Sum of All Combinations of Products of Two Matrices
$begingroup$
Suppose that $mathbf A$ and $mathbf B$ are square, diagonalizable matrices. Consider the following infinite sum of all combinations of these two matrices:
begin{align} mathbf S = mathbf I &+mathbf A + mathbf B +\
&+mathbf A^2 + mathbf Amathbf B + mathbf Bmathbf A + mathbf B^2+ \
&+mathbf A^3 + mathbf A^2mathbf B + mathbf Amathbf Bmathbf A+ mathbf Amathbf B^2 + mathbf Bmathbf A^2+ mathbf Bmathbf Amathbf B+ mathbf B^2mathbf A + mathbf B^3+\
&+cdots
end{align}
Let $mathbf A=mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1}$, and $mathbf B=mathbf Q_Bmathbf Lambda_Bmathbf Q_B^{-1}$ be the eigendecompositions of $mathbf A$ and $mathbf B$.
If $mathbf B=mathbf 0$, then the sum would be equal to
$$mathbf S = sum_{k=0}^infty mathbf A^k=sum_{k=0}^infty(mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1})^k=mathbf Q_A left(sum_{k=0}^inftymathbfLambda_A^kright)mathbf Q_A^{-1}.$$
Is there a general way to express $mathbf S$ as a function of $mathbf Q_A$, $mathbf Q_B$, and sums that involve $mathbf Lambda_A$ and $mathbf Lambda_B$?
linear-algebra combinatorics matrices summation
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
$endgroup$
add a comment |
$begingroup$
Suppose that $mathbf A$ and $mathbf B$ are square, diagonalizable matrices. Consider the following infinite sum of all combinations of these two matrices:
begin{align} mathbf S = mathbf I &+mathbf A + mathbf B +\
&+mathbf A^2 + mathbf Amathbf B + mathbf Bmathbf A + mathbf B^2+ \
&+mathbf A^3 + mathbf A^2mathbf B + mathbf Amathbf Bmathbf A+ mathbf Amathbf B^2 + mathbf Bmathbf A^2+ mathbf Bmathbf Amathbf B+ mathbf B^2mathbf A + mathbf B^3+\
&+cdots
end{align}
Let $mathbf A=mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1}$, and $mathbf B=mathbf Q_Bmathbf Lambda_Bmathbf Q_B^{-1}$ be the eigendecompositions of $mathbf A$ and $mathbf B$.
If $mathbf B=mathbf 0$, then the sum would be equal to
$$mathbf S = sum_{k=0}^infty mathbf A^k=sum_{k=0}^infty(mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1})^k=mathbf Q_A left(sum_{k=0}^inftymathbfLambda_A^kright)mathbf Q_A^{-1}.$$
Is there a general way to express $mathbf S$ as a function of $mathbf Q_A$, $mathbf Q_B$, and sums that involve $mathbf Lambda_A$ and $mathbf Lambda_B$?
linear-algebra combinatorics matrices summation
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
$endgroup$
3
$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
2
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55
add a comment |
$begingroup$
Suppose that $mathbf A$ and $mathbf B$ are square, diagonalizable matrices. Consider the following infinite sum of all combinations of these two matrices:
begin{align} mathbf S = mathbf I &+mathbf A + mathbf B +\
&+mathbf A^2 + mathbf Amathbf B + mathbf Bmathbf A + mathbf B^2+ \
&+mathbf A^3 + mathbf A^2mathbf B + mathbf Amathbf Bmathbf A+ mathbf Amathbf B^2 + mathbf Bmathbf A^2+ mathbf Bmathbf Amathbf B+ mathbf B^2mathbf A + mathbf B^3+\
&+cdots
end{align}
Let $mathbf A=mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1}$, and $mathbf B=mathbf Q_Bmathbf Lambda_Bmathbf Q_B^{-1}$ be the eigendecompositions of $mathbf A$ and $mathbf B$.
If $mathbf B=mathbf 0$, then the sum would be equal to
$$mathbf S = sum_{k=0}^infty mathbf A^k=sum_{k=0}^infty(mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1})^k=mathbf Q_A left(sum_{k=0}^inftymathbfLambda_A^kright)mathbf Q_A^{-1}.$$
Is there a general way to express $mathbf S$ as a function of $mathbf Q_A$, $mathbf Q_B$, and sums that involve $mathbf Lambda_A$ and $mathbf Lambda_B$?
linear-algebra combinatorics matrices summation
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
$endgroup$
Suppose that $mathbf A$ and $mathbf B$ are square, diagonalizable matrices. Consider the following infinite sum of all combinations of these two matrices:
begin{align} mathbf S = mathbf I &+mathbf A + mathbf B +\
&+mathbf A^2 + mathbf Amathbf B + mathbf Bmathbf A + mathbf B^2+ \
&+mathbf A^3 + mathbf A^2mathbf B + mathbf Amathbf Bmathbf A+ mathbf Amathbf B^2 + mathbf Bmathbf A^2+ mathbf Bmathbf Amathbf B+ mathbf B^2mathbf A + mathbf B^3+\
&+cdots
end{align}
Let $mathbf A=mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1}$, and $mathbf B=mathbf Q_Bmathbf Lambda_Bmathbf Q_B^{-1}$ be the eigendecompositions of $mathbf A$ and $mathbf B$.
If $mathbf B=mathbf 0$, then the sum would be equal to
$$mathbf S = sum_{k=0}^infty mathbf A^k=sum_{k=0}^infty(mathbf Q_Amathbf Lambda_Amathbf Q_A^{-1})^k=mathbf Q_A left(sum_{k=0}^inftymathbfLambda_A^kright)mathbf Q_A^{-1}.$$
Is there a general way to express $mathbf S$ as a function of $mathbf Q_A$, $mathbf Q_B$, and sums that involve $mathbf Lambda_A$ and $mathbf Lambda_B$?
linear-algebra combinatorics matrices summation
linear-algebra combinatorics matrices summation
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
edited Dec 12 '18 at 14:09
mzp
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
asked Dec 12 '18 at 13:56
mzpmzp
1,24721234
1,24721234
3
$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
2
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55
add a comment |
3
$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
2
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55
3
3
$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
2
2
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55
add a comment |
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$begingroup$
Formally, the sum of all products of length $k$ is $({bf A} + {bf B})^k$. In particular, if ${bf A}, {bf B}$ commute, they are simultaneously diagonalizable, so we can take ${bf Q}_A = {bf Q}_B$, and then $S$ Is given by the same formula as above but replacing ${bf Lambda}_A^k$ with $({bf Lambda}_A + {bf Lambda}_B)^k$.
$endgroup$
– Travis
Dec 12 '18 at 14:24
2
$begingroup$
In general, though, I don't think one can expect a formula in terms of ${bf Q}_A, {bf Q}_B$, and sums that involve ${bf Lambda}_A$ and ${bf Lambda}_B$ but not ${bf Q}_A$ or ${bf Q}_B$, since they contain the data of the relative positions of eigenspaces of ${bf A}, {bf B}$. In general, the eigenvalues of a sum ${bf A} + {bf B}$ of two matrices cannot be determined (completely) from the eigenvalues of ${bf A}$ and ${bf B}$.
$endgroup$
– Travis
Dec 12 '18 at 14:34
$begingroup$
Just a thought: is $S = I + AS + BS$?
$endgroup$
– player100
Dec 12 '18 at 14:40
$begingroup$
following up on @Travis comments: in fact the key is that $S = sum^infty_{k=0} C^k$ where $C = A+B$, and so there is a nice formula depending on the decomposition of $C$. However, there is no guarantee $C$ is diagonalizable, so you might need the Jordan form, etc.
$endgroup$
– antkam
Dec 12 '18 at 16:55