Create a symmetric positive definite matrix (but not diagonal) with eigenvalues 1, 2, 4 - how to approach this problem?linear transformation of real symmetric square positive semi-definite matrixHow is the normalized trace defined?How to construct a $4 times 4$ symmetric, positive definite matrix with integer eigenvaluesProve that determinant of a 2x2 symmetric positive definite matrix is positive by “completing the square” method.Eigenvalues of [symmetric] matrix A whose product $AA$ is diagonalHow can I find two symmetric positive definite roots of a symmetric positive definite matrix?Numerically determine eigenvalues of real non-symmetric matrix known to have positive eigenvalues$LDL^top$ for symmetric positive semidefinite matrices that are not positive definiteI know symmetric matrix $S = QDQ^T$, but how can matrices with form ADA be symmetric?Show that $ S( a times b) +(Sa) times b + a times (Sb) =0 $
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Create a symmetric positive definite matrix (but not diagonal) with eigenvalues 1, 2, 4 - how to approach this problem?
linear transformation of real symmetric square positive semi-definite matrixHow is the normalized trace defined?How to construct a $4 times 4$ symmetric, positive definite matrix with integer eigenvaluesProve that determinant of a 2x2 symmetric positive definite matrix is positive by “completing the square” method.Eigenvalues of [symmetric] matrix A whose product $AA$ is diagonalHow can I find two symmetric positive definite roots of a symmetric positive definite matrix?Numerically determine eigenvalues of real non-symmetric matrix known to have positive eigenvalues$LDL^top$ for symmetric positive semidefinite matrices that are not positive definiteI know symmetric matrix $S = QDQ^T$, but how can matrices with form ADA be symmetric?Show that $ S( a times b) +(Sa) times b + a times (Sb) =0 $
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Trace of the matrix is obviously 7 and determinant 8, a 3x3 matrix should be easiest to construct, but where to go from here?Any brute force method simply won't work.
linear-algebra symmetric-matrices
$endgroup$
add a comment |
$begingroup$
Trace of the matrix is obviously 7 and determinant 8, a 3x3 matrix should be easiest to construct, but where to go from here?Any brute force method simply won't work.
linear-algebra symmetric-matrices
$endgroup$
add a comment |
$begingroup$
Trace of the matrix is obviously 7 and determinant 8, a 3x3 matrix should be easiest to construct, but where to go from here?Any brute force method simply won't work.
linear-algebra symmetric-matrices
$endgroup$
Trace of the matrix is obviously 7 and determinant 8, a 3x3 matrix should be easiest to construct, but where to go from here?Any brute force method simply won't work.
linear-algebra symmetric-matrices
linear-algebra symmetric-matrices
edited 8 hours ago


José Carlos Santos
198k25 gold badges157 silver badges275 bronze badges
198k25 gold badges157 silver badges275 bronze badges
asked 9 hours ago
user3711671user3711671
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808 bronze badges
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3 Answers
3
active
oldest
votes
$begingroup$
Hint: Choose an appropriate orthogonal matrix $M$ and take$$M^T.beginbmatrix1&0&0\0&2&0\0&0&4endbmatrix.M.$$
$endgroup$
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Why not pick any orthogonal transformation matrix $P$ and easily construct its inverse $P^T = P^-1$, with your matrix being $A = PDP^T$, where $D$ is the diagonal matrix of eigenvalues?
$endgroup$
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
add a comment |
$begingroup$
You can try with this block-diagonal matrix:
$$beginbmatrix frac32 & -frac12 & 0 \ -frac12 & frac32 & 0 \ 0 & 0 & 4endbmatrix$$
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Choose an appropriate orthogonal matrix $M$ and take$$M^T.beginbmatrix1&0&0\0&2&0\0&0&4endbmatrix.M.$$
$endgroup$
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Hint: Choose an appropriate orthogonal matrix $M$ and take$$M^T.beginbmatrix1&0&0\0&2&0\0&0&4endbmatrix.M.$$
$endgroup$
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Hint: Choose an appropriate orthogonal matrix $M$ and take$$M^T.beginbmatrix1&0&0\0&2&0\0&0&4endbmatrix.M.$$
$endgroup$
Hint: Choose an appropriate orthogonal matrix $M$ and take$$M^T.beginbmatrix1&0&0\0&2&0\0&0&4endbmatrix.M.$$
answered 8 hours ago


José Carlos SantosJosé Carlos Santos
198k25 gold badges157 silver badges275 bronze badges
198k25 gold badges157 silver badges275 bronze badges
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
How does one construct a random orthogonal matrix?
$endgroup$
– user3711671
8 hours ago
1
1
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Take a vector $u$ with norm $1$. Then a vector $v$ orthogonal to $u$, also with norm $1$. Now, take the matrix $M$ such that the entries of its first column are the coordinates of $u$, the entries of its second column are the coordinates of $v$, and the entries of its third column are the coordinates of $utimes v$.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Why does that method work?
$endgroup$
– user3711671
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
$begingroup$
Because then the columns of your matrix will be vectors with norm $1$ which will be orthogonal to each other. And a matrix is orthogonal if and only if this condition holds.
$endgroup$
– José Carlos Santos
8 hours ago
add a comment |
$begingroup$
Why not pick any orthogonal transformation matrix $P$ and easily construct its inverse $P^T = P^-1$, with your matrix being $A = PDP^T$, where $D$ is the diagonal matrix of eigenvalues?
$endgroup$
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
add a comment |
$begingroup$
Why not pick any orthogonal transformation matrix $P$ and easily construct its inverse $P^T = P^-1$, with your matrix being $A = PDP^T$, where $D$ is the diagonal matrix of eigenvalues?
$endgroup$
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
add a comment |
$begingroup$
Why not pick any orthogonal transformation matrix $P$ and easily construct its inverse $P^T = P^-1$, with your matrix being $A = PDP^T$, where $D$ is the diagonal matrix of eigenvalues?
$endgroup$
Why not pick any orthogonal transformation matrix $P$ and easily construct its inverse $P^T = P^-1$, with your matrix being $A = PDP^T$, where $D$ is the diagonal matrix of eigenvalues?
answered 8 hours ago
gt6989bgt6989b
38.3k2 gold badges25 silver badges57 bronze badges
38.3k2 gold badges25 silver badges57 bronze badges
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
add a comment |
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
What is orthogonal transformation matrix?How do you construct such random matrix?
$endgroup$
– user3711671
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
$begingroup$
@user3711671 Why not Google it or use Wikipedia?
$endgroup$
– gt6989b
8 hours ago
add a comment |
$begingroup$
You can try with this block-diagonal matrix:
$$beginbmatrix frac32 & -frac12 & 0 \ -frac12 & frac32 & 0 \ 0 & 0 & 4endbmatrix$$
$endgroup$
add a comment |
$begingroup$
You can try with this block-diagonal matrix:
$$beginbmatrix frac32 & -frac12 & 0 \ -frac12 & frac32 & 0 \ 0 & 0 & 4endbmatrix$$
$endgroup$
add a comment |
$begingroup$
You can try with this block-diagonal matrix:
$$beginbmatrix frac32 & -frac12 & 0 \ -frac12 & frac32 & 0 \ 0 & 0 & 4endbmatrix$$
$endgroup$
You can try with this block-diagonal matrix:
$$beginbmatrix frac32 & -frac12 & 0 \ -frac12 & frac32 & 0 \ 0 & 0 & 4endbmatrix$$
answered 8 hours ago
mechanodroidmechanodroid
30.3k6 gold badges27 silver badges48 bronze badges
30.3k6 gold badges27 silver badges48 bronze badges
add a comment |
add a comment |
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