Mixing
Uniqueness of a $2times 2$ real symmetric matrix under certain conditions.
$begingroup$
Let $a$ and $b$ be real numbers. We are to show that there exists a unique $2times 2$ real symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ if and only if $a^2=4b$.
Now I know that if $lambda_1$ and $lambda_2$ are eigenvalues of a real $2times 2$ symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ then $lambda_1+lambda_2=a$ and $lambda_1lambda_2=b$. But I am absolutely lost while proving uniqueness when $a^2=4b$. I am also unable to use the property that $A$ is symmetric. Please help. This is a Masters level entrance examination question.
linear-algebra
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
$endgroup$
add a comment |
$begingroup$
Let $a$ and $b$ be real numbers. We are to show that there exists a unique $2times 2$ real symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ if and only if $a^2=4b$.
Now I know that if $lambda_1$ and $lambda_2$ are eigenvalues of a real $2times 2$ symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ then $lambda_1+lambda_2=a$ and $lambda_1lambda_2=b$. But I am absolutely lost while proving uniqueness when $a^2=4b$. I am also unable to use the property that $A$ is symmetric. Please help. This is a Masters level entrance examination question.
linear-algebra
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
$endgroup$
add a comment |
$begingroup$
Let $a$ and $b$ be real numbers. We are to show that there exists a unique $2times 2$ real symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ if and only if $a^2=4b$.
Now I know that if $lambda_1$ and $lambda_2$ are eigenvalues of a real $2times 2$ symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ then $lambda_1+lambda_2=a$ and $lambda_1lambda_2=b$. But I am absolutely lost while proving uniqueness when $a^2=4b$. I am also unable to use the property that $A$ is symmetric. Please help. This is a Masters level entrance examination question.
linear-algebra
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
$endgroup$
Let $a$ and $b$ be real numbers. We are to show that there exists a unique $2times 2$ real symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ if and only if $a^2=4b$.
Now I know that if $lambda_1$ and $lambda_2$ are eigenvalues of a real $2times 2$ symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ then $lambda_1+lambda_2=a$ and $lambda_1lambda_2=b$. But I am absolutely lost while proving uniqueness when $a^2=4b$. I am also unable to use the property that $A$ is symmetric. Please help. This is a Masters level entrance examination question.
linear-algebra
linear-algebra
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
edited Apr 26 '18 at 14:25
Adrian Keister
5,26971933
5,26971933
asked Apr 26 '18 at 14:06
NeelNeel
636
asked Apr 26 '18 at 14:06
NeelNeel
636
asked Apr 26 '18 at 14:06
NeelNeel
636
636
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.
$A^2-text{tr}(A)A+det(A)I=0$
With $a^2=b/4$ we have
$A^2-a A+(a^2/4)I=0$
$4A^2- 4aA +a^2I=(2A-aI)^2=0$
Now we know that $A$ is symmetric so $2A-aI$.
Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.
Solution is not unique.. there are plenty of such symmetric matrices type $cI$.
But if $a$ is specified there is of course only one matrix of type $cI$.
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
$endgroup$
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
add a comment |
$begingroup$
Here is a simpler but not enlightening approach. Let $A=pmatrix{x&y\ y&z}$. For the on "only if" part, note that $B=pmatrix{z&-y\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $operatorname{tr}(A)^2=4det(A)$, what are the implications for the values of $x,y,z$?
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
$endgroup$
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
add a comment |
$begingroup$
Hint: Since $A$ is symmetric, it is diagonalizable.
Write
$$A=PDP^{-1}$$
with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
$endgroup$
add a comment |
$begingroup$
The "only if" part (That is, we are considering that $A$ is unique)
To Show: $a^2=4b$ (under the conditions given)
Consider $A$ to be a real symmetric matrix such that $$ A = left[ begin{array}{cccc}
alpha_1 & alpha_2 \
alpha_2 & alpha_3 \
end{array} right] $$
($alpha_1,alpha_2,alpha_3,alpha_4 $ are all real)
Now, $trace(A)=lambda_1+lambda_3=a$
$|A|=alpha_1timesalpha_3-{alpha_2}^2=b$
If, $lambda_1,lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form
$(lambda-lambda_1)(lambda-lambda_1)=0$
$Rightarrow lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=0$
$Rightarrow lambda^2-tr(A)lambda+|A|=0$
$Rightarrow lambda^2-alambda+ b=0$
By Cayley-Hamilton Theorem,
$A^2-tr(A)lambda+|A|=0$
$Rightarrow A^2-(lambda_1+lambda_2)A+|A|=0$
$Rightarrow (A-lambda_1I)(A-lambda_2I)=0$
Since $A$ is unique we must have $lambda_1=lambda_2$
That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$Rightarrow a^2=4b$
The if part:
Suppose $a^2=4b$
To show, $A$ is unique
Now, solving the characteristic equation, we have
$lambda^2-tr(A)lambda+|A|=0$
$lambda=frac{a pm sqrt{a^2-4b}}{2}$
So, $lambda_1= frac{a + sqrt{a^2-4b}}{2}$
$lambda_2= frac{a - sqrt{a^2-4b}}{2}$
But, since $a^2=4b$, so
$lambda_1=frac{a}{2}=lambda_2$
Again, $A=lambda_1I$, or $A=lambda_2I$
Since, $lambda_1=lambda_2=frac{a}{2}$,
$A=frac{a}{2}I$
$Rightarrow$ $A$ is unique
$endgroup$
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.
$A^2-text{tr}(A)A+det(A)I=0$
With $a^2=b/4$ we have
$A^2-a A+(a^2/4)I=0$
$4A^2- 4aA +a^2I=(2A-aI)^2=0$
Now we know that $A$ is symmetric so $2A-aI$.
Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.
Solution is not unique.. there are plenty of such symmetric matrices type $cI$.
But if $a$ is specified there is of course only one matrix of type $cI$.
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
$endgroup$
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
add a comment |
$begingroup$
We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.
$A^2-text{tr}(A)A+det(A)I=0$
With $a^2=b/4$ we have
$A^2-a A+(a^2/4)I=0$
$4A^2- 4aA +a^2I=(2A-aI)^2=0$
Now we know that $A$ is symmetric so $2A-aI$.
Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.
Solution is not unique.. there are plenty of such symmetric matrices type $cI$.
But if $a$ is specified there is of course only one matrix of type $cI$.
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
$endgroup$
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
add a comment |
$begingroup$
We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.
$A^2-text{tr}(A)A+det(A)I=0$
With $a^2=b/4$ we have
$A^2-a A+(a^2/4)I=0$
$4A^2- 4aA +a^2I=(2A-aI)^2=0$
Now we know that $A$ is symmetric so $2A-aI$.
Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.
Solution is not unique.. there are plenty of such symmetric matrices type $cI$.
But if $a$ is specified there is of course only one matrix of type $cI$.
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
$endgroup$
We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.
$A^2-text{tr}(A)A+det(A)I=0$
With $a^2=b/4$ we have
$A^2-a A+(a^2/4)I=0$
$4A^2- 4aA +a^2I=(2A-aI)^2=0$
Now we know that $A$ is symmetric so $2A-aI$.
Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.
Solution is not unique.. there are plenty of such symmetric matrices type $cI$.
But if $a$ is specified there is of course only one matrix of type $cI$.
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
edited Apr 26 '18 at 17:25
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
answered Apr 26 '18 at 16:41
WidawensenWidawensen
4,77831447
4,77831447
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
add a comment |
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
I am having a slight doubt. For a symmetric matrix $A$ with $operatorname{tr}(A)=a$ and $det(A)=b$ that satisfies $a^2=4b$, the solution for $A$ provided by you becomes unique as $a$ and $b$ are specified.
$endgroup$
– Neel
Apr 26 '18 at 17:17
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
in question $a^2 =4 b$. This is satisfied generally by matrix $cI$. Check for example for $2I$ or $3I$, but if $a$ is specified then $cI$ is unique.
$endgroup$
– Widawensen
Apr 26 '18 at 17:28
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
You are right!!
$endgroup$
– Neel
Apr 26 '18 at 17:29
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
$begingroup$
@Neel I'm glad that we have solution :)
$endgroup$
– Widawensen
Apr 26 '18 at 17:30
add a comment |
$begingroup$
Here is a simpler but not enlightening approach. Let $A=pmatrix{x&y\ y&z}$. For the on "only if" part, note that $B=pmatrix{z&-y\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $operatorname{tr}(A)^2=4det(A)$, what are the implications for the values of $x,y,z$?
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
$endgroup$
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
add a comment |
$begingroup$
Here is a simpler but not enlightening approach. Let $A=pmatrix{x&y\ y&z}$. For the on "only if" part, note that $B=pmatrix{z&-y\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $operatorname{tr}(A)^2=4det(A)$, what are the implications for the values of $x,y,z$?
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
$endgroup$
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
add a comment |
$begingroup$
Here is a simpler but not enlightening approach. Let $A=pmatrix{x&y\ y&z}$. For the on "only if" part, note that $B=pmatrix{z&-y\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $operatorname{tr}(A)^2=4det(A)$, what are the implications for the values of $x,y,z$?
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
$endgroup$
Here is a simpler but not enlightening approach. Let $A=pmatrix{x&y\ y&z}$. For the on "only if" part, note that $B=pmatrix{z&-y\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $operatorname{tr}(A)^2=4det(A)$, what are the implications for the values of $x,y,z$?
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
answered Apr 26 '18 at 16:08
user1551user1551
74k566129
74k566129
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
add a comment |
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
I am getting something absurd as follows: $$(x-z)^2=-4y^2$$
$endgroup$
– Neel
Apr 26 '18 at 16:41
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
$begingroup$
@Neel You have received not absurd but $y=0$ and $x=z$, the same you can get from my solution..
$endgroup$
– Widawensen
Apr 26 '18 at 16:58
add a comment |
$begingroup$
Hint: Since $A$ is symmetric, it is diagonalizable.
Write
$$A=PDP^{-1}$$
with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
$endgroup$
add a comment |
$begingroup$
Hint: Since $A$ is symmetric, it is diagonalizable.
Write
$$A=PDP^{-1}$$
with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
$endgroup$
add a comment |
$begingroup$
Hint: Since $A$ is symmetric, it is diagonalizable.
Write
$$A=PDP^{-1}$$
with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
$endgroup$
Hint: Since $A$ is symmetric, it is diagonalizable.
Write
$$A=PDP^{-1}$$
with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
answered Apr 26 '18 at 14:10
N. S.N. S.
105k7114210
105k7114210
add a comment |
add a comment |
$begingroup$
The "only if" part (That is, we are considering that $A$ is unique)
To Show: $a^2=4b$ (under the conditions given)
Consider $A$ to be a real symmetric matrix such that $$ A = left[ begin{array}{cccc}
alpha_1 & alpha_2 \
alpha_2 & alpha_3 \
end{array} right] $$
($alpha_1,alpha_2,alpha_3,alpha_4 $ are all real)
Now, $trace(A)=lambda_1+lambda_3=a$
$|A|=alpha_1timesalpha_3-{alpha_2}^2=b$
If, $lambda_1,lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form
$(lambda-lambda_1)(lambda-lambda_1)=0$
$Rightarrow lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=0$
$Rightarrow lambda^2-tr(A)lambda+|A|=0$
$Rightarrow lambda^2-alambda+ b=0$
By Cayley-Hamilton Theorem,
$A^2-tr(A)lambda+|A|=0$
$Rightarrow A^2-(lambda_1+lambda_2)A+|A|=0$
$Rightarrow (A-lambda_1I)(A-lambda_2I)=0$
Since $A$ is unique we must have $lambda_1=lambda_2$
That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$Rightarrow a^2=4b$
The if part:
Suppose $a^2=4b$
To show, $A$ is unique
Now, solving the characteristic equation, we have
$lambda^2-tr(A)lambda+|A|=0$
$lambda=frac{a pm sqrt{a^2-4b}}{2}$
So, $lambda_1= frac{a + sqrt{a^2-4b}}{2}$
$lambda_2= frac{a - sqrt{a^2-4b}}{2}$
But, since $a^2=4b$, so
$lambda_1=frac{a}{2}=lambda_2$
Again, $A=lambda_1I$, or $A=lambda_2I$
Since, $lambda_1=lambda_2=frac{a}{2}$,
$A=frac{a}{2}I$
$Rightarrow$ $A$ is unique
$endgroup$
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
add a comment |
$begingroup$
The "only if" part (That is, we are considering that $A$ is unique)
To Show: $a^2=4b$ (under the conditions given)
Consider $A$ to be a real symmetric matrix such that $$ A = left[ begin{array}{cccc}
alpha_1 & alpha_2 \
alpha_2 & alpha_3 \
end{array} right] $$
($alpha_1,alpha_2,alpha_3,alpha_4 $ are all real)
Now, $trace(A)=lambda_1+lambda_3=a$
$|A|=alpha_1timesalpha_3-{alpha_2}^2=b$
If, $lambda_1,lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form
$(lambda-lambda_1)(lambda-lambda_1)=0$
$Rightarrow lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=0$
$Rightarrow lambda^2-tr(A)lambda+|A|=0$
$Rightarrow lambda^2-alambda+ b=0$
By Cayley-Hamilton Theorem,
$A^2-tr(A)lambda+|A|=0$
$Rightarrow A^2-(lambda_1+lambda_2)A+|A|=0$
$Rightarrow (A-lambda_1I)(A-lambda_2I)=0$
Since $A$ is unique we must have $lambda_1=lambda_2$
That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$Rightarrow a^2=4b$
The if part:
Suppose $a^2=4b$
To show, $A$ is unique
Now, solving the characteristic equation, we have
$lambda^2-tr(A)lambda+|A|=0$
$lambda=frac{a pm sqrt{a^2-4b}}{2}$
So, $lambda_1= frac{a + sqrt{a^2-4b}}{2}$
$lambda_2= frac{a - sqrt{a^2-4b}}{2}$
But, since $a^2=4b$, so
$lambda_1=frac{a}{2}=lambda_2$
Again, $A=lambda_1I$, or $A=lambda_2I$
Since, $lambda_1=lambda_2=frac{a}{2}$,
$A=frac{a}{2}I$
$Rightarrow$ $A$ is unique
$endgroup$
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
add a comment |
$begingroup$
The "only if" part (That is, we are considering that $A$ is unique)
To Show: $a^2=4b$ (under the conditions given)
Consider $A$ to be a real symmetric matrix such that $$ A = left[ begin{array}{cccc}
alpha_1 & alpha_2 \
alpha_2 & alpha_3 \
end{array} right] $$
($alpha_1,alpha_2,alpha_3,alpha_4 $ are all real)
Now, $trace(A)=lambda_1+lambda_3=a$
$|A|=alpha_1timesalpha_3-{alpha_2}^2=b$
If, $lambda_1,lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form
$(lambda-lambda_1)(lambda-lambda_1)=0$
$Rightarrow lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=0$
$Rightarrow lambda^2-tr(A)lambda+|A|=0$
$Rightarrow lambda^2-alambda+ b=0$
By Cayley-Hamilton Theorem,
$A^2-tr(A)lambda+|A|=0$
$Rightarrow A^2-(lambda_1+lambda_2)A+|A|=0$
$Rightarrow (A-lambda_1I)(A-lambda_2I)=0$
Since $A$ is unique we must have $lambda_1=lambda_2$
That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$Rightarrow a^2=4b$
The if part:
Suppose $a^2=4b$
To show, $A$ is unique
Now, solving the characteristic equation, we have
$lambda^2-tr(A)lambda+|A|=0$
$lambda=frac{a pm sqrt{a^2-4b}}{2}$
So, $lambda_1= frac{a + sqrt{a^2-4b}}{2}$
$lambda_2= frac{a - sqrt{a^2-4b}}{2}$
But, since $a^2=4b$, so
$lambda_1=frac{a}{2}=lambda_2$
Again, $A=lambda_1I$, or $A=lambda_2I$
Since, $lambda_1=lambda_2=frac{a}{2}$,
$A=frac{a}{2}I$
$Rightarrow$ $A$ is unique
$endgroup$
The "only if" part (That is, we are considering that $A$ is unique)
To Show: $a^2=4b$ (under the conditions given)
Consider $A$ to be a real symmetric matrix such that $$ A = left[ begin{array}{cccc}
alpha_1 & alpha_2 \
alpha_2 & alpha_3 \
end{array} right] $$
($alpha_1,alpha_2,alpha_3,alpha_4 $ are all real)
Now, $trace(A)=lambda_1+lambda_3=a$
$|A|=alpha_1timesalpha_3-{alpha_2}^2=b$
If, $lambda_1,lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form
$(lambda-lambda_1)(lambda-lambda_1)=0$
$Rightarrow lambda^2-(lambda_1+lambda_2)lambda+lambda_1lambda_2=0$
$Rightarrow lambda^2-tr(A)lambda+|A|=0$
$Rightarrow lambda^2-alambda+ b=0$
By Cayley-Hamilton Theorem,
$A^2-tr(A)lambda+|A|=0$
$Rightarrow A^2-(lambda_1+lambda_2)A+|A|=0$
$Rightarrow (A-lambda_1I)(A-lambda_2I)=0$
Since $A$ is unique we must have $lambda_1=lambda_2$
That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$Rightarrow a^2=4b$
The if part:
Suppose $a^2=4b$
To show, $A$ is unique
Now, solving the characteristic equation, we have
$lambda^2-tr(A)lambda+|A|=0$
$lambda=frac{a pm sqrt{a^2-4b}}{2}$
So, $lambda_1= frac{a + sqrt{a^2-4b}}{2}$
$lambda_2= frac{a - sqrt{a^2-4b}}{2}$
But, since $a^2=4b$, so
$lambda_1=frac{a}{2}=lambda_2$
Again, $A=lambda_1I$, or $A=lambda_2I$
Since, $lambda_1=lambda_2=frac{a}{2}$,
$A=frac{a}{2}I$
$Rightarrow$ $A$ is unique
edited Jan 30 at 17:38
answered Apr 27 '18 at 14:02
user440191
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
add a comment |
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
1
1
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
$begingroup$
Your answer is a lot more complete, I congratulate you for your good work..!!
$endgroup$
– Neel
Apr 28 '18 at 7:59
add a comment |
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