Mixing
Decomposition of matrix into sum of product of vectors
$begingroup$
If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that
$$X = sum_{i=1}^nv_iv_i^T$$
I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!
linear-algebra eigenvalues-eigenvectors matrix-decomposition
edited Jan 25 at 20:53
mechanodroid
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that
$$X = sum_{i=1}^nv_iv_i^T$$
I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!
linear-algebra eigenvalues-eigenvectors matrix-decomposition
edited Jan 25 at 20:53
mechanodroid
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
$endgroup$
add a comment |
$begingroup$
If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that
$$X = sum_{i=1}^nv_iv_i^T$$
I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!
linear-algebra eigenvalues-eigenvectors matrix-decomposition
edited Jan 25 at 20:53
mechanodroid
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
$endgroup$
If I have an $n times n$ matrix $X$ that is symmetric and positive semi-definite, how do I prove that there exists vectors $v_1,ldots,v_n$ such that
$$X = sum_{i=1}^nv_iv_i^T$$
I know this is some form of factorization and know of how to find the eigenvectors of a matrix but am a little stuck on how to approach this. Any hints would be great!
linear-algebra eigenvalues-eigenvectors matrix-decomposition
linear-algebra eigenvalues-eigenvectors matrix-decomposition
edited Jan 25 at 20:53
mechanodroid
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
edited Jan 25 at 20:53
mechanodroid
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
edited Jan 25 at 20:53
mechanodroid
28.9k62548
edited Jan 25 at 20:53
mechanodroid
28.9k62548
edited Jan 25 at 20:53
mechanodroid
28.9k62548
28.9k62548
asked Jan 25 at 20:14
AnthonyAnthony
35519
asked Jan 25 at 20:14
AnthonyAnthony
35519
asked Jan 25 at 20:14
AnthonyAnthony
35519
35519
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.
Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.
Note that we have the equality
$$V = sum_{i=1}^n lambda_iP_i$$
Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
$$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
$endgroup$
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
add a comment |
$begingroup$
Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.
Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.
Note that we have the equality
$$V = sum_{i=1}^n lambda_iP_i$$
Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
$$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
$endgroup$
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
add a comment |
$begingroup$
Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.
Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.
Note that we have the equality
$$V = sum_{i=1}^n lambda_iP_i$$
Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
$$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
$endgroup$
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
add a comment |
$begingroup$
Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.
Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.
Note that we have the equality
$$V = sum_{i=1}^n lambda_iP_i$$
Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
$$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
$endgroup$
Since $V$ is symmetric and positive definite, there exists an orthonormal basis ${e_1, ldots e_n}$ such that $Ve_i = lambda_ie_i$ for some $lambda_i > 0$.
Denote $P_i$ the orthogonal projection onto $operatorname{span}{e_i}$, i.e. $P_ix = langle x, e_irangle e_i$.
Note that we have the equality
$$V = sum_{i=1}^n lambda_iP_i$$
Now verify that the matrix of $P_i$ w.r.t. the standard basis is $e_ie_i^T$. Hence
$$V = sum_{i=1}^n lambda_ie_ie_i^T = sum_{i=1}^n left(sqrt{lambda_i}e_iright)left(sqrt{lambda_i}e_iright)^T$$
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
edited Jan 26 at 21:45
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
answered Jan 25 at 20:50
mechanodroidmechanodroid
28.9k62548
28.9k62548
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
add a comment |
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Thank you! This makes perfect sense
$endgroup$
– Anthony
Jan 25 at 22:10
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
$begingroup$
Hi, sorry one question. Are the radicals placed correctly in your last line? Like is one of the eigenvectors not square-rooted?
$endgroup$
– Anthony
Jan 26 at 20:59
1
1
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
@Anthony It was a typo, thanks for noticing.
$endgroup$
– mechanodroid
Jan 26 at 21:45
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
$begingroup$
sorry one last question, $P_i$ is the orthogonal projection of what onto the span?
$endgroup$
– Anthony
Jan 27 at 16:51
1
1
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
$begingroup$
@Anthony $P_i$ is the linear map which sends a vector $x$ to its orthogonal projection onto the subspace $operatorname{span}{e_i}$, which is equal to $langle x, e_irangle e_i$. The matrix of $P_i$ is $e_ie_i^T$ because $$(e_ie_i^T)x = e_i(e_i^Tx) = e_ilangle x, e_irangle$$
$endgroup$
– mechanodroid
Jan 27 at 17:25
add a comment |
$begingroup$
Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
$endgroup$
add a comment |
$begingroup$
Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
$endgroup$
add a comment |
$begingroup$
Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
$endgroup$
Hint: The matrix being symmetric, there exists an orthonormal basis which consists of eigenvectors of the matrix (see Spectral theorem).
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
answered Jan 25 at 20:43
ScientificaScientifica
6,82941335
6,82941335
add a comment |
add a comment |
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