$$
beginequationbeginaligned
C &=AB-BA\ &=
beginbmatrix
a_12b_21-a_21b_12 & a_11b_12+a_12b_22 - (a_12b_11+a_22b_12) \
a_21b_11+a_22b_21-(a_11b_21+a_21b_22) & a_21b_12-a_12b_21 \
endbmatrix;
endalignedendequationtag3labeleq3
$$
From this, it is obvious that $C_11 = -C_22 implies C_11 + C_22 = 0$.
But here (pg. 19) they go on to further solve $C=AB-BA$ by forming an augmented matrix of $AB-BA$ and coefficients of $C$ and solving that to show that $C_11 + C_22 = 0$. Can anyone please explain to me why this step was necessary when one can clearly see in $eqrefeq3$ that $C_11 + C_22 = 0$? To me it is obvious (or seems to be obvious in any case) that $eqrefeq3$ gives the general form of $C$ and all $2 times 2$ matrices $C$ such that $C=AB - BA$ will be of this form so why then go through the trouble of solving the augmented matrix to show that $C_11 +C_22 = 0$? What am I missing? Thanks.
linear-algebra matrices
linear-algebra matrices
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
edited Mar 22 at 2:51
qwerjkhaskdjf
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
asked Mar 22 at 2:20
qwerjkhaskdjfqwerjkhaskdjf
33
33
$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59
add a comment |
$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59
$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you're missing is that you were asked to prove that each statement implies the other. So if you're given a $2 times 2$ matrix $C$ with diagonal elements that add up to $0$, you're asked to prove that there are some matrices $A$ and $B$ such that $C=AB-BA$
Otherwise, even though you know that every matrix of the form $AB-BA$ has that property (the diagonal elements add to $0$), it's possible that there are some other matrices also having that property that can't be expressed as $AB-BA$. The second half of the proof rules out that possibility.
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
$endgroup$
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
What you're missing is that you were asked to prove that each statement implies the other. So if you're given a $2 times 2$ matrix $C$ with diagonal elements that add up to $0$, you're asked to prove that there are some matrices $A$ and $B$ such that $C=AB-BA$
Otherwise, even though you know that every matrix of the form $AB-BA$ has that property (the diagonal elements add to $0$), it's possible that there are some other matrices also having that property that can't be expressed as $AB-BA$. The second half of the proof rules out that possibility.
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
$endgroup$
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
add a comment |
$begingroup$
What you're missing is that you were asked to prove that each statement implies the other. So if you're given a $2 times 2$ matrix $C$ with diagonal elements that add up to $0$, you're asked to prove that there are some matrices $A$ and $B$ such that $C=AB-BA$
Otherwise, even though you know that every matrix of the form $AB-BA$ has that property (the diagonal elements add to $0$), it's possible that there are some other matrices also having that property that can't be expressed as $AB-BA$. The second half of the proof rules out that possibility.
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
$endgroup$
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
add a comment |
$begingroup$
What you're missing is that you were asked to prove that each statement implies the other. So if you're given a $2 times 2$ matrix $C$ with diagonal elements that add up to $0$, you're asked to prove that there are some matrices $A$ and $B$ such that $C=AB-BA$
Otherwise, even though you know that every matrix of the form $AB-BA$ has that property (the diagonal elements add to $0$), it's possible that there are some other matrices also having that property that can't be expressed as $AB-BA$. The second half of the proof rules out that possibility.
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
$endgroup$
What you're missing is that you were asked to prove that each statement implies the other. So if you're given a $2 times 2$ matrix $C$ with diagonal elements that add up to $0$, you're asked to prove that there are some matrices $A$ and $B$ such that $C=AB-BA$
Otherwise, even though you know that every matrix of the form $AB-BA$ has that property (the diagonal elements add to $0$), it's possible that there are some other matrices also having that property that can't be expressed as $AB-BA$. The second half of the proof rules out that possibility.
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
answered Mar 22 at 3:03
Robert ShoreRobert Shore
3,611324
3,611324
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
add a comment |
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
1
1
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
So just to make sure I have understood correctly, what I have done is prove that if $AB-BA$ exists, it satisfies $C$ but not that all matrices $C$ such that $C_11 + C_22 = 0$ can be expressed in the form $C=AB-BA$? So my working shows $C=AB-BA implies C_11 +C_22 = 0$ but not $C_11 +C_22 = 0 implies C=AB-BA$ when both implications are necessary for if and only if?
$endgroup$
– qwerjkhaskdjf
Mar 22 at 3:13
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
$endgroup$
– Robert Shore
Mar 22 at 3:16
$begingroup$
Yes, that's exactly right. The second half seeks to prove $C_11+C_22=0 Rightarrow exists A, B$ such that $C=AB-BA$.
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– Robert Shore
Mar 22 at 3:16
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$begingroup$
This holds not just for $2 times 2$-matrices, but for square matrices over any size. See William Kahan, Only Commutators Have Trace Zero.
$endgroup$
– darij grinberg
Mar 22 at 2:26
$begingroup$
Ah, thanks for the additional information.
$endgroup$
– qwerjkhaskdjf
Mar 22 at 2:59