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1

$begingroup$

Let $a$ and $b$ be real numbers. Show that there exists a unique $2 times 2$
real symmetric matrix $A$ with $mathrm{trace}(A) = a$ and $det(A) = b$ if and only if $a^2=4b$.

linear-algebra

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edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

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put on hold as off-topic by Saad, José Carlos Santos, Arnaud D., Morgan Rodgers, onurcanbektas 20 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, Arnaud D., Morgan Rodgers, onurcanbektas

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm... Does $a^2 - 4b = 0$ remind you of anything? Perhaps if you swap around the names of the two letters? That's where I would look first if I were trying to solve this problem.
  $endgroup$
  – Arthur
  21 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – Brian S
  21 hours ago
  

  
  
  
  

add a comment | 

-2

1

$begingroup$

Let $a$ and $b$ be real numbers. Show that there exists a unique $2 times 2$
real symmetric matrix $A$ with $mathrm{trace}(A) = a$ and $det(A) = b$ if and only if $a^2=4b$.

linear-algebra

share|cite|improve this question

edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

parth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Saad, José Carlos Santos, Arnaud D., Morgan Rodgers, onurcanbektas 20 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, Arnaud D., Morgan Rodgers, onurcanbektas

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hmm... Does $a^2 - 4b = 0$ remind you of anything? Perhaps if you swap around the names of the two letters? That's where I would look first if I were trying to solve this problem.
  $endgroup$
  – Arthur
  21 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math.SE! Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
  $endgroup$
  – Brian S
  21 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $a$ and $b$ be real numbers. Show that there exists a unique $2 times 2$
real symmetric matrix $A$ with $mathrm{trace}(A) = a$ and $det(A) = b$ if and only if $a^2=4b$.

linear-algebra

share|cite|improve this question

edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

parth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

parth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

parth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 21 hours ago

Brian S

19610

edited 21 hours ago

Brian S

19610

asked 21 hours ago

parthparth

32

New contributor

parth is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 21 hours ago

parthparth

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1 Answer
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0

$begingroup$

Here is a non-sophisticated approach.

A $2 times 2$ non-symmetric real matrix is of the form

$$left(
begin{matrix}
p & q \
q & s \
end{matrix}
right)
$$
with $p,q,s in mathbb R$.

Its determinant is $ps-q^2=b,$ and its trace is $p+s=a$.

$q$ must be $0$, or else $-q$ would be another solution and it would not be unique.

We therefore have two equations in two unknowns: $ps=b$ and $p+s=a.$

Solving these involves a quadratic equation with discriminant $a^2-4b,$

so there is a unique solution iff $a^2=4b.$

share|cite|improve this answer

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

$endgroup$

add a comment | 

0

$begingroup$

Here is a non-sophisticated approach.

A $2 times 2$ non-symmetric real matrix is of the form

$$left(
begin{matrix}
p & q \
q & s \
end{matrix}
right)
$$
with $p,q,s in mathbb R$.

Its determinant is $ps-q^2=b,$ and its trace is $p+s=a$.

$q$ must be $0$, or else $-q$ would be another solution and it would not be unique.

We therefore have two equations in two unknowns: $ps=b$ and $p+s=a.$

Solving these involves a quadratic equation with discriminant $a^2-4b,$

so there is a unique solution iff $a^2=4b.$

share|cite|improve this answer

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

$endgroup$

add a comment | 

0

$begingroup$

Here is a non-sophisticated approach.

A $2 times 2$ non-symmetric real matrix is of the form

$$left(
begin{matrix}
p & q \
q & s \
end{matrix}
right)
$$
with $p,q,s in mathbb R$.

Its determinant is $ps-q^2=b,$ and its trace is $p+s=a$.

$q$ must be $0$, or else $-q$ would be another solution and it would not be unique.

We therefore have two equations in two unknowns: $ps=b$ and $p+s=a.$

Solving these involves a quadratic equation with discriminant $a^2-4b,$

so there is a unique solution iff $a^2=4b.$

share|cite|improve this answer

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

$endgroup$

add a comment | 

$begingroup$

Here is a non-sophisticated approach.

A $2 times 2$ non-symmetric real matrix is of the form

$$left(
begin{matrix}
p & q \
q & s \
end{matrix}
right)
$$
with $p,q,s in mathbb R$.

Its determinant is $ps-q^2=b,$ and its trace is $p+s=a$.

$q$ must be $0$, or else $-q$ would be another solution and it would not be unique.

We therefore have two equations in two unknowns: $ps=b$ and $p+s=a.$

Solving these involves a quadratic equation with discriminant $a^2-4b,$

so there is a unique solution iff $a^2=4b.$

share|cite|improve this answer

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

$endgroup$

share|cite|improve this answer

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217

answered 20 hours ago

J. W. TannerJ. W. Tanner

2,8671217