Mixing
Prove that $A^2 + B^2 = O_2$ given conditions
$begingroup$
Show that, given $A,B$ second order matrices with real entries, such that $AB = BA$, $det{(A + iB)} = 0$ and $4 det{A} >( text{tr}{A} )^2$, then $A^2 + B^2 = O_2$.
My progress:
Considering the polynomial $det(A + xB)$, since $i$ is a root, $-i$ is also a root, and thus $det(A + xB) = x^2 + 1$. From this we deduce that $det(A) = det(B) = 1$.
Since $AB = BA$, we have that $det(A^2 + B^2) = det(A + iB) det(A - iB) = 0$.
Since we know that $det(A^2 - B^2) = 4$, we can also deduce the form of the polynomial $det(A^2 + xB^2)$, but that doesn't seem to help.
I know that the condition $4 det{A} >( text{tr}{A} )^2$ implies that $A$ has two distinct complex eigenvalues, but I don't know how to use that.
Any ideas?
linear-algebra
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
$endgroup$
add a comment |
$begingroup$
Show that, given $A,B$ second order matrices with real entries, such that $AB = BA$, $det{(A + iB)} = 0$ and $4 det{A} >( text{tr}{A} )^2$, then $A^2 + B^2 = O_2$.
My progress:
Considering the polynomial $det(A + xB)$, since $i$ is a root, $-i$ is also a root, and thus $det(A + xB) = x^2 + 1$. From this we deduce that $det(A) = det(B) = 1$.
Since $AB = BA$, we have that $det(A^2 + B^2) = det(A + iB) det(A - iB) = 0$.
Since we know that $det(A^2 - B^2) = 4$, we can also deduce the form of the polynomial $det(A^2 + xB^2)$, but that doesn't seem to help.
I know that the condition $4 det{A} >( text{tr}{A} )^2$ implies that $A$ has two distinct complex eigenvalues, but I don't know how to use that.
Any ideas?
linear-algebra
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
$endgroup$
$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51
add a comment |
$begingroup$
Show that, given $A,B$ second order matrices with real entries, such that $AB = BA$, $det{(A + iB)} = 0$ and $4 det{A} >( text{tr}{A} )^2$, then $A^2 + B^2 = O_2$.
My progress:
Considering the polynomial $det(A + xB)$, since $i$ is a root, $-i$ is also a root, and thus $det(A + xB) = x^2 + 1$. From this we deduce that $det(A) = det(B) = 1$.
Since $AB = BA$, we have that $det(A^2 + B^2) = det(A + iB) det(A - iB) = 0$.
Since we know that $det(A^2 - B^2) = 4$, we can also deduce the form of the polynomial $det(A^2 + xB^2)$, but that doesn't seem to help.
I know that the condition $4 det{A} >( text{tr}{A} )^2$ implies that $A$ has two distinct complex eigenvalues, but I don't know how to use that.
Any ideas?
linear-algebra
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
$endgroup$
Show that, given $A,B$ second order matrices with real entries, such that $AB = BA$, $det{(A + iB)} = 0$ and $4 det{A} >( text{tr}{A} )^2$, then $A^2 + B^2 = O_2$.
My progress:
Considering the polynomial $det(A + xB)$, since $i$ is a root, $-i$ is also a root, and thus $det(A + xB) = x^2 + 1$. From this we deduce that $det(A) = det(B) = 1$.
Since $AB = BA$, we have that $det(A^2 + B^2) = det(A + iB) det(A - iB) = 0$.
Since we know that $det(A^2 - B^2) = 4$, we can also deduce the form of the polynomial $det(A^2 + xB^2)$, but that doesn't seem to help.
I know that the condition $4 det{A} >( text{tr}{A} )^2$ implies that $A$ has two distinct complex eigenvalues, but I don't know how to use that.
Any ideas?
linear-algebra
linear-algebra
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
asked Jan 17 at 1:20
Davidmath7Davidmath7
915
915
$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51
add a comment |
$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51
$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51
$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51
add a comment |
2 Answers
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$begingroup$
$det(A+iB)=0$ implies that there exists a non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
$endgroup$
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$begingroup$
The condition $4det A>(operatorname{tr} A)^2$ implies that $x^2-(operatorname{tr} A)x+det A$, the characteristic polynomial of $A$, has not any real root. Therefore $A$ is non-singular. Let $X=A^{-1}B$. Since $AB=BA$, the statements $A^2+B^2=0$ and $I+X^2=0$ are equivalent. Also, $det(A+iB)=0$ implies that $det(I+X^2)=|det(I+iX)|^2=0$. Thus it suffices to prove that
$$
I+X^2=0,text{ if },Xin M_2(mathbb R),text{ and },I+X^2,text{ is singular}.
$$
Suppose the premises hold. Then $(I+X^2)u=0$ for some nonzero real vector $u$. If you can show that $u$ and $Xu$ are linearly independent and $(I+X^2)(Xu)=0$, you are done because $I+X^2=0$ now maps both basis vectors $u$ and $Xu$ of $mathbb R^2$ to zero.
answered Jan 17 at 6:55
user1551user1551
72.9k566128
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$det(A+iB)=0$ implies that there exists a non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
$endgroup$
add a comment |
$begingroup$
$det(A+iB)=0$ implies that there exists a non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
$endgroup$
add a comment |
$begingroup$
$det(A+iB)=0$ implies that there exists a non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
$endgroup$
$det(A+iB)=0$ implies that there exists a non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
answered Jan 17 at 1:52
Tsemo AristideTsemo Aristide
58.6k11445
58.6k11445
add a comment |
add a comment |
$begingroup$
The condition $4det A>(operatorname{tr} A)^2$ implies that $x^2-(operatorname{tr} A)x+det A$, the characteristic polynomial of $A$, has not any real root. Therefore $A$ is non-singular. Let $X=A^{-1}B$. Since $AB=BA$, the statements $A^2+B^2=0$ and $I+X^2=0$ are equivalent. Also, $det(A+iB)=0$ implies that $det(I+X^2)=|det(I+iX)|^2=0$. Thus it suffices to prove that
$$
I+X^2=0,text{ if },Xin M_2(mathbb R),text{ and },I+X^2,text{ is singular}.
$$
Suppose the premises hold. Then $(I+X^2)u=0$ for some nonzero real vector $u$. If you can show that $u$ and $Xu$ are linearly independent and $(I+X^2)(Xu)=0$, you are done because $I+X^2=0$ now maps both basis vectors $u$ and $Xu$ of $mathbb R^2$ to zero.
answered Jan 17 at 6:55
user1551user1551
72.9k566128
$endgroup$
add a comment |
$begingroup$
The condition $4det A>(operatorname{tr} A)^2$ implies that $x^2-(operatorname{tr} A)x+det A$, the characteristic polynomial of $A$, has not any real root. Therefore $A$ is non-singular. Let $X=A^{-1}B$. Since $AB=BA$, the statements $A^2+B^2=0$ and $I+X^2=0$ are equivalent. Also, $det(A+iB)=0$ implies that $det(I+X^2)=|det(I+iX)|^2=0$. Thus it suffices to prove that
$$
I+X^2=0,text{ if },Xin M_2(mathbb R),text{ and },I+X^2,text{ is singular}.
$$
Suppose the premises hold. Then $(I+X^2)u=0$ for some nonzero real vector $u$. If you can show that $u$ and $Xu$ are linearly independent and $(I+X^2)(Xu)=0$, you are done because $I+X^2=0$ now maps both basis vectors $u$ and $Xu$ of $mathbb R^2$ to zero.
answered Jan 17 at 6:55
user1551user1551
72.9k566128
$endgroup$
add a comment |
$begingroup$
The condition $4det A>(operatorname{tr} A)^2$ implies that $x^2-(operatorname{tr} A)x+det A$, the characteristic polynomial of $A$, has not any real root. Therefore $A$ is non-singular. Let $X=A^{-1}B$. Since $AB=BA$, the statements $A^2+B^2=0$ and $I+X^2=0$ are equivalent. Also, $det(A+iB)=0$ implies that $det(I+X^2)=|det(I+iX)|^2=0$. Thus it suffices to prove that
$$
I+X^2=0,text{ if },Xin M_2(mathbb R),text{ and },I+X^2,text{ is singular}.
$$
Suppose the premises hold. Then $(I+X^2)u=0$ for some nonzero real vector $u$. If you can show that $u$ and $Xu$ are linearly independent and $(I+X^2)(Xu)=0$, you are done because $I+X^2=0$ now maps both basis vectors $u$ and $Xu$ of $mathbb R^2$ to zero.
answered Jan 17 at 6:55
user1551user1551
72.9k566128
$endgroup$
The condition $4det A>(operatorname{tr} A)^2$ implies that $x^2-(operatorname{tr} A)x+det A$, the characteristic polynomial of $A$, has not any real root. Therefore $A$ is non-singular. Let $X=A^{-1}B$. Since $AB=BA$, the statements $A^2+B^2=0$ and $I+X^2=0$ are equivalent. Also, $det(A+iB)=0$ implies that $det(I+X^2)=|det(I+iX)|^2=0$. Thus it suffices to prove that
$$
I+X^2=0,text{ if },Xin M_2(mathbb R),text{ and },I+X^2,text{ is singular}.
$$
Suppose the premises hold. Then $(I+X^2)u=0$ for some nonzero real vector $u$. If you can show that $u$ and $Xu$ are linearly independent and $(I+X^2)(Xu)=0$, you are done because $I+X^2=0$ now maps both basis vectors $u$ and $Xu$ of $mathbb R^2$ to zero.
answered Jan 17 at 6:55
user1551user1551
72.9k566128
answered Jan 17 at 6:55
user1551user1551
72.9k566128
answered Jan 17 at 6:55
user1551user1551
72.9k566128
answered Jan 17 at 6:55
user1551user1551
72.9k566128
72.9k566128
add a comment |
add a comment |
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$begingroup$
Your argument doesn't sound right. All conditions involved here are homogeneous. If they are true for $A$ and $B$, they are also true for $cA$ and $cB$ for any $c>0$. Therefore, you cannot argue that $det(A)=det(B)=1$ without imposing any additional conditions.
$endgroup$
– user1551
Jan 17 at 6:51