Mixing
A very interesting property of symmetric positive definite matrix. Need proof! (Citation needed)
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let P be a symmetric positive definite matrix. Without losing generality, assume P is of dimension three, i.e.,
$$mathbfP = beginbmatrixP_1&P_2&P_3\P_2&P_4&P_5\P_3&P_5&P_6 endbmatrix $$.
Let I be the identity matrix of the same dimension as P. Let
$$mathbfI_1 = beginbmatrix1 & 0 & 0\0 & 0 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_2 = beginbmatrix0 & 0 & 0\0 & 1 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_3 = beginbmatrix0 & 0 & 0\0 & 0 & 0\ 0 & 0 & 1endbmatrix.$$
Then, $$lambda((mathbfI+mathbfPmathbfI_1)^n
(mathbfI+mathbfPmathbfI_2)^n(mathbfI+mathbfPmathbfI_3)^n) >1 forall ngeq 0,$$ where $lambda$ is an arbitrary eigenvalue of the matrix in the parentheses.
The observation above is actually valid for all symmetric positive definite matrices of all dimensions. This conclusion is arrived by running more than 20,000 random matrix tests, i.e., more than 20,000 random P matrix is generated.
I tried to prove this property by using the properties of positive definite matrix and by connecting the two facts that
- $lambda(mathbfI+P)>1$
- $mathbfI+mathbfPmathbfI_1+mathbfI+mathbfPmathbfI_2+mathbfI+mathbfPmathbfI_3 = 3mathbfI+mathbfP$.
However, I did not have any progress. Please share me with your idea once you find any approach which may probably prove that.
matrices spectral-theory
asked Jul 21 at 2:42
He Li
61
add a comment |Â
up vote
1
down vote
favorite
Let P be a symmetric positive definite matrix. Without losing generality, assume P is of dimension three, i.e.,
$$mathbfP = beginbmatrixP_1&P_2&P_3\P_2&P_4&P_5\P_3&P_5&P_6 endbmatrix $$.
Let I be the identity matrix of the same dimension as P. Let
$$mathbfI_1 = beginbmatrix1 & 0 & 0\0 & 0 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_2 = beginbmatrix0 & 0 & 0\0 & 1 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_3 = beginbmatrix0 & 0 & 0\0 & 0 & 0\ 0 & 0 & 1endbmatrix.$$
Then, $$lambda((mathbfI+mathbfPmathbfI_1)^n
(mathbfI+mathbfPmathbfI_2)^n(mathbfI+mathbfPmathbfI_3)^n) >1 forall ngeq 0,$$ where $lambda$ is an arbitrary eigenvalue of the matrix in the parentheses.
The observation above is actually valid for all symmetric positive definite matrices of all dimensions. This conclusion is arrived by running more than 20,000 random matrix tests, i.e., more than 20,000 random P matrix is generated.
I tried to prove this property by using the properties of positive definite matrix and by connecting the two facts that
- $lambda(mathbfI+P)>1$
- $mathbfI+mathbfPmathbfI_1+mathbfI+mathbfPmathbfI_2+mathbfI+mathbfPmathbfI_3 = 3mathbfI+mathbfP$.
However, I did not have any progress. Please share me with your idea once you find any approach which may probably prove that.
matrices spectral-theory
asked Jul 21 at 2:42
He Li
61
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let P be a symmetric positive definite matrix. Without losing generality, assume P is of dimension three, i.e.,
$$mathbfP = beginbmatrixP_1&P_2&P_3\P_2&P_4&P_5\P_3&P_5&P_6 endbmatrix $$.
Let I be the identity matrix of the same dimension as P. Let
$$mathbfI_1 = beginbmatrix1 & 0 & 0\0 & 0 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_2 = beginbmatrix0 & 0 & 0\0 & 1 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_3 = beginbmatrix0 & 0 & 0\0 & 0 & 0\ 0 & 0 & 1endbmatrix.$$
Then, $$lambda((mathbfI+mathbfPmathbfI_1)^n
(mathbfI+mathbfPmathbfI_2)^n(mathbfI+mathbfPmathbfI_3)^n) >1 forall ngeq 0,$$ where $lambda$ is an arbitrary eigenvalue of the matrix in the parentheses.
The observation above is actually valid for all symmetric positive definite matrices of all dimensions. This conclusion is arrived by running more than 20,000 random matrix tests, i.e., more than 20,000 random P matrix is generated.
I tried to prove this property by using the properties of positive definite matrix and by connecting the two facts that
- $lambda(mathbfI+P)>1$
- $mathbfI+mathbfPmathbfI_1+mathbfI+mathbfPmathbfI_2+mathbfI+mathbfPmathbfI_3 = 3mathbfI+mathbfP$.
However, I did not have any progress. Please share me with your idea once you find any approach which may probably prove that.
matrices spectral-theory
asked Jul 21 at 2:42
He Li
61
Let P be a symmetric positive definite matrix. Without losing generality, assume P is of dimension three, i.e.,
$$mathbfP = beginbmatrixP_1&P_2&P_3\P_2&P_4&P_5\P_3&P_5&P_6 endbmatrix $$.
Let I be the identity matrix of the same dimension as P. Let
$$mathbfI_1 = beginbmatrix1 & 0 & 0\0 & 0 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_2 = beginbmatrix0 & 0 & 0\0 & 1 & 0\ 0 & 0 & 0endbmatrix,$$
$$mathbfI_3 = beginbmatrix0 & 0 & 0\0 & 0 & 0\ 0 & 0 & 1endbmatrix.$$
Then, $$lambda((mathbfI+mathbfPmathbfI_1)^n
(mathbfI+mathbfPmathbfI_2)^n(mathbfI+mathbfPmathbfI_3)^n) >1 forall ngeq 0,$$ where $lambda$ is an arbitrary eigenvalue of the matrix in the parentheses.
The observation above is actually valid for all symmetric positive definite matrices of all dimensions. This conclusion is arrived by running more than 20,000 random matrix tests, i.e., more than 20,000 random P matrix is generated.
I tried to prove this property by using the properties of positive definite matrix and by connecting the two facts that
- $lambda(mathbfI+P)>1$
- $mathbfI+mathbfPmathbfI_1+mathbfI+mathbfPmathbfI_2+mathbfI+mathbfPmathbfI_3 = 3mathbfI+mathbfP$.
However, I did not have any progress. Please share me with your idea once you find any approach which may probably prove that.
matrices spectral-theory
asked Jul 21 at 2:42
He Li
61
asked Jul 21 at 2:42
He Li
61
asked Jul 21 at 2:42
He Li
61
asked Jul 21 at 2:42
He Li
61
61
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34
add a comment |Â
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858194%2fa-very-interesting-property-of-symmetric-positive-definite-matrix-need-proof%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Your notation here is somewhat ambiguous. What exactly is meant by $lambdaleft((mathbfI + mathbfPmathbfI_1)^n(mathbfI + mathbfPmathbfI_2)^n(mathbfI + mathbfPmathbfI_3)^nright) > 1$. The left hand side is a matrix but the right hand side looks like a scalar. Please clarify. Thanks.
– stats_model
Jul 21 at 3:15
It is clarified in the question that $lambda(mathbf(A))$ refers to the eigenvalue of the matrix A and an eigenvalue is a scalar.
– He Li
Jul 21 at 14:02
Oh misread that, thought $lambda$ was the eigenvalue of P itself.
– stats_model
Jul 21 at 18:34