Inverse of a triangle matrix The 2019 Stack Overflow Developer Survey Results Are InInverse of $(A + B)$ and $(A + BCD)$?How to invert a matrixmatrix inverse and limitName for 'flipped' triangular matrixCalculation of $trace(L^THL)$, L is lower triangular, H is symmetric.Solving a matrix equation problemDerivative of Symmetric Positive Definite Matrix w.r.t. to its Lower Triangular Cholesky FactorHow to decompose a square matrix into two non-square matrices?Cholesky decomposition of the inverse of a matrixUsing Cholesky decomposition to compute covariance matrix determinant
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Inverse of a triangle matrix
The 2019 Stack Overflow Developer Survey Results Are InInverse of $(A + B)$ and $(A + BCD)$?How to invert a matrixmatrix inverse and limitName for 'flipped' triangular matrixCalculation of $trace(L^THL)$, L is lower triangular, H is symmetric.Solving a matrix equation problemDerivative of Symmetric Positive Definite Matrix w.r.t. to its Lower Triangular Cholesky FactorHow to decompose a square matrix into two non-square matrices?Cholesky decomposition of the inverse of a matrixUsing Cholesky decomposition to compute covariance matrix determinant
$begingroup$
I have been given an assignment where I need to find the inverse of an upper triangular matrix. I have been given these two formulas and am to use Lemma $2$. Can someone explain to me how this works. I understand Lemma $1$ its pretty simple but the different powers in lemma two lost me. Why would it go from first power to squared to suddenly fourth and then to logarithms? Below are the two lemmas I have been given.
I Triangular matrices for linear systems with special structure
$1$. Triangular matrices and back substitution
$A$ - lower triangular square matrix $n times n$; $a_ij = 0$ for $i<j$
$det A neq 0$ nonsingular
First $a_ii=1$ for all $i$
Split $A=I-L$, $l_ij$ for $i leq j$
$L^n=0$ easy to verify
Lemma $mathbf 1$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =I +L +L^2 +ldots +L^n-1.$$
Proof: Let $B$ be equal to the right-hand side of equation ($1$). Then $$B(I-L)=I-L^n=I$$ because $L^n=0$ and therefore $B=A^-1$.
Lemma $mathbf 2$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =(I+L^2^[log n]-1) (I+L^2^[log n]-2) cdot ldots cdot (I+L^4) (I+L^2) (I+L). tag2$$
Proof: Expand the product in the right-hand side of ($2$). Since $L^n=0$, we are left with $I +L +L^2 +ldots +L^n-1$ which is equal to $A^-1$ by Lemma $1$. $$tagQ.E.D.$$
I am working with an original matrix of size $1473 times 1473$. Was told to do a Cholesky decomposition on it and then invert it using Lemma $2$.
Maybe even a name for these two lemmas so I can do more reading would help too.
Any help is greatly appreciated.
inverse matrix-equations matrix-decomposition
$endgroup$
add a comment |
$begingroup$
I have been given an assignment where I need to find the inverse of an upper triangular matrix. I have been given these two formulas and am to use Lemma $2$. Can someone explain to me how this works. I understand Lemma $1$ its pretty simple but the different powers in lemma two lost me. Why would it go from first power to squared to suddenly fourth and then to logarithms? Below are the two lemmas I have been given.
I Triangular matrices for linear systems with special structure
$1$. Triangular matrices and back substitution
$A$ - lower triangular square matrix $n times n$; $a_ij = 0$ for $i<j$
$det A neq 0$ nonsingular
First $a_ii=1$ for all $i$
Split $A=I-L$, $l_ij$ for $i leq j$
$L^n=0$ easy to verify
Lemma $mathbf 1$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =I +L +L^2 +ldots +L^n-1.$$
Proof: Let $B$ be equal to the right-hand side of equation ($1$). Then $$B(I-L)=I-L^n=I$$ because $L^n=0$ and therefore $B=A^-1$.
Lemma $mathbf 2$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =(I+L^2^[log n]-1) (I+L^2^[log n]-2) cdot ldots cdot (I+L^4) (I+L^2) (I+L). tag2$$
Proof: Expand the product in the right-hand side of ($2$). Since $L^n=0$, we are left with $I +L +L^2 +ldots +L^n-1$ which is equal to $A^-1$ by Lemma $1$. $$tagQ.E.D.$$
I am working with an original matrix of size $1473 times 1473$. Was told to do a Cholesky decomposition on it and then invert it using Lemma $2$.
Maybe even a name for these two lemmas so I can do more reading would help too.
Any help is greatly appreciated.
inverse matrix-equations matrix-decomposition
$endgroup$
$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48
add a comment |
$begingroup$
I have been given an assignment where I need to find the inverse of an upper triangular matrix. I have been given these two formulas and am to use Lemma $2$. Can someone explain to me how this works. I understand Lemma $1$ its pretty simple but the different powers in lemma two lost me. Why would it go from first power to squared to suddenly fourth and then to logarithms? Below are the two lemmas I have been given.
I Triangular matrices for linear systems with special structure
$1$. Triangular matrices and back substitution
$A$ - lower triangular square matrix $n times n$; $a_ij = 0$ for $i<j$
$det A neq 0$ nonsingular
First $a_ii=1$ for all $i$
Split $A=I-L$, $l_ij$ for $i leq j$
$L^n=0$ easy to verify
Lemma $mathbf 1$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =I +L +L^2 +ldots +L^n-1.$$
Proof: Let $B$ be equal to the right-hand side of equation ($1$). Then $$B(I-L)=I-L^n=I$$ because $L^n=0$ and therefore $B=A^-1$.
Lemma $mathbf 2$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =(I+L^2^[log n]-1) (I+L^2^[log n]-2) cdot ldots cdot (I+L^4) (I+L^2) (I+L). tag2$$
Proof: Expand the product in the right-hand side of ($2$). Since $L^n=0$, we are left with $I +L +L^2 +ldots +L^n-1$ which is equal to $A^-1$ by Lemma $1$. $$tagQ.E.D.$$
I am working with an original matrix of size $1473 times 1473$. Was told to do a Cholesky decomposition on it and then invert it using Lemma $2$.
Maybe even a name for these two lemmas so I can do more reading would help too.
Any help is greatly appreciated.
inverse matrix-equations matrix-decomposition
$endgroup$
I have been given an assignment where I need to find the inverse of an upper triangular matrix. I have been given these two formulas and am to use Lemma $2$. Can someone explain to me how this works. I understand Lemma $1$ its pretty simple but the different powers in lemma two lost me. Why would it go from first power to squared to suddenly fourth and then to logarithms? Below are the two lemmas I have been given.
I Triangular matrices for linear systems with special structure
$1$. Triangular matrices and back substitution
$A$ - lower triangular square matrix $n times n$; $a_ij = 0$ for $i<j$
$det A neq 0$ nonsingular
First $a_ii=1$ for all $i$
Split $A=I-L$, $l_ij$ for $i leq j$
$L^n=0$ easy to verify
Lemma $mathbf 1$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =I +L +L^2 +ldots +L^n-1.$$
Proof: Let $B$ be equal to the right-hand side of equation ($1$). Then $$B(I-L)=I-L^n=I$$ because $L^n=0$ and therefore $B=A^-1$.
Lemma $mathbf 2$. If $A=I-L$, where $L$ is strictly lower triangular then $$A^-1 =(I+L^2^[log n]-1) (I+L^2^[log n]-2) cdot ldots cdot (I+L^4) (I+L^2) (I+L). tag2$$
Proof: Expand the product in the right-hand side of ($2$). Since $L^n=0$, we are left with $I +L +L^2 +ldots +L^n-1$ which is equal to $A^-1$ by Lemma $1$. $$tagQ.E.D.$$
I am working with an original matrix of size $1473 times 1473$. Was told to do a Cholesky decomposition on it and then invert it using Lemma $2$.
Maybe even a name for these two lemmas so I can do more reading would help too.
Any help is greatly appreciated.
inverse matrix-equations matrix-decomposition
inverse matrix-equations matrix-decomposition
edited Mar 23 at 17:18
Rócherz
3,0263823
3,0263823
asked Jun 8 '17 at 20:18
Mnkyskilz 21Mnkyskilz 21
1
1
$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48
add a comment |
$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48
$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48
add a comment |
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$begingroup$
The exponents are successive powers of two. Consider the binary representation of a number.
$endgroup$
– amd
Jun 8 '17 at 21:47
$begingroup$
In the future, please take the time to type in important parts of your questions instead of posting images. Images take longer to download, are not searchable and are inaccessible to people using screen readers.
$endgroup$
– amd
Jun 8 '17 at 21:48