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Show that $A$ is similar to this matrix
Similarity of real matrices over $mathbb{C}$Finding the eigenvalues of a $3N times 3N$ block matrixCan I find the minimal polynomial by using the characteristic polynomial?A problem related to diagonalizable matrixCyclic subspace problemFinding the characteristic and minimal polynomials of this block matrix.The set of cyclic endomorphisms is openQuestion related to Unitary congruence and complex symmetric matricesCalculating the determinant of a matrix.Finding Jordan canonical form of a matrix given the characteristic polynomialcharacteristic polynomial of skew Hermitian matrix coefficients are real??
$begingroup$
Let $A in M_n(mathbb{R})$ such that $A^2 = -I_n$. Show that there exists a non singular $S in M_n(mathbb{R})$ such that
$$S^{-1}AS =
begin{bmatrix}
0 && -I_{frac{n}{2}} \
I_{frac{n}{2}} && 0 \
end{bmatrix}$$
So far what I have done is noted that $A^2 +I_n = 0$ and so $p(t)=t^2+1$ is its minimal polynomial. Therefore the eigenvalues of $A$ are ${i,-i}$. Thus the characteristic polynomial is
$$(x-i)^r(x+i)^s$$
such that $r+s=n$. Now if $r>s$ or $s>r$ the characteristic polynomial would not be real which is a contradiction. Thus $r=s$ so $2r=n$ which is to say that $n$ is even. Now by jordans real form we can find a nonsingular $X in M_n(mathbb{R})$ such that
$$X^{-1}AX = J_n(0)^2 + (J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$$
From here I have no idea how to get to the desired form, thanks in advance for any suggestions!
linear-algebra
edited Mar 12 at 22:59
mechanodroid
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
$endgroup$
add a comment |
$begingroup$
Let $A in M_n(mathbb{R})$ such that $A^2 = -I_n$. Show that there exists a non singular $S in M_n(mathbb{R})$ such that
$$S^{-1}AS =
begin{bmatrix}
0 && -I_{frac{n}{2}} \
I_{frac{n}{2}} && 0 \
end{bmatrix}$$
So far what I have done is noted that $A^2 +I_n = 0$ and so $p(t)=t^2+1$ is its minimal polynomial. Therefore the eigenvalues of $A$ are ${i,-i}$. Thus the characteristic polynomial is
$$(x-i)^r(x+i)^s$$
such that $r+s=n$. Now if $r>s$ or $s>r$ the characteristic polynomial would not be real which is a contradiction. Thus $r=s$ so $2r=n$ which is to say that $n$ is even. Now by jordans real form we can find a nonsingular $X in M_n(mathbb{R})$ such that
$$X^{-1}AX = J_n(0)^2 + (J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$$
From here I have no idea how to get to the desired form, thanks in advance for any suggestions!
linear-algebra
edited Mar 12 at 22:59
mechanodroid
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
$endgroup$
$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
$endgroup$
– copper.hat
Mar 12 at 19:46
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
$endgroup$
– Justin Stevenson
Mar 12 at 19:56
$begingroup$
Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
$endgroup$
– copper.hat
Mar 12 at 19:59
$begingroup$
I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
$endgroup$
– Justin Stevenson
Mar 12 at 21:57
$begingroup$
Only two hours?
$endgroup$
– copper.hat
Mar 12 at 22:18
add a comment |
$begingroup$
Let $A in M_n(mathbb{R})$ such that $A^2 = -I_n$. Show that there exists a non singular $S in M_n(mathbb{R})$ such that
$$S^{-1}AS =
begin{bmatrix}
0 && -I_{frac{n}{2}} \
I_{frac{n}{2}} && 0 \
end{bmatrix}$$
So far what I have done is noted that $A^2 +I_n = 0$ and so $p(t)=t^2+1$ is its minimal polynomial. Therefore the eigenvalues of $A$ are ${i,-i}$. Thus the characteristic polynomial is
$$(x-i)^r(x+i)^s$$
such that $r+s=n$. Now if $r>s$ or $s>r$ the characteristic polynomial would not be real which is a contradiction. Thus $r=s$ so $2r=n$ which is to say that $n$ is even. Now by jordans real form we can find a nonsingular $X in M_n(mathbb{R})$ such that
$$X^{-1}AX = J_n(0)^2 + (J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$$
From here I have no idea how to get to the desired form, thanks in advance for any suggestions!
linear-algebra
edited Mar 12 at 22:59
mechanodroid
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
$endgroup$
Let $A in M_n(mathbb{R})$ such that $A^2 = -I_n$. Show that there exists a non singular $S in M_n(mathbb{R})$ such that
$$S^{-1}AS =
begin{bmatrix}
0 && -I_{frac{n}{2}} \
I_{frac{n}{2}} && 0 \
end{bmatrix}$$
So far what I have done is noted that $A^2 +I_n = 0$ and so $p(t)=t^2+1$ is its minimal polynomial. Therefore the eigenvalues of $A$ are ${i,-i}$. Thus the characteristic polynomial is
$$(x-i)^r(x+i)^s$$
such that $r+s=n$. Now if $r>s$ or $s>r$ the characteristic polynomial would not be real which is a contradiction. Thus $r=s$ so $2r=n$ which is to say that $n$ is even. Now by jordans real form we can find a nonsingular $X in M_n(mathbb{R})$ such that
$$X^{-1}AX = J_n(0)^2 + (J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$$
From here I have no idea how to get to the desired form, thanks in advance for any suggestions!
linear-algebra
linear-algebra
edited Mar 12 at 22:59
mechanodroid
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
edited Mar 12 at 22:59
mechanodroid
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
edited Mar 12 at 22:59
mechanodroid
28.9k62548
edited Mar 12 at 22:59
mechanodroid
28.9k62548
edited Mar 12 at 22:59
mechanodroid
28.9k62548
28.9k62548
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
asked Mar 12 at 19:34
Justin StevensonJustin Stevenson
957519
957519
$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
$endgroup$
– copper.hat
Mar 12 at 19:46
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
$endgroup$
– Justin Stevenson
Mar 12 at 19:56
$begingroup$
Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
$endgroup$
– copper.hat
Mar 12 at 19:59
$begingroup$
I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
$endgroup$
– Justin Stevenson
Mar 12 at 21:57
$begingroup$
Only two hours?
$endgroup$
– copper.hat
Mar 12 at 22:18
add a comment |
$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
$endgroup$
– copper.hat
Mar 12 at 19:46
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
$endgroup$
– Justin Stevenson
Mar 12 at 19:56
$begingroup$
Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
$endgroup$
– copper.hat
Mar 12 at 19:59
$begingroup$
I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
$endgroup$
– Justin Stevenson
Mar 12 at 21:57
$begingroup$
Only two hours?
$endgroup$
– copper.hat
Mar 12 at 22:18
$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
$endgroup$
– copper.hat
Mar 12 at 19:46
$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
$endgroup$
– copper.hat
Mar 12 at 19:46
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
$endgroup$
– Justin Stevenson
Mar 12 at 19:56
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
$endgroup$
– Justin Stevenson
Mar 12 at 19:56
$begingroup$
Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
$endgroup$
– copper.hat
Mar 12 at 19:59
$begingroup$
Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
$endgroup$
– copper.hat
Mar 12 at 19:59
$begingroup$
I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
$endgroup$
– Justin Stevenson
Mar 12 at 21:57
$begingroup$
I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
$endgroup$
– Justin Stevenson
Mar 12 at 21:57
$begingroup$
Only two hours?
$endgroup$
– copper.hat
Mar 12 at 22:18
$begingroup$
Only two hours?
$endgroup$
– copper.hat
Mar 12 at 22:18
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First note that $n$ must be even as $det A^2 > 0$ and $det (-I) = (-1)^n$.
Since $A^2+I =0$ you know that the eigenvalues are $pm i$ and there is a (complex) basis such that $A$ is
diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From
this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the
vectors $u_1,...,u_{n over 2}, v_1,..., v_{n over 2}$ are linearly independent.
In this basis we see that $A$ has the desired form.
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
$endgroup$
add a comment |
$begingroup$
Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $mathbb{C}$.
You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{frac{n}2}(x-i)^{frac{n}2}$ so $A$ diagonalizes to
$$D = begin{bmatrix}
iI_{frac{n}{2}} & 0 \
0 & -iI_{frac{n}{2}} \
end{bmatrix}$$
Notice that the matrix $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $mathbb{C}$.
Hence, $A$ and $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ are similar over $mathbb{C}$. Since both are real, they are also similar over $mathbb{R}$.
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
You can give the $Bbb R$-vector space $Bbb R^n$ the structure of a complex
vector space by having the imaginary unit $defii{mathbf i}ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $Bbb C$-vector space is $n/2$. Now let $b_1,ldots,b_{n/2}$ be a basis of that $Bbb C$-vector space, and $b_{i+n/2}=ii,b_i$ (which means that $b_{i+n/2}=Acdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,ldots,n/2$. Then $b_1,ldots,b_n$ is an $Bbb R$-basis of$~Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
$endgroup$
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
$begingroup$
First note that $n$ must be even as $det A^2 > 0$ and $det (-I) = (-1)^n$.
Since $A^2+I =0$ you know that the eigenvalues are $pm i$ and there is a (complex) basis such that $A$ is
diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From
this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the
vectors $u_1,...,u_{n over 2}, v_1,..., v_{n over 2}$ are linearly independent.
In this basis we see that $A$ has the desired form.
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
$endgroup$
add a comment |
$begingroup$
First note that $n$ must be even as $det A^2 > 0$ and $det (-I) = (-1)^n$.
Since $A^2+I =0$ you know that the eigenvalues are $pm i$ and there is a (complex) basis such that $A$ is
diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From
this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the
vectors $u_1,...,u_{n over 2}, v_1,..., v_{n over 2}$ are linearly independent.
In this basis we see that $A$ has the desired form.
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
$endgroup$
add a comment |
$begingroup$
First note that $n$ must be even as $det A^2 > 0$ and $det (-I) = (-1)^n$.
Since $A^2+I =0$ you know that the eigenvalues are $pm i$ and there is a (complex) basis such that $A$ is
diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From
this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the
vectors $u_1,...,u_{n over 2}, v_1,..., v_{n over 2}$ are linearly independent.
In this basis we see that $A$ has the desired form.
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
$endgroup$
First note that $n$ must be even as $det A^2 > 0$ and $det (-I) = (-1)^n$.
Since $A^2+I =0$ you know that the eigenvalues are $pm i$ and there is a (complex) basis such that $A$ is
diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From
this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the
vectors $u_1,...,u_{n over 2}, v_1,..., v_{n over 2}$ are linearly independent.
In this basis we see that $A$ has the desired form.
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
answered Mar 12 at 22:44
copper.hatcopper.hat
128k559160
128k559160
add a comment |
add a comment |
$begingroup$
Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $mathbb{C}$.
You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{frac{n}2}(x-i)^{frac{n}2}$ so $A$ diagonalizes to
$$D = begin{bmatrix}
iI_{frac{n}{2}} & 0 \
0 & -iI_{frac{n}{2}} \
end{bmatrix}$$
Notice that the matrix $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $mathbb{C}$.
Hence, $A$ and $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ are similar over $mathbb{C}$. Since both are real, they are also similar over $mathbb{R}$.
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $mathbb{C}$.
You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{frac{n}2}(x-i)^{frac{n}2}$ so $A$ diagonalizes to
$$D = begin{bmatrix}
iI_{frac{n}{2}} & 0 \
0 & -iI_{frac{n}{2}} \
end{bmatrix}$$
Notice that the matrix $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $mathbb{C}$.
Hence, $A$ and $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ are similar over $mathbb{C}$. Since both are real, they are also similar over $mathbb{R}$.
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
$endgroup$
add a comment |
$begingroup$
Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $mathbb{C}$.
You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{frac{n}2}(x-i)^{frac{n}2}$ so $A$ diagonalizes to
$$D = begin{bmatrix}
iI_{frac{n}{2}} & 0 \
0 & -iI_{frac{n}{2}} \
end{bmatrix}$$
Notice that the matrix $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $mathbb{C}$.
Hence, $A$ and $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ are similar over $mathbb{C}$. Since both are real, they are also similar over $mathbb{R}$.
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
$endgroup$
Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $mathbb{C}$.
You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{frac{n}2}(x-i)^{frac{n}2}$ so $A$ diagonalizes to
$$D = begin{bmatrix}
iI_{frac{n}{2}} & 0 \
0 & -iI_{frac{n}{2}} \
end{bmatrix}$$
Notice that the matrix $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $mathbb{C}$.
Hence, $A$ and $begin{bmatrix}
0 & -I_{frac{n}{2}} \
I_{frac{n}{2}} & 0 \
end{bmatrix}$ are similar over $mathbb{C}$. Since both are real, they are also similar over $mathbb{R}$.
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
answered Mar 12 at 22:56
mechanodroidmechanodroid
28.9k62548
28.9k62548
add a comment |
add a comment |
$begingroup$
You can give the $Bbb R$-vector space $Bbb R^n$ the structure of a complex
vector space by having the imaginary unit $defii{mathbf i}ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $Bbb C$-vector space is $n/2$. Now let $b_1,ldots,b_{n/2}$ be a basis of that $Bbb C$-vector space, and $b_{i+n/2}=ii,b_i$ (which means that $b_{i+n/2}=Acdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,ldots,n/2$. Then $b_1,ldots,b_n$ is an $Bbb R$-basis of$~Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
$endgroup$
add a comment |
$begingroup$
You can give the $Bbb R$-vector space $Bbb R^n$ the structure of a complex
vector space by having the imaginary unit $defii{mathbf i}ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $Bbb C$-vector space is $n/2$. Now let $b_1,ldots,b_{n/2}$ be a basis of that $Bbb C$-vector space, and $b_{i+n/2}=ii,b_i$ (which means that $b_{i+n/2}=Acdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,ldots,n/2$. Then $b_1,ldots,b_n$ is an $Bbb R$-basis of$~Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
$endgroup$
add a comment |
$begingroup$
You can give the $Bbb R$-vector space $Bbb R^n$ the structure of a complex
vector space by having the imaginary unit $defii{mathbf i}ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $Bbb C$-vector space is $n/2$. Now let $b_1,ldots,b_{n/2}$ be a basis of that $Bbb C$-vector space, and $b_{i+n/2}=ii,b_i$ (which means that $b_{i+n/2}=Acdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,ldots,n/2$. Then $b_1,ldots,b_n$ is an $Bbb R$-basis of$~Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
$endgroup$
You can give the $Bbb R$-vector space $Bbb R^n$ the structure of a complex
vector space by having the imaginary unit $defii{mathbf i}ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $Bbb C$-vector space is $n/2$. Now let $b_1,ldots,b_{n/2}$ be a basis of that $Bbb C$-vector space, and $b_{i+n/2}=ii,b_i$ (which means that $b_{i+n/2}=Acdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,ldots,n/2$. Then $b_1,ldots,b_n$ is an $Bbb R$-basis of$~Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
answered Mar 13 at 7:42
Marc van LeeuwenMarc van Leeuwen
88.4k5111228
88.4k5111228
add a comment |
add a comment |
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$begingroup$
Since $A^2+I=0$ you know that the Jordan form has blocks of size 1. Since $A$ is real you know the eigenvalues occur in conjugate pairs.
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– copper.hat
Mar 12 at 19:46
$begingroup$
Right I overlooked that. So is that to say that my Jordan form is actually just: $(J_2(0) - J_2(0)^T) oplus dots oplus(J_2(0) - J_2(0)^T)$
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– Justin Stevenson
Mar 12 at 19:56
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Well, there is a (complex) set of eigenvectors that can be used to get a real matrix in the form desired.
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– copper.hat
Mar 12 at 19:59
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I've thought about it for the past 2 hours, I can't seem to figure out what the eigenvectors are and how to use them to get the desired form.
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– Justin Stevenson
Mar 12 at 21:57
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Only two hours?
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– copper.hat
Mar 12 at 22:18