Find $min(operatorname{trace}(AA^T))$ for invertible $A_{ntimes n}$
3
$begingroup$
For invertible $A_{ntimes n}$ find $min(operatorname{trace}(AA^T))$
(a) $0$
(b) $1$
(c) $n$
(d) $n^2$
Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.
linear-algebra matrices trace
edited Jan 1 at 8:20
user26857
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
$endgroup$
|
show 2 more comments
3
$begingroup$
For invertible $A_{ntimes n}$ find $min(operatorname{trace}(AA^T))$
(a) $0$
(b) $1$
(c) $n$
(d) $n^2$
Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.
linear-algebra matrices trace
edited Jan 1 at 8:20
user26857
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
$endgroup$
2
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
1
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
1
$begingroup$
If the entries must be integers, $n$ is correct.
$endgroup$
– Will Jagy
May 12 '14 at 2:36
$begingroup$
Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
$endgroup$
– Morty
May 12 '14 at 2:47
|
show 2 more comments
3
3
3
5
$begingroup$
For invertible $A_{ntimes n}$ find $min(operatorname{trace}(AA^T))$
(a) $0$
(b) $1$
(c) $n$
(d) $n^2$
Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.
linear-algebra matrices trace
edited Jan 1 at 8:20
user26857
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
$endgroup$
For invertible $A_{ntimes n}$ find $min(operatorname{trace}(AA^T))$
(a) $0$
(b) $1$
(c) $n$
(d) $n^2$
Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it.
linear-algebra matrices trace
linear-algebra matrices trace
edited Jan 1 at 8:20
user26857
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
edited Jan 1 at 8:20
user26857
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
edited Jan 1 at 8:20
user26857
39.3k124183
edited Jan 1 at 8:20
user26857
39.3k124183
edited Jan 1 at 8:20
user26857
39.3k124183
39.3k124183
asked May 12 '14 at 2:30
MortyMorty
337111
asked May 12 '14 at 2:30
MortyMorty
337111
asked May 12 '14 at 2:30
MortyMorty
337111
337111
2
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
1
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
1
$begingroup$
If the entries must be integers, $n$ is correct.
$endgroup$
– Will Jagy
May 12 '14 at 2:36
$begingroup$
Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
$endgroup$
– Morty
May 12 '14 at 2:47
|
show 2 more comments
2
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
1
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
1
$begingroup$
If the entries must be integers, $n$ is correct.
$endgroup$
– Will Jagy
May 12 '14 at 2:36
$begingroup$
Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
$endgroup$
– Morty
May 12 '14 at 2:47
2
2
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
1
1
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
1
1
$begingroup$
If the entries must be integers, $n$ is correct.
$endgroup$
– Will Jagy
May 12 '14 at 2:36
$begingroup$
If the entries must be integers, $n$ is correct.
$endgroup$
– Will Jagy
May 12 '14 at 2:36
$begingroup$
Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
$endgroup$
– Morty
May 12 '14 at 2:47
$begingroup$
Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
$endgroup$
– Morty
May 12 '14 at 2:47
|
show 2 more comments
1 Answer
1
active
oldest
votes
7
$begingroup$
I will assume all the entries are integers. By multiplying out the matrices, you see that
$$ text{trace}(A^TA) = sum_{i,j} a_{i,j}^2 .$$
Suppose $text{trace}(A^TA) < n$. Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros. Hence $A$ is not invertible.
Hence $text{trace}(A^TA) ge n$. We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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oldest
votes
7
$begingroup$
I will assume all the entries are integers. By multiplying out the matrices, you see that
$$ text{trace}(A^TA) = sum_{i,j} a_{i,j}^2 .$$
Suppose $text{trace}(A^TA) < n$. Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros. Hence $A$ is not invertible.
Hence $text{trace}(A^TA) ge n$. We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
add a comment |
7
$begingroup$
I will assume all the entries are integers. By multiplying out the matrices, you see that
$$ text{trace}(A^TA) = sum_{i,j} a_{i,j}^2 .$$
Suppose $text{trace}(A^TA) < n$. Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros. Hence $A$ is not invertible.
Hence $text{trace}(A^TA) ge n$. We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
add a comment |
7
7
7
$begingroup$
I will assume all the entries are integers. By multiplying out the matrices, you see that
$$ text{trace}(A^TA) = sum_{i,j} a_{i,j}^2 .$$
Suppose $text{trace}(A^TA) < n$. Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros. Hence $A$ is not invertible.
Hence $text{trace}(A^TA) ge n$. We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
$endgroup$
I will assume all the entries are integers. By multiplying out the matrices, you see that
$$ text{trace}(A^TA) = sum_{i,j} a_{i,j}^2 .$$
Suppose $text{trace}(A^TA) < n$. Then at most $n-1$ entries must be non-zero, meaning that at least one column of the matrix is all zeros. Hence $A$ is not invertible.
Hence $text{trace}(A^TA) ge n$. We know that the value $n$ can be obtained (e.g. by the identity, or any permutation matrix).
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
answered May 12 '14 at 3:28
Stephen Montgomery-SmithStephen Montgomery-Smith
17.8k12247
17.8k12247
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
add a comment |
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Nice proof by the way. I recently discussed this question with my guide, it is a bit trickier than it seems. Since nothing is said about the entries in $A$, and the trace is the sum of inner products of $i$th row with itself (as Siddharth has pointed out), considering entries as integers, we get the answer to be $n$; considering entries are real, there doesn't exist any minimum (since we can get $epsilon$, $epsilon rightarrow 0$) and (here is the punch), considering complex entries, we can get $-infty$. So this question is better suited for a subjective answer than objective.
$endgroup$
– Morty
May 12 '14 at 8:03
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
Anyhow among the options $0$ can also be achieved (with complex entries). So it comes down to what the question setter thought of while framing this :).
$endgroup$
– Morty
May 12 '14 at 8:05
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@Morty, in what possible way doe this question have a «subjective answer»?! That the minimum depends on the field is a quite objective fact.
$endgroup$
– Mariano Suárez-Álvarez
May 14 '14 at 0:36
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
$begingroup$
@MarianoSuarez-Alvarez I meant that you can check a box since the field is not clear, and the option for the most general field, i.e. complex, is missing.
$endgroup$
– Morty
May 14 '14 at 4:55
add a comment |
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2
$begingroup$
Well if you take $varepsiloncdot I_n$ for small $varepsilon$ then you get a lower value, to begin with.
$endgroup$
– Ian Coley
May 12 '14 at 2:32
1
$begingroup$
and in that way you get all positive values for the function. Since it only takes positive values, this gives you the answer.
$endgroup$
– Mariano Suárez-Álvarez
May 12 '14 at 2:33
$begingroup$
If $A$ is real, there is no minimum value. As Mariano mentioned, if you compute the $i$th diagonal entry of $AA^{T}$, this is the norm squared of the $i$th row of $A$ and is hence positive. Thus, the trace is positive. But Ian's answer shows that you can make the trace as close to 0 as you like.
$endgroup$
– Siddharth Venkatesh
May 12 '14 at 2:35
1
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If the entries must be integers, $n$ is correct.
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– Will Jagy
May 12 '14 at 2:36
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Well thanks to all of you. This is a multiple choice question with options $0$,$1$, $n$ and $n^2$. But clearly $0$ cant be attained, but $0^+$ can be. I guess the question was set keeping integers in mind. @WillJagy But how do I prove it formally (for integral values)? I mean why cant I have $n-1$ or less?
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– Morty
May 12 '14 at 2:47