True or False: Entries on the main diagonal of matrix A
$begingroup$
Q. If $A = [a_{ij}]$ is an $m times n$ matrix which satisfies $A^T = -A$, then the entries on the main diagonal of $A$ are all equal to $0$.
I don't see how $A^T = -A$ can be true for a $m times n$ matrix. Also, two matrices are only equal if they have the same size (dimensions) and have the same entries. If A is a m x n matrix, then so will -A. However, AT would be a n x m matrix, so $A^T neq -A$.
So I'm not sure on how to proceed with this question. Any help would be appreciated.
linear-algebra matrices
edited Jan 19 at 18:19
Omnomnomnom
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
$endgroup$
add a comment |
$begingroup$
Q. If $A = [a_{ij}]$ is an $m times n$ matrix which satisfies $A^T = -A$, then the entries on the main diagonal of $A$ are all equal to $0$.
I don't see how $A^T = -A$ can be true for a $m times n$ matrix. Also, two matrices are only equal if they have the same size (dimensions) and have the same entries. If A is a m x n matrix, then so will -A. However, AT would be a n x m matrix, so $A^T neq -A$.
So I'm not sure on how to proceed with this question. Any help would be appreciated.
linear-algebra matrices
edited Jan 19 at 18:19
Omnomnomnom
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
$endgroup$
$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
1
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24
add a comment |
$begingroup$
Q. If $A = [a_{ij}]$ is an $m times n$ matrix which satisfies $A^T = -A$, then the entries on the main diagonal of $A$ are all equal to $0$.
I don't see how $A^T = -A$ can be true for a $m times n$ matrix. Also, two matrices are only equal if they have the same size (dimensions) and have the same entries. If A is a m x n matrix, then so will -A. However, AT would be a n x m matrix, so $A^T neq -A$.
So I'm not sure on how to proceed with this question. Any help would be appreciated.
linear-algebra matrices
edited Jan 19 at 18:19
Omnomnomnom
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
$endgroup$
Q. If $A = [a_{ij}]$ is an $m times n$ matrix which satisfies $A^T = -A$, then the entries on the main diagonal of $A$ are all equal to $0$.
I don't see how $A^T = -A$ can be true for a $m times n$ matrix. Also, two matrices are only equal if they have the same size (dimensions) and have the same entries. If A is a m x n matrix, then so will -A. However, AT would be a n x m matrix, so $A^T neq -A$.
So I'm not sure on how to proceed with this question. Any help would be appreciated.
linear-algebra matrices
linear-algebra matrices
edited Jan 19 at 18:19
Omnomnomnom
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
edited Jan 19 at 18:19
Omnomnomnom
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
edited Jan 19 at 18:19
Omnomnomnom
128k791184
edited Jan 19 at 18:19
Omnomnomnom
128k791184
edited Jan 19 at 18:19
Omnomnomnom
128k791184
128k791184
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
asked Jan 19 at 18:05
PeraltaLearnsPeraltaLearns
205
205
$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
1
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24
add a comment |
$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
1
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24
$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
1
1
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As said in the comments, we need $m = n$ for this question to make sense. Because if not, then $A^T$ would have dimension $n space x space m$ which would equal the matrix $-A$ which would have dimension $m space x space n$. This is a contradiction since matrices of different sizes cannot equal each other.
Now, if we do assume the correct dimensions, note that when you transpose a matrix, the entries along the diagonal stay the same. So by our assumption, we have that for a diagonal entry $a_{ii}$, it must equal $-a_{ii}$. The only number to have such a property is $0$.
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
$endgroup$
add a comment |
$begingroup$
First, given a matrix $A$ of size $m$ by $n$, its transpose $A^T$ has size $n$ by $m$. So $A^T=-A$ implies $n=m$.
Further, $(A)_{ii}=(A^T)_{ii}=(-A)_{ii}=-A_{ii}$, for any $i$. I have used the definition of the transpose for the first equality (the coefficients on the diagonal don't change when you transpose a matrix), and your hypothesis on the second.
So you have that $A_{ii}=-A_{ii}$, which can only happen if $A_{ii}=0$. Meaning that your diagonal coefficents are all $0$.
answered Jan 19 at 18:39
J.FJ.F
33812
$endgroup$
add a comment |
$begingroup$
Firstly, for this question to make any sense (that is, that the statement $A^T=-A$ is meaningful) we must have $m=n$. Now let's examine the diagonal entries, $a_{ii}$. The $i, j$-th element of $A^t$ is the $j, i$-th element of $A$. In particular (for $i=j$) the diagonal entries of $A^t$ are the same as the diagonal entries of $A$. So for any $1le ile n$, we have $a_{ii}=-a_{ii}$, since $A^T=-A$, which implies $a_{ii}=0$, as desired.
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As said in the comments, we need $m = n$ for this question to make sense. Because if not, then $A^T$ would have dimension $n space x space m$ which would equal the matrix $-A$ which would have dimension $m space x space n$. This is a contradiction since matrices of different sizes cannot equal each other.
Now, if we do assume the correct dimensions, note that when you transpose a matrix, the entries along the diagonal stay the same. So by our assumption, we have that for a diagonal entry $a_{ii}$, it must equal $-a_{ii}$. The only number to have such a property is $0$.
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
$endgroup$
add a comment |
$begingroup$
As said in the comments, we need $m = n$ for this question to make sense. Because if not, then $A^T$ would have dimension $n space x space m$ which would equal the matrix $-A$ which would have dimension $m space x space n$. This is a contradiction since matrices of different sizes cannot equal each other.
Now, if we do assume the correct dimensions, note that when you transpose a matrix, the entries along the diagonal stay the same. So by our assumption, we have that for a diagonal entry $a_{ii}$, it must equal $-a_{ii}$. The only number to have such a property is $0$.
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
$endgroup$
add a comment |
$begingroup$
As said in the comments, we need $m = n$ for this question to make sense. Because if not, then $A^T$ would have dimension $n space x space m$ which would equal the matrix $-A$ which would have dimension $m space x space n$. This is a contradiction since matrices of different sizes cannot equal each other.
Now, if we do assume the correct dimensions, note that when you transpose a matrix, the entries along the diagonal stay the same. So by our assumption, we have that for a diagonal entry $a_{ii}$, it must equal $-a_{ii}$. The only number to have such a property is $0$.
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
$endgroup$
As said in the comments, we need $m = n$ for this question to make sense. Because if not, then $A^T$ would have dimension $n space x space m$ which would equal the matrix $-A$ which would have dimension $m space x space n$. This is a contradiction since matrices of different sizes cannot equal each other.
Now, if we do assume the correct dimensions, note that when you transpose a matrix, the entries along the diagonal stay the same. So by our assumption, we have that for a diagonal entry $a_{ii}$, it must equal $-a_{ii}$. The only number to have such a property is $0$.
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
edited Jan 19 at 18:48
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
answered Jan 19 at 18:43
Nicholas RobertsNicholas Roberts
117112
117112
add a comment |
add a comment |
$begingroup$
First, given a matrix $A$ of size $m$ by $n$, its transpose $A^T$ has size $n$ by $m$. So $A^T=-A$ implies $n=m$.
Further, $(A)_{ii}=(A^T)_{ii}=(-A)_{ii}=-A_{ii}$, for any $i$. I have used the definition of the transpose for the first equality (the coefficients on the diagonal don't change when you transpose a matrix), and your hypothesis on the second.
So you have that $A_{ii}=-A_{ii}$, which can only happen if $A_{ii}=0$. Meaning that your diagonal coefficents are all $0$.
answered Jan 19 at 18:39
J.FJ.F
33812
$endgroup$
add a comment |
$begingroup$
First, given a matrix $A$ of size $m$ by $n$, its transpose $A^T$ has size $n$ by $m$. So $A^T=-A$ implies $n=m$.
Further, $(A)_{ii}=(A^T)_{ii}=(-A)_{ii}=-A_{ii}$, for any $i$. I have used the definition of the transpose for the first equality (the coefficients on the diagonal don't change when you transpose a matrix), and your hypothesis on the second.
So you have that $A_{ii}=-A_{ii}$, which can only happen if $A_{ii}=0$. Meaning that your diagonal coefficents are all $0$.
answered Jan 19 at 18:39
J.FJ.F
33812
$endgroup$
add a comment |
$begingroup$
First, given a matrix $A$ of size $m$ by $n$, its transpose $A^T$ has size $n$ by $m$. So $A^T=-A$ implies $n=m$.
Further, $(A)_{ii}=(A^T)_{ii}=(-A)_{ii}=-A_{ii}$, for any $i$. I have used the definition of the transpose for the first equality (the coefficients on the diagonal don't change when you transpose a matrix), and your hypothesis on the second.
So you have that $A_{ii}=-A_{ii}$, which can only happen if $A_{ii}=0$. Meaning that your diagonal coefficents are all $0$.
answered Jan 19 at 18:39
J.FJ.F
33812
$endgroup$
First, given a matrix $A$ of size $m$ by $n$, its transpose $A^T$ has size $n$ by $m$. So $A^T=-A$ implies $n=m$.
Further, $(A)_{ii}=(A^T)_{ii}=(-A)_{ii}=-A_{ii}$, for any $i$. I have used the definition of the transpose for the first equality (the coefficients on the diagonal don't change when you transpose a matrix), and your hypothesis on the second.
So you have that $A_{ii}=-A_{ii}$, which can only happen if $A_{ii}=0$. Meaning that your diagonal coefficents are all $0$.
answered Jan 19 at 18:39
J.FJ.F
33812
answered Jan 19 at 18:39
J.FJ.F
33812
answered Jan 19 at 18:39
J.FJ.F
33812
answered Jan 19 at 18:39
J.FJ.F
33812
33812
add a comment |
add a comment |
$begingroup$
Firstly, for this question to make any sense (that is, that the statement $A^T=-A$ is meaningful) we must have $m=n$. Now let's examine the diagonal entries, $a_{ii}$. The $i, j$-th element of $A^t$ is the $j, i$-th element of $A$. In particular (for $i=j$) the diagonal entries of $A^t$ are the same as the diagonal entries of $A$. So for any $1le ile n$, we have $a_{ii}=-a_{ii}$, since $A^T=-A$, which implies $a_{ii}=0$, as desired.
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
$endgroup$
add a comment |
$begingroup$
Firstly, for this question to make any sense (that is, that the statement $A^T=-A$ is meaningful) we must have $m=n$. Now let's examine the diagonal entries, $a_{ii}$. The $i, j$-th element of $A^t$ is the $j, i$-th element of $A$. In particular (for $i=j$) the diagonal entries of $A^t$ are the same as the diagonal entries of $A$. So for any $1le ile n$, we have $a_{ii}=-a_{ii}$, since $A^T=-A$, which implies $a_{ii}=0$, as desired.
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
$endgroup$
add a comment |
$begingroup$
Firstly, for this question to make any sense (that is, that the statement $A^T=-A$ is meaningful) we must have $m=n$. Now let's examine the diagonal entries, $a_{ii}$. The $i, j$-th element of $A^t$ is the $j, i$-th element of $A$. In particular (for $i=j$) the diagonal entries of $A^t$ are the same as the diagonal entries of $A$. So for any $1le ile n$, we have $a_{ii}=-a_{ii}$, since $A^T=-A$, which implies $a_{ii}=0$, as desired.
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
$endgroup$
Firstly, for this question to make any sense (that is, that the statement $A^T=-A$ is meaningful) we must have $m=n$. Now let's examine the diagonal entries, $a_{ii}$. The $i, j$-th element of $A^t$ is the $j, i$-th element of $A$. In particular (for $i=j$) the diagonal entries of $A^t$ are the same as the diagonal entries of $A$. So for any $1le ile n$, we have $a_{ii}=-a_{ii}$, since $A^T=-A$, which implies $a_{ii}=0$, as desired.
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
answered Jan 19 at 18:40
Yuval GatYuval Gat
570213
570213
add a comment |
add a comment |
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$begingroup$
For your condition to have any sort of meaning, you have to have $m=n$.
$endgroup$
– J.F
Jan 19 at 18:06
$begingroup$
Is that because A^T = -A only when A is a n x n matrix?
$endgroup$
– PeraltaLearns
Jan 19 at 18:10
1
$begingroup$
It is because $2$ matrices are equal if they have the same size AND the same coefficients. Very much like two functions are equal if they have the same domains, codomains and the same values.
$endgroup$
– J.F
Jan 19 at 18:12
$begingroup$
Ok given the facts I think this is true, but I'm not sure on how to go about and explain it. I think it has to do with the fact that for A^T you are turning rows into columns, and for -A you are scaling all entries by -1. Overall, the dimensions of the matrices are equal but the entries (coefficients) are not and so A^T != -A, but this changes if you let the main diagonal of A be 0's.
$endgroup$
– PeraltaLearns
Jan 19 at 18:24