Finding a formula of a power of a matrix
up vote
2
down vote
favorite
Part of a solution I came across of calculating the following matrix:
$$begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $M^2,M^3,M^4$ but I can seem to see the pattern. How should I approach this issue?
linear-algebra
asked Nov 20 at 15:35
vesii
635
add a comment |
up vote
2
down vote
favorite
Part of a solution I came across of calculating the following matrix:
$$begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $M^2,M^3,M^4$ but I can seem to see the pattern. How should I approach this issue?
linear-algebra
asked Nov 20 at 15:35
vesii
635
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Part of a solution I came across of calculating the following matrix:
$$begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $M^2,M^3,M^4$ but I can seem to see the pattern. How should I approach this issue?
linear-algebra
asked Nov 20 at 15:35
vesii
635
Part of a solution I came across of calculating the following matrix:
$$begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}^n$$
I'm trying to find a formula for this matrix so I can prove it using induction. I tried to calculate $M^2,M^3,M^4$ but I can seem to see the pattern. How should I approach this issue?
linear-algebra
linear-algebra
asked Nov 20 at 15:35
vesii
635
asked Nov 20 at 15:35
vesii
635
asked Nov 20 at 15:35
vesii
635
asked Nov 20 at 15:35
vesii
635
asked Nov 20 at 15:35
vesii
635
635
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44
add a comment |
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $theta$ is given by
$$
R_theta = pmatrix{cos theta & -sin theta\ sin theta & cos theta}
$$
Your matrix is simply $R_{45^circ}$. You should find, then, that $(R_{45^circ})^n = R_{(45n)^circ}$.
answered Nov 20 at 15:39
Omnomnomnom
125k788176
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
add a comment |
up vote
2
down vote
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.6k22351
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
add a comment |
up vote
0
down vote
Idea
$$
\begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}=begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}cdotfrac{sqrt{2}}{2}
\begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}^ncdotBig(frac{1}{sqrt{2}}Big)^n=begin{pmatrix}-i & i\
1 & 1
end{pmatrix}cdotbegin{pmatrix}(1-i)^n & 0\
0 & (1+i)^n
end{pmatrix}cdotbegin{pmatrix}-i & i\
1 & 1
end{pmatrix}^{-1}cdotfrac{1}{2^{frac n 2}}
$$
answered Nov 20 at 15:40
Samvel Safaryan
493111
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $theta$ is given by
$$
R_theta = pmatrix{cos theta & -sin theta\ sin theta & cos theta}
$$
Your matrix is simply $R_{45^circ}$. You should find, then, that $(R_{45^circ})^n = R_{(45n)^circ}$.
answered Nov 20 at 15:39
Omnomnomnom
125k788176
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
add a comment |
up vote
4
down vote
It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $theta$ is given by
$$
R_theta = pmatrix{cos theta & -sin theta\ sin theta & cos theta}
$$
Your matrix is simply $R_{45^circ}$. You should find, then, that $(R_{45^circ})^n = R_{(45n)^circ}$.
answered Nov 20 at 15:39
Omnomnomnom
125k788176
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
add a comment |
up vote
4
down vote
up vote
4
down vote
It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $theta$ is given by
$$
R_theta = pmatrix{cos theta & -sin theta\ sin theta & cos theta}
$$
Your matrix is simply $R_{45^circ}$. You should find, then, that $(R_{45^circ})^n = R_{(45n)^circ}$.
answered Nov 20 at 15:39
Omnomnomnom
125k788176
It helps to interpret the matrix geometrically. The matrix of a counterclockwise rotation about the origin by angle $theta$ is given by
$$
R_theta = pmatrix{cos theta & -sin theta\ sin theta & cos theta}
$$
Your matrix is simply $R_{45^circ}$. You should find, then, that $(R_{45^circ})^n = R_{(45n)^circ}$.
answered Nov 20 at 15:39
Omnomnomnom
125k788176
answered Nov 20 at 15:39
Omnomnomnom
125k788176
answered Nov 20 at 15:39
Omnomnomnom
125k788176
answered Nov 20 at 15:39
Omnomnomnom
125k788176
125k788176
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
add a comment |
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
What is the order of $R_{45n}$ (when speaking of groups)?
– vesii
Nov 20 at 15:47
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
$R_{45^circ}$ has order $8$, so that should tell you everything you need to know
– Omnomnomnom
Nov 20 at 19:49
add a comment |
up vote
2
down vote
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.6k22351
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
add a comment |
up vote
2
down vote
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.6k22351
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.6k22351
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.6k22351
answered Nov 20 at 15:38
gt6989b
32.6k22351
answered Nov 20 at 15:38
gt6989b
32.6k22351
answered Nov 20 at 15:38
gt6989b
32.6k22351
32.6k22351
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
add a comment |
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
can it be diagonalized? I get the characteristic polynomial: $p(lambda)=lambda^2-sqrt{2} lambda+1$.
– vesii
Nov 20 at 16:01
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
@vesii - It can be diagonalized in $Bbb C$ just fine. You will find that even though $V$ and $D$ are complex matrices, $A^n$ is real for all $n$.
– Paul Sinclair
Nov 20 at 21:09
add a comment |
up vote
0
down vote
Idea
$$
\begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}=begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}cdotfrac{sqrt{2}}{2}
\begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}^ncdotBig(frac{1}{sqrt{2}}Big)^n=begin{pmatrix}-i & i\
1 & 1
end{pmatrix}cdotbegin{pmatrix}(1-i)^n & 0\
0 & (1+i)^n
end{pmatrix}cdotbegin{pmatrix}-i & i\
1 & 1
end{pmatrix}^{-1}cdotfrac{1}{2^{frac n 2}}
$$
answered Nov 20 at 15:40
Samvel Safaryan
493111
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
add a comment |
up vote
0
down vote
Idea
$$
\begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}=begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}cdotfrac{sqrt{2}}{2}
\begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}^ncdotBig(frac{1}{sqrt{2}}Big)^n=begin{pmatrix}-i & i\
1 & 1
end{pmatrix}cdotbegin{pmatrix}(1-i)^n & 0\
0 & (1+i)^n
end{pmatrix}cdotbegin{pmatrix}-i & i\
1 & 1
end{pmatrix}^{-1}cdotfrac{1}{2^{frac n 2}}
$$
answered Nov 20 at 15:40
Samvel Safaryan
493111
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
add a comment |
up vote
0
down vote
up vote
0
down vote
Idea
$$
\begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}=begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}cdotfrac{sqrt{2}}{2}
\begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}^ncdotBig(frac{1}{sqrt{2}}Big)^n=begin{pmatrix}-i & i\
1 & 1
end{pmatrix}cdotbegin{pmatrix}(1-i)^n & 0\
0 & (1+i)^n
end{pmatrix}cdotbegin{pmatrix}-i & i\
1 & 1
end{pmatrix}^{-1}cdotfrac{1}{2^{frac n 2}}
$$
answered Nov 20 at 15:40
Samvel Safaryan
493111
Idea
$$
\begin{pmatrix}frac{sqrt{2}}{2} & -frac{sqrt{2}}{2}\
frac{sqrt{2}}{2} & frac{sqrt{2}}{2}
end{pmatrix}=begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}cdotfrac{sqrt{2}}{2}
\begin{pmatrix}1 & -1\
1 & 1
end{pmatrix}^ncdotBig(frac{1}{sqrt{2}}Big)^n=begin{pmatrix}-i & i\
1 & 1
end{pmatrix}cdotbegin{pmatrix}(1-i)^n & 0\
0 & (1+i)^n
end{pmatrix}cdotbegin{pmatrix}-i & i\
1 & 1
end{pmatrix}^{-1}cdotfrac{1}{2^{frac n 2}}
$$
answered Nov 20 at 15:40
Samvel Safaryan
493111
edited Nov 20 at 15:52
answered Nov 20 at 15:40
Samvel Safaryan
493111
answered Nov 20 at 15:40
Samvel Safaryan
493111
answered Nov 20 at 15:40
Samvel Safaryan
493111
493111
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
add a comment |
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
Yes I see it but using Wolfram ([link][1]) it gives me a matrix which is depend on $i$ [1]: wolframalpha.com/input/?i=%7B%7B1,-1%7D,%7B1,1%7D%7D%5En
– vesii
Nov 20 at 15:50
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
$(cosphi+icdotsinphi)^n=cos(ncdotphi)+icdotsin(ncdotphi)$
– Samvel Safaryan
Nov 20 at 15:55
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006466%2ffinding-a-formula-of-a-power-of-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Could you give us a bit more context here? What class is this for? Are you familiar with general linear-algebra techniques such as diagonalization? Are you comfortable using complex numbers?
– Omnomnomnom
Nov 20 at 15:44