Proving that $text{tr}^2(A) - 4det(A) geq 0$
I'm trying to figure out how to prove that when given a diagonal matrix
begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.
linear-algebra determinant
|
show 4 more comments
I'm trying to figure out how to prove that when given a diagonal matrix
begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.
linear-algebra determinant
3
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
1
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
1
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
2
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59
|
show 4 more comments
I'm trying to figure out how to prove that when given a diagonal matrix
begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.
linear-algebra determinant
I'm trying to figure out how to prove that when given a diagonal matrix
begin{align}
A=
begin{bmatrix}
lambda & 0\
0 & mu
end{bmatrix},
end{align}
with a positive determinant $det(A) > 0$, the following statement is true
$$text{tr}^2(A) - 4det(A) geq 0$$
Thank you in advance.
linear-algebra determinant
linear-algebra determinant
edited Nov 30 '18 at 11:43
Rebellos
14.5k31246
14.5k31246
asked Nov 30 '18 at 11:34
Jens KramerJens Kramer
557
557
3
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
1
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
1
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
2
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59
|
show 4 more comments
3
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
1
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
1
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
2
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59
3
3
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
1
1
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
1
1
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
2
2
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59
|
show 4 more comments
3 Answers
3
active
oldest
votes
More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.
add a comment |
Hint :
Simply using the definitions of $text{tr}$ and $det$ :
The trace of the matrix $A$, is :
$$text{tr}(A) = lambda + mu $$
The determinant of $A$, simply is :
$$det(A) = lambda mu$$
Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.
add a comment |
$begin{align}
tr(A)^2-4 det(A)geq 0
end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$.
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
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
More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.
add a comment |
More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.
add a comment |
More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.
More generally, the characteristic polynomial of a $2 times 2$ matrix $A$ is $x^2-operatorname{tr}(A)x + det(A)$. The roots of this polynomial are the eigenvalues of $A$ and so are real iff the discriminant $operatorname{tr}(A)^2 -4det(A)$ is nonnegative.
answered Nov 30 '18 at 12:07
lhflhf
163k10167388
163k10167388
add a comment |
add a comment |
Hint :
Simply using the definitions of $text{tr}$ and $det$ :
The trace of the matrix $A$, is :
$$text{tr}(A) = lambda + mu $$
The determinant of $A$, simply is :
$$det(A) = lambda mu$$
Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.
add a comment |
Hint :
Simply using the definitions of $text{tr}$ and $det$ :
The trace of the matrix $A$, is :
$$text{tr}(A) = lambda + mu $$
The determinant of $A$, simply is :
$$det(A) = lambda mu$$
Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.
add a comment |
Hint :
Simply using the definitions of $text{tr}$ and $det$ :
The trace of the matrix $A$, is :
$$text{tr}(A) = lambda + mu $$
The determinant of $A$, simply is :
$$det(A) = lambda mu$$
Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.
Hint :
Simply using the definitions of $text{tr}$ and $det$ :
The trace of the matrix $A$, is :
$$text{tr}(A) = lambda + mu $$
The determinant of $A$, simply is :
$$det(A) = lambda mu$$
Can you now follow a simple algebraic path to $text{tr}^2(A) - 4det(A)$ and conclude about its sign ? Use also the condition that as you stated, the determinant is positive.
answered Nov 30 '18 at 11:38
RebellosRebellos
14.5k31246
14.5k31246
add a comment |
add a comment |
$begin{align}
tr(A)^2-4 det(A)geq 0
end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$.
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
add a comment |
$begin{align}
tr(A)^2-4 det(A)geq 0
end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$.
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
add a comment |
$begin{align}
tr(A)^2-4 det(A)geq 0
end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$.
$begin{align}
tr(A)^2-4 det(A)geq 0
end{align} iff (lambda + mu)^2 ge 4 lambda mu iff lambda^2 -2 lambda mu + mu^2 ge 0 iff (lambda- mu)^2 ge 0$.
answered Nov 30 '18 at 11:38
FredFred
44.3k1845
44.3k1845
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
add a comment |
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
2
2
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
Simply elaborated, but why give a thorough solution to a simple problem of which the OP have showed zero attempts ?
– Rebellos
Nov 30 '18 at 11:39
3
3
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
Not leaving much for OP to do, Fred.
– Gerry Myerson
Nov 30 '18 at 11:42
add a comment |
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3
Have you tried writing the trace and determinant in terms of $lambda$ and $mu$ and then doing a little algebra?
– Gerry Myerson
Nov 30 '18 at 11:36
I have, I get $frac{lambda^2+mu^2-2lambdamu}{4}>lambdamu.$... I can't really figure what to do from here.
– Jens Kramer
Nov 30 '18 at 11:39
1
I think you've slipped up in your algebra. Also, that $lambda^2+mu^2-2lambdamu$ looks familiar....
– Gerry Myerson
Nov 30 '18 at 11:41
1
Algebra, algebra! Please, please, please check your algebra!
– Gerry Myerson
Nov 30 '18 at 11:49
2
The expansion of the quadratic term tr(A)^2.. So $lambda^2+mu^2+2lambdamugeq 4lambdamu$ which the implies $lambda^2+mu^2-2lambdamugeq 0$. Simplifying further gets $(lambda-mu)^2geq0$. I'm not quite sure what this is telling me.
– Jens Kramer
Nov 30 '18 at 11:59