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
32k22351
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
417111
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
32k22351
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
32k22351
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
32k22351
HINT
If you diagonalize and write $A = VDV^{-1}$ then $A^n = VD^n V^{-1}$.
answered Nov 20 at 15:38
gt6989b
32k22351
answered Nov 20 at 15:38
gt6989b
32k22351
answered Nov 20 at 15:38
gt6989b
32k22351
answered Nov 20 at 15:38
gt6989b
32k22351
32k22351
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
417111
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
417111
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
417111
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
417111
edited Nov 20 at 15:52
answered Nov 20 at 15:40
Samvel Safaryan
417111
answered Nov 20 at 15:40
Samvel Safaryan
417111
answered Nov 20 at 15:40
Samvel Safaryan
417111
417111
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 |
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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