Solve for the exponent of a matrix
$begingroup$
we discussed matrices in class and had the following task: Given
$$U=begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\1 & frac{1}{15} & 0 & 0 & 0 &0\0 & frac{8}{15} & frac{3}{15} & 0 & 0 & 0\0 & frac{6}{15} & frac{9}{15} & frac{6}{15} & 0 & 0\ 0 & 0 & frac{3}{15} & frac{8}{15} & frac{10}{15} & 0\0 & 0 & 0 & frac{1}{15} & frac{5}{15} & 1end{bmatrix},
quad overrightarrow{s}=begin{bmatrix}1\0\0\0\0\0end{bmatrix},$$
solve for $x$ such that $U^xoverrightarrow{s}=overrightarrow{s_x}$ where the last element (Row) of $overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$overrightarrow{s_{15}}=begin{bmatrix}0\0\0\0\0,0137\0,9863end{bmatrix},quad overrightarrow{s_{16}}=begin{bmatrix}0\0\0\0\0,0091\0,9909end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
linear-algebra matrices inequality matrix-equations
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
$endgroup$
|
show 1 more comment
$begingroup$
we discussed matrices in class and had the following task: Given
$$U=begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\1 & frac{1}{15} & 0 & 0 & 0 &0\0 & frac{8}{15} & frac{3}{15} & 0 & 0 & 0\0 & frac{6}{15} & frac{9}{15} & frac{6}{15} & 0 & 0\ 0 & 0 & frac{3}{15} & frac{8}{15} & frac{10}{15} & 0\0 & 0 & 0 & frac{1}{15} & frac{5}{15} & 1end{bmatrix},
quad overrightarrow{s}=begin{bmatrix}1\0\0\0\0\0end{bmatrix},$$
solve for $x$ such that $U^xoverrightarrow{s}=overrightarrow{s_x}$ where the last element (Row) of $overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$overrightarrow{s_{15}}=begin{bmatrix}0\0\0\0\0,0137\0,9863end{bmatrix},quad overrightarrow{s_{16}}=begin{bmatrix}0\0\0\0\0,0091\0,9909end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
linear-algebra matrices inequality matrix-equations
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
$endgroup$
$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10
|
show 1 more comment
$begingroup$
we discussed matrices in class and had the following task: Given
$$U=begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\1 & frac{1}{15} & 0 & 0 & 0 &0\0 & frac{8}{15} & frac{3}{15} & 0 & 0 & 0\0 & frac{6}{15} & frac{9}{15} & frac{6}{15} & 0 & 0\ 0 & 0 & frac{3}{15} & frac{8}{15} & frac{10}{15} & 0\0 & 0 & 0 & frac{1}{15} & frac{5}{15} & 1end{bmatrix},
quad overrightarrow{s}=begin{bmatrix}1\0\0\0\0\0end{bmatrix},$$
solve for $x$ such that $U^xoverrightarrow{s}=overrightarrow{s_x}$ where the last element (Row) of $overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$overrightarrow{s_{15}}=begin{bmatrix}0\0\0\0\0,0137\0,9863end{bmatrix},quad overrightarrow{s_{16}}=begin{bmatrix}0\0\0\0\0,0091\0,9909end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
linear-algebra matrices inequality matrix-equations
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
$endgroup$
we discussed matrices in class and had the following task: Given
$$U=begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0\1 & frac{1}{15} & 0 & 0 & 0 &0\0 & frac{8}{15} & frac{3}{15} & 0 & 0 & 0\0 & frac{6}{15} & frac{9}{15} & frac{6}{15} & 0 & 0\ 0 & 0 & frac{3}{15} & frac{8}{15} & frac{10}{15} & 0\0 & 0 & 0 & frac{1}{15} & frac{5}{15} & 1end{bmatrix},
quad overrightarrow{s}=begin{bmatrix}1\0\0\0\0\0end{bmatrix},$$
solve for $x$ such that $U^xoverrightarrow{s}=overrightarrow{s_x}$ where the last element (Row) of $overrightarrow{s_x}$ should be equal or greater than 0.99.
We were told that the only way to get $x$ is by inserting random numbers and "search" for it. By doing this we actually found:
$$overrightarrow{s_{15}}=begin{bmatrix}0\0\0\0\0,0137\0,9863end{bmatrix},quad overrightarrow{s_{16}}=begin{bmatrix}0\0\0\0\0,0091\0,9909end{bmatrix}.$$
So the answer is pretty much $x=16$, but is there no way to solve for $x$ instead of inserting random numbers until you find the answer?
Thanks for any answers
linear-algebra matrices inequality matrix-equations
linear-algebra matrices inequality matrix-equations
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
edited Dec 5 '18 at 14:07
Semiclassical
11.1k32465
11.1k32465
asked Dec 5 '18 at 13:25
Alex FAlex F
132
asked Dec 5 '18 at 13:25
Alex FAlex F
132
asked Dec 5 '18 at 13:25
Alex FAlex F
132
132
$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10
|
show 1 more comment
$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10
$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
$endgroup$
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
$endgroup$
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
add a comment |
$begingroup$
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
$endgroup$
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
add a comment |
$begingroup$
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
$endgroup$
Hint :
Using this you get the diagonalization of $U=PDP^{-1}$, then $U^x = PDP^{-1} dots PDP^{-1} = P D^x P^{-1}$ where $D$ is the matrix with the elements of $D$ elevated to power $x$. Using this you can solve for the last element being greater to $0.99$
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
answered Dec 5 '18 at 13:55
P. QuintonP. Quinton
1,7261213
1,7261213
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
add a comment |
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
$begingroup$
i actually found a function that does exactly what i want with your approach. Thank you very much for this answer! really useful
$endgroup$
– Alex F
Dec 5 '18 at 16:32
add a comment |
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$begingroup$
Is $overrightarrow{s}$ a column vector? If so how can the top element of the product be $1,$ as all entries of top row of matrix are $0$ ?
$endgroup$
– coffeemath
Dec 5 '18 at 13:36
$begingroup$
yes $overrightarrow{s}$ is a column vector. Can you clarify what you mean with the top element of the product is 1? The product of $U^{15}*overrightarrow{s}=overrightarrow{s_{15}}$
$endgroup$
– Alex F
Dec 5 '18 at 13:42
$begingroup$
Oh I get it... the definitions were so close in question it looked like the matrix was multiplying the column vector. But it's just two separate equations, one for what U is, and one for what s is.
$endgroup$
– coffeemath
Dec 5 '18 at 14:06
$begingroup$
On that note, I've edited your question so as to make it more clear visually.
$endgroup$
– Semiclassical
Dec 5 '18 at 14:08
$begingroup$
Thank you very much @Semiclassical Iam new to this page, so Iam just getting used to the formatting, but that's really nice of you! :) Thanks for taking that time
$endgroup$
– Alex F
Dec 5 '18 at 14:10