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Did author omit trace in this problem statement?
Symplectic structures on Hermitian matricesIs it possible to compute row and column sums of $A^{-1}$ given row and column sums of $A$?Kronecker Products and Powers notationRow vectors with basis vectors as their elements.Linear algebra exercise on the vanishing of certain sums of tracesInvariant sum of matrix entries squared under any orthonormal basisLinear combination of column vectorsMust the following correlation matrix have a nonzero eigenvalue $le 1$?Identity matrix when collecting column vector$hat{Y} = X^That{beta}$ Matrix Dimension For Linear Regression Coefficients $beta$
$begingroup$
I have this problem:
Prove that the following identity is true:
$$boldsymbol{A} boldsymbol{A^T} = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
I assumed the notation $a_i$ means ith column vector of the matrix $boldsymbol{A}$.
I attempted to solve it and I believe the author of the problem meant to prove that:
$$ text{tr}( boldsymbol{A} boldsymbol{A^T}) = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
where $text{tr}$ is trace of a matrix. Otherwise, I get an incorrect answer.
Am I correct?
linear-algebra matrices trace
asked 15 hours ago
MattMatt
17418
$endgroup$
add a comment |
$begingroup$
I have this problem:
Prove that the following identity is true:
$$boldsymbol{A} boldsymbol{A^T} = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
I assumed the notation $a_i$ means ith column vector of the matrix $boldsymbol{A}$.
I attempted to solve it and I believe the author of the problem meant to prove that:
$$ text{tr}( boldsymbol{A} boldsymbol{A^T}) = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
where $text{tr}$ is trace of a matrix. Otherwise, I get an incorrect answer.
Am I correct?
linear-algebra matrices trace
asked 15 hours ago
MattMatt
17418
$endgroup$
$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
1
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago
add a comment |
$begingroup$
I have this problem:
Prove that the following identity is true:
$$boldsymbol{A} boldsymbol{A^T} = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
I assumed the notation $a_i$ means ith column vector of the matrix $boldsymbol{A}$.
I attempted to solve it and I believe the author of the problem meant to prove that:
$$ text{tr}( boldsymbol{A} boldsymbol{A^T}) = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
where $text{tr}$ is trace of a matrix. Otherwise, I get an incorrect answer.
Am I correct?
linear-algebra matrices trace
asked 15 hours ago
MattMatt
17418
$endgroup$
I have this problem:
Prove that the following identity is true:
$$boldsymbol{A} boldsymbol{A^T} = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
I assumed the notation $a_i$ means ith column vector of the matrix $boldsymbol{A}$.
I attempted to solve it and I believe the author of the problem meant to prove that:
$$ text{tr}( boldsymbol{A} boldsymbol{A^T}) = sum_{i} boldsymbol{a_i} boldsymbol{a_i^T}$$
where $text{tr}$ is trace of a matrix. Otherwise, I get an incorrect answer.
Am I correct?
linear-algebra matrices trace
linear-algebra matrices trace
asked 15 hours ago
MattMatt
17418
asked 15 hours ago
MattMatt
17418
edited 15 hours ago
Matt
asked 15 hours ago
MattMatt
17418
asked 15 hours ago
MattMatt
17418
asked 15 hours ago
MattMatt
17418
17418
$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
1
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago
add a comment |
$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
1
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago
$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
1
1
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With the current equation ($AA^{T} = sumlimits_{i}a_{i}a_{i}^{T}$ where $a_{i}$ is the $i$-th column of $A$), everything is fine. This is a correct equation (note that if $A$ is $mtimes n$, then $a_{i}a_{i}^{T}$ is a matrix of size $mtimes m$, as is $AA^{T}$, so the dimensions match up).
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
$endgroup$
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
add a comment |
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1 Answer
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votes
$begingroup$
With the current equation ($AA^{T} = sumlimits_{i}a_{i}a_{i}^{T}$ where $a_{i}$ is the $i$-th column of $A$), everything is fine. This is a correct equation (note that if $A$ is $mtimes n$, then $a_{i}a_{i}^{T}$ is a matrix of size $mtimes m$, as is $AA^{T}$, so the dimensions match up).
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
$endgroup$
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
add a comment |
$begingroup$
With the current equation ($AA^{T} = sumlimits_{i}a_{i}a_{i}^{T}$ where $a_{i}$ is the $i$-th column of $A$), everything is fine. This is a correct equation (note that if $A$ is $mtimes n$, then $a_{i}a_{i}^{T}$ is a matrix of size $mtimes m$, as is $AA^{T}$, so the dimensions match up).
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
$endgroup$
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
add a comment |
$begingroup$
With the current equation ($AA^{T} = sumlimits_{i}a_{i}a_{i}^{T}$ where $a_{i}$ is the $i$-th column of $A$), everything is fine. This is a correct equation (note that if $A$ is $mtimes n$, then $a_{i}a_{i}^{T}$ is a matrix of size $mtimes m$, as is $AA^{T}$, so the dimensions match up).
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
$endgroup$
With the current equation ($AA^{T} = sumlimits_{i}a_{i}a_{i}^{T}$ where $a_{i}$ is the $i$-th column of $A$), everything is fine. This is a correct equation (note that if $A$ is $mtimes n$, then $a_{i}a_{i}^{T}$ is a matrix of size $mtimes m$, as is $AA^{T}$, so the dimensions match up).
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
answered 15 hours ago
Minus One-TwelfthMinus One-Twelfth
1,93719
1,93719
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
add a comment |
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
Thanks, there was a mistake in my reasoning, I will attempt to solve it again.
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
$begingroup$
I solved it successfully, it was rather easy, I just messed up the dimensions as you suggested. Thanks for help again!
$endgroup$
– Matt
15 hours ago
add a comment |
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$begingroup$
$newcommand{tr}{operatorname{tr}}$I don't think they would have meant to show that $trleft(AA^{T}right) = sumlimits_{i} color{red}{a_{i}a_{i}^{T}}$, because the left-hand side is a scalar but the right-hand side is a matrix. But if you make the right-hand side summand instead be $a_{i}^{T}a_{i}$ (OR if you take $a_{i}$ as the $i$-th row of $A$), then the equation would be correct.
$endgroup$
– Minus One-Twelfth
15 hours ago
$begingroup$
Another possibility is that the $color{red}>$ sign is a pure typo and should not be there; if that is the case, then perhaps they are asking you to show that $color{blue}{AA^{T} = sumlimits_{i} a_{i}a_{i}^{T}}$, where $a_{i}$ is the $i$-th column of $A$ (this is in fact true as well).
$endgroup$
– Minus One-Twelfth
15 hours ago
1
$begingroup$
The > sign is my typo, I surrounded the expression in block quotes for nicer formatting which I shouldn't do because it messed up the expression by adding the > sign.
$endgroup$
– Matt
15 hours ago