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
525
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
525
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
525
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
525
asked Nov 20 at 15:35
vesii
525
asked Nov 20 at 15:35
vesii
525
asked Nov 20 at 15:35
vesii
525
asked Nov 20 at 15:35
vesii
525
525
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
32.2k22351
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
467111
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
32.2k22351
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
32.2k22351
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
32.2k22351
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32.2k22351
answered Nov 20 at 15:38
gt6989b
32.2k22351
answered Nov 20 at 15:38
gt6989b
32.2k22351
answered Nov 20 at 15:38
gt6989b
32.2k22351
32.2k22351
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
467111
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
467111
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
467111
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
467111
edited Nov 20 at 15:52
answered Nov 20 at 15:40
Samvel Safaryan
467111
answered Nov 20 at 15:40
Samvel Safaryan
467111
answered Nov 20 at 15:40
Samvel Safaryan
467111
467111
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 |
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