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Solving matrix exponential of 4x4 matrix
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Wondering how to find the matrix exponential of the following matrix without having to do through the long process of finding eigenvalues/eigenvectors and Jordan forms. Is there a quicker way to do it using Sine/Cosine ?
If
A = $$beginpmatrix0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
1 & 1 & 0 & 1 \
1 & 1 & 1 & 0 \
endpmatrix$$
Find beginequation e^tA endequation
matrices differential-equations matrix-exponential
asked 12 hours ago
Culliton Andrew
111
add a comment |Â
up vote
2
down vote
favorite
Wondering how to find the matrix exponential of the following matrix without having to do through the long process of finding eigenvalues/eigenvectors and Jordan forms. Is there a quicker way to do it using Sine/Cosine ?
If
A = $$beginpmatrix0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
1 & 1 & 0 & 1 \
1 & 1 & 1 & 0 \
endpmatrix$$
Find beginequation e^tA endequation
matrices differential-equations matrix-exponential
asked 12 hours ago
Culliton Andrew
111
I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Wondering how to find the matrix exponential of the following matrix without having to do through the long process of finding eigenvalues/eigenvectors and Jordan forms. Is there a quicker way to do it using Sine/Cosine ?
If
A = $$beginpmatrix0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
1 & 1 & 0 & 1 \
1 & 1 & 1 & 0 \
endpmatrix$$
Find beginequation e^tA endequation
matrices differential-equations matrix-exponential
asked 12 hours ago
Culliton Andrew
111
Wondering how to find the matrix exponential of the following matrix without having to do through the long process of finding eigenvalues/eigenvectors and Jordan forms. Is there a quicker way to do it using Sine/Cosine ?
If
A = $$beginpmatrix0 & 1 & 1 & 1 \
1 & 0 & 1 & 1 \
1 & 1 & 0 & 1 \
1 & 1 & 1 & 0 \
endpmatrix$$
Find beginequation e^tA endequation
matrices differential-equations matrix-exponential
asked 12 hours ago
Culliton Andrew
111
asked 12 hours ago
Culliton Andrew
111
asked 12 hours ago
Culliton Andrew
111
asked 12 hours ago
Culliton Andrew
111
111
I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago
add a comment |Â
I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago
I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
Let $boldsymbol A = boldsymbol U - boldsymbol I_4$. Simple calculations show that $boldsymbol U^n = 4^n-1 boldsymbol U$. Now for every $n in mathbb N^*$:
beginalign*
boldsymbol A^n =( boldsymbol U - boldsymbol I_4)^n &= sum_0^n binom n j boldsymbol U^j (-1)^n-j boldsymbol I \
&=(-1)^n boldsymbol I + sum_1^n binom n j 4^j-1(-1)^n-j boldsymbol U \
&= (-1)^nboldsymbol I + frac 14 sum_0^n binom n j 4^j (-1)^n-j boldsymbol U - (-1)^n boldsymbol U /4 \
&= frac 14 (4-1)^n boldsymbol U + (-1)^nboldsymbol I - (-1)^n boldsymbol U/4 \
&= frac 14 cdot (3^n - (-1)^n) boldsymbol U + (-1)^nboldsymbol I
endalign*
Therefore,
beginalign*
exp(tboldsymbol A) &= sum_0^infty frac 1 n! t^nboldsymbol A^n \
&= sum_0^infty frac t^n4n! (3^n boldsymbol U - (-1)^n boldsymbol U + 4(-1)^nboldsymbol I) \
&= frac 14 (mathrm e^3t - mathrm e^-t) boldsymbol U + mathrm e^-tboldsymbol I.
endalign*
answered 12 hours ago
xbh
7656
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
add a comment |Â
up vote
4
down vote
Let $u$ be the $4times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then
$$
beginaligned
A &=uu'-I ,\
A+I &=uu' ,\
4 &=u'u .
endaligned
$$
Then
$$
e^tA=e^-tI+t(A+I)=e^-tIcdot e^t(A+I)=e^-tcdot e^t,uu' .
$$
Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu' uu'=u u'u u'=4uu'$, inductively $(uu')^n+1=4^n, uu'$. The exponential of $t,uu'$ is then
$$
beginaligned
exp(tuu')
&=
I
+frac 11!t, uu'
+frac 12!t^2, 4, uu'
+frac 13!t^3, 4^2, uu'
+frac 14!t^4, 4^3, uu'
+frac 15!t^5, 4^4, uu'
+dots
\
&=
I
+
frac 14left(
frac 11!t, 4
+frac 12!t^2, 4^2
+frac 13!t^3, 4^3
+frac 14!t^4, 4^4
+frac 15!t^5, 4^5
+dots
right), uu'
\
&=
I +
frac 14(e^4t-1),uu'
.
endaligned
$$
The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$
left(beginarrayrrrr
frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright)
endarrayright)
$$
answered 11 hours ago
dan_fulea
3,8471211
add a comment |Â
up vote
1
down vote
Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $pi_A(A)=0$. Let's decompose the function $e^tx$ as
$$
e^tx=q(x)pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $pi_A$) then
$$
e^tA=q(A)pi_A(A)+r(A)=q(A)cdot 0+r(A)=r(A).
$$
Since $pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^-t=r(-1)=-a+b,qquad e^3t=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=frace^3t-e^-t4,qquad b=frace^3t-e^-t4+e^-t,
$$
that is
$$
r(x)=ax+b=frace^3t-e^-t4(x+1)+e^-t.
$$
Finally (with $x=A$)
$$
e^tA=frace^3t-e^-t4(A+I)+e^-tI.
$$
answered 10 hours ago
A.Γ.
20.1k22253
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Let $boldsymbol A = boldsymbol U - boldsymbol I_4$. Simple calculations show that $boldsymbol U^n = 4^n-1 boldsymbol U$. Now for every $n in mathbb N^*$:
beginalign*
boldsymbol A^n =( boldsymbol U - boldsymbol I_4)^n &= sum_0^n binom n j boldsymbol U^j (-1)^n-j boldsymbol I \
&=(-1)^n boldsymbol I + sum_1^n binom n j 4^j-1(-1)^n-j boldsymbol U \
&= (-1)^nboldsymbol I + frac 14 sum_0^n binom n j 4^j (-1)^n-j boldsymbol U - (-1)^n boldsymbol U /4 \
&= frac 14 (4-1)^n boldsymbol U + (-1)^nboldsymbol I - (-1)^n boldsymbol U/4 \
&= frac 14 cdot (3^n - (-1)^n) boldsymbol U + (-1)^nboldsymbol I
endalign*
Therefore,
beginalign*
exp(tboldsymbol A) &= sum_0^infty frac 1 n! t^nboldsymbol A^n \
&= sum_0^infty frac t^n4n! (3^n boldsymbol U - (-1)^n boldsymbol U + 4(-1)^nboldsymbol I) \
&= frac 14 (mathrm e^3t - mathrm e^-t) boldsymbol U + mathrm e^-tboldsymbol I.
endalign*
answered 12 hours ago
xbh
7656
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
add a comment |Â
up vote
5
down vote
Let $boldsymbol A = boldsymbol U - boldsymbol I_4$. Simple calculations show that $boldsymbol U^n = 4^n-1 boldsymbol U$. Now for every $n in mathbb N^*$:
beginalign*
boldsymbol A^n =( boldsymbol U - boldsymbol I_4)^n &= sum_0^n binom n j boldsymbol U^j (-1)^n-j boldsymbol I \
&=(-1)^n boldsymbol I + sum_1^n binom n j 4^j-1(-1)^n-j boldsymbol U \
&= (-1)^nboldsymbol I + frac 14 sum_0^n binom n j 4^j (-1)^n-j boldsymbol U - (-1)^n boldsymbol U /4 \
&= frac 14 (4-1)^n boldsymbol U + (-1)^nboldsymbol I - (-1)^n boldsymbol U/4 \
&= frac 14 cdot (3^n - (-1)^n) boldsymbol U + (-1)^nboldsymbol I
endalign*
Therefore,
beginalign*
exp(tboldsymbol A) &= sum_0^infty frac 1 n! t^nboldsymbol A^n \
&= sum_0^infty frac t^n4n! (3^n boldsymbol U - (-1)^n boldsymbol U + 4(-1)^nboldsymbol I) \
&= frac 14 (mathrm e^3t - mathrm e^-t) boldsymbol U + mathrm e^-tboldsymbol I.
endalign*
answered 12 hours ago
xbh
7656
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Let $boldsymbol A = boldsymbol U - boldsymbol I_4$. Simple calculations show that $boldsymbol U^n = 4^n-1 boldsymbol U$. Now for every $n in mathbb N^*$:
beginalign*
boldsymbol A^n =( boldsymbol U - boldsymbol I_4)^n &= sum_0^n binom n j boldsymbol U^j (-1)^n-j boldsymbol I \
&=(-1)^n boldsymbol I + sum_1^n binom n j 4^j-1(-1)^n-j boldsymbol U \
&= (-1)^nboldsymbol I + frac 14 sum_0^n binom n j 4^j (-1)^n-j boldsymbol U - (-1)^n boldsymbol U /4 \
&= frac 14 (4-1)^n boldsymbol U + (-1)^nboldsymbol I - (-1)^n boldsymbol U/4 \
&= frac 14 cdot (3^n - (-1)^n) boldsymbol U + (-1)^nboldsymbol I
endalign*
Therefore,
beginalign*
exp(tboldsymbol A) &= sum_0^infty frac 1 n! t^nboldsymbol A^n \
&= sum_0^infty frac t^n4n! (3^n boldsymbol U - (-1)^n boldsymbol U + 4(-1)^nboldsymbol I) \
&= frac 14 (mathrm e^3t - mathrm e^-t) boldsymbol U + mathrm e^-tboldsymbol I.
endalign*
answered 12 hours ago
xbh
7656
Let $boldsymbol A = boldsymbol U - boldsymbol I_4$. Simple calculations show that $boldsymbol U^n = 4^n-1 boldsymbol U$. Now for every $n in mathbb N^*$:
beginalign*
boldsymbol A^n =( boldsymbol U - boldsymbol I_4)^n &= sum_0^n binom n j boldsymbol U^j (-1)^n-j boldsymbol I \
&=(-1)^n boldsymbol I + sum_1^n binom n j 4^j-1(-1)^n-j boldsymbol U \
&= (-1)^nboldsymbol I + frac 14 sum_0^n binom n j 4^j (-1)^n-j boldsymbol U - (-1)^n boldsymbol U /4 \
&= frac 14 (4-1)^n boldsymbol U + (-1)^nboldsymbol I - (-1)^n boldsymbol U/4 \
&= frac 14 cdot (3^n - (-1)^n) boldsymbol U + (-1)^nboldsymbol I
endalign*
Therefore,
beginalign*
exp(tboldsymbol A) &= sum_0^infty frac 1 n! t^nboldsymbol A^n \
&= sum_0^infty frac t^n4n! (3^n boldsymbol U - (-1)^n boldsymbol U + 4(-1)^nboldsymbol I) \
&= frac 14 (mathrm e^3t - mathrm e^-t) boldsymbol U + mathrm e^-tboldsymbol I.
endalign*
answered 12 hours ago
xbh
7656
edited 11 hours ago
answered 12 hours ago
xbh
7656
answered 12 hours ago
xbh
7656
answered 12 hours ago
xbh
7656
7656
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
add a comment |Â
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
1
1
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
I don't think this can be right---$frac14 e^3t bf U$ is degenerate, but the exponential of a matrix is always invertible.
– Travis
12 hours ago
1
1
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
$U^n=4^n-1U$ is not true for $n=0$ (in the summation $sum_0^n$).
– A.Γ.
12 hours ago
1
1
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@Travis Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
@A.Γ. Thanks, i fixed it!
– xbh
11 hours ago
add a comment |Â
up vote
4
down vote
Let $u$ be the $4times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then
$$
beginaligned
A &=uu'-I ,\
A+I &=uu' ,\
4 &=u'u .
endaligned
$$
Then
$$
e^tA=e^-tI+t(A+I)=e^-tIcdot e^t(A+I)=e^-tcdot e^t,uu' .
$$
Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu' uu'=u u'u u'=4uu'$, inductively $(uu')^n+1=4^n, uu'$. The exponential of $t,uu'$ is then
$$
beginaligned
exp(tuu')
&=
I
+frac 11!t, uu'
+frac 12!t^2, 4, uu'
+frac 13!t^3, 4^2, uu'
+frac 14!t^4, 4^3, uu'
+frac 15!t^5, 4^4, uu'
+dots
\
&=
I
+
frac 14left(
frac 11!t, 4
+frac 12!t^2, 4^2
+frac 13!t^3, 4^3
+frac 14!t^4, 4^4
+frac 15!t^5, 4^5
+dots
right), uu'
\
&=
I +
frac 14(e^4t-1),uu'
.
endaligned
$$
The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$
left(beginarrayrrrr
frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright)
endarrayright)
$$
answered 11 hours ago
dan_fulea
3,8471211
add a comment |Â
up vote
4
down vote
Let $u$ be the $4times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then
$$
beginaligned
A &=uu'-I ,\
A+I &=uu' ,\
4 &=u'u .
endaligned
$$
Then
$$
e^tA=e^-tI+t(A+I)=e^-tIcdot e^t(A+I)=e^-tcdot e^t,uu' .
$$
Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu' uu'=u u'u u'=4uu'$, inductively $(uu')^n+1=4^n, uu'$. The exponential of $t,uu'$ is then
$$
beginaligned
exp(tuu')
&=
I
+frac 11!t, uu'
+frac 12!t^2, 4, uu'
+frac 13!t^3, 4^2, uu'
+frac 14!t^4, 4^3, uu'
+frac 15!t^5, 4^4, uu'
+dots
\
&=
I
+
frac 14left(
frac 11!t, 4
+frac 12!t^2, 4^2
+frac 13!t^3, 4^3
+frac 14!t^4, 4^4
+frac 15!t^5, 4^5
+dots
right), uu'
\
&=
I +
frac 14(e^4t-1),uu'
.
endaligned
$$
The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$
left(beginarrayrrrr
frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright)
endarrayright)
$$
answered 11 hours ago
dan_fulea
3,8471211
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let $u$ be the $4times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then
$$
beginaligned
A &=uu'-I ,\
A+I &=uu' ,\
4 &=u'u .
endaligned
$$
Then
$$
e^tA=e^-tI+t(A+I)=e^-tIcdot e^t(A+I)=e^-tcdot e^t,uu' .
$$
Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu' uu'=u u'u u'=4uu'$, inductively $(uu')^n+1=4^n, uu'$. The exponential of $t,uu'$ is then
$$
beginaligned
exp(tuu')
&=
I
+frac 11!t, uu'
+frac 12!t^2, 4, uu'
+frac 13!t^3, 4^2, uu'
+frac 14!t^4, 4^3, uu'
+frac 15!t^5, 4^4, uu'
+dots
\
&=
I
+
frac 14left(
frac 11!t, 4
+frac 12!t^2, 4^2
+frac 13!t^3, 4^3
+frac 14!t^4, 4^4
+frac 15!t^5, 4^5
+dots
right), uu'
\
&=
I +
frac 14(e^4t-1),uu'
.
endaligned
$$
The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$
left(beginarrayrrrr
frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright)
endarrayright)
$$
answered 11 hours ago
dan_fulea
3,8471211
Let $u$ be the $4times 1$ matrix with all entries equal to $1$. Let $u'$ be its transpose. Then
$$
beginaligned
A &=uu'-I ,\
A+I &=uu' ,\
4 &=u'u .
endaligned
$$
Then
$$
e^tA=e^-tI+t(A+I)=e^-tIcdot e^t(A+I)=e^-tcdot e^t,uu' .
$$
Now the powers of $uu'$ are easily computed. For instance $(uu')^2=uu' uu'=u u'u u'=4uu'$, inductively $(uu')^n+1=4^n, uu'$. The exponential of $t,uu'$ is then
$$
beginaligned
exp(tuu')
&=
I
+frac 11!t, uu'
+frac 12!t^2, 4, uu'
+frac 13!t^3, 4^2, uu'
+frac 14!t^4, 4^3, uu'
+frac 15!t^5, 4^4, uu'
+dots
\
&=
I
+
frac 14left(
frac 11!t, 4
+frac 12!t^2, 4^2
+frac 13!t^3, 4^3
+frac 14!t^4, 4^4
+frac 15!t^5, 4^5
+dots
right), uu'
\
&=
I +
frac 14(e^4t-1),uu'
.
endaligned
$$
The above $uu'$ is the matrix with ones as entries.
From here we get the needed formula, as confirmed by sage:
sage: A = matrix( QQ, 4,4, [0,1,1,1, 1,0,1,1, 1,1,0,1, 1,1,1,0] )
sage: A
[0 1 1 1]
[1 0 1 1]
[1 1 0 1]
[1 1 1 0]
sage: var('t');
sage: exp(t*A)
[1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t)]
[1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) - 1)*e^(-t) 1/4*(e^(4*t) + 3)*e^(-t)]
sage: latex(_)
$$
left(beginarrayrrrr
frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) \
frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) - 1right) e^left(-tright) & frac14 , left(e^left(4 , tright) + 3right) e^left(-tright)
endarrayright)
$$
answered 11 hours ago
dan_fulea
3,8471211
answered 11 hours ago
dan_fulea
3,8471211
answered 11 hours ago
dan_fulea
3,8471211
answered 11 hours ago
dan_fulea
3,8471211
3,8471211
add a comment |Â
add a comment |Â
up vote
1
down vote
Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $pi_A(A)=0$. Let's decompose the function $e^tx$ as
$$
e^tx=q(x)pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $pi_A$) then
$$
e^tA=q(A)pi_A(A)+r(A)=q(A)cdot 0+r(A)=r(A).
$$
Since $pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^-t=r(-1)=-a+b,qquad e^3t=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=frace^3t-e^-t4,qquad b=frace^3t-e^-t4+e^-t,
$$
that is
$$
r(x)=ax+b=frace^3t-e^-t4(x+1)+e^-t.
$$
Finally (with $x=A$)
$$
e^tA=frace^3t-e^-t4(A+I)+e^-tI.
$$
answered 10 hours ago
A.Γ.
20.1k22253
add a comment |Â
up vote
1
down vote
Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $pi_A(A)=0$. Let's decompose the function $e^tx$ as
$$
e^tx=q(x)pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $pi_A$) then
$$
e^tA=q(A)pi_A(A)+r(A)=q(A)cdot 0+r(A)=r(A).
$$
Since $pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^-t=r(-1)=-a+b,qquad e^3t=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=frace^3t-e^-t4,qquad b=frace^3t-e^-t4+e^-t,
$$
that is
$$
r(x)=ax+b=frace^3t-e^-t4(x+1)+e^-t.
$$
Finally (with $x=A$)
$$
e^tA=frace^3t-e^-t4(A+I)+e^-tI.
$$
answered 10 hours ago
A.Γ.
20.1k22253
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $pi_A(A)=0$. Let's decompose the function $e^tx$ as
$$
e^tx=q(x)pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $pi_A$) then
$$
e^tA=q(A)pi_A(A)+r(A)=q(A)cdot 0+r(A)=r(A).
$$
Since $pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^-t=r(-1)=-a+b,qquad e^3t=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=frace^3t-e^-t4,qquad b=frace^3t-e^-t4+e^-t,
$$
that is
$$
r(x)=ax+b=frace^3t-e^-t4(x+1)+e^-t.
$$
Finally (with $x=A$)
$$
e^tA=frace^3t-e^-t4(A+I)+e^-tI.
$$
answered 10 hours ago
A.Γ.
20.1k22253
Another way to do it is to use the minimal polynomial and Lagrange interpolation. Calculating $A^2$ we notice that $A^2=3I+2A$, hence, $A^2-2A-3I=0$. It means that
$$
pi_A(x)=x^2-2x-3=(x+1)(x-3)
$$
annihilates $A$, i.e. $pi_A(A)=0$. Let's decompose the function $e^tx$ as
$$
e^tx=q(x)pi_A(x)+r(x).
$$
If we manage to do that (with $q$, $r$ analytical near zeros of $pi_A$) then
$$
e^tA=q(A)pi_A(A)+r(A)=q(A)cdot 0+r(A)=r(A).
$$
Since $pi_A$ is of degree 2 it is enough to search for $r$ of degree 1 in the form $r(x)=ax+b$. To find $a,b$ we can use interpolation conditions at $x=-1$ and $x=3$:
$$
e^-t=r(-1)=-a+b,qquad e^3t=r(3)=3a+b.
$$
Solving the system of 2 equations and 2 unknowns gives
$$
a=frace^3t-e^-t4,qquad b=frace^3t-e^-t4+e^-t,
$$
that is
$$
r(x)=ax+b=frace^3t-e^-t4(x+1)+e^-t.
$$
Finally (with $x=A$)
$$
e^tA=frace^3t-e^-t4(A+I)+e^-tI.
$$
answered 10 hours ago
A.Γ.
20.1k22253
edited 3 hours ago
answered 10 hours ago
A.Γ.
20.1k22253
answered 10 hours ago
A.Γ.
20.1k22253
answered 10 hours ago
A.Γ.
20.1k22253
20.1k22253
add a comment |Â
add a comment |Â
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I'm afraid I am unfamiliar
– Culliton Andrew
12 hours ago
you want a quicker way involving only pen and paper or computer based would also work for you?
– pointguard0
12 hours ago
Apologies, I should have specified this is a question that appeared on my Differential Equations Exam so it would have to be pen and paper. I am looking for a different way to do it because the 'long way' calculating the characteristic polynomial etc. took too long in the exam causing me to run out of time.
– Culliton Andrew
12 hours ago
Pen and paper work implies that such matrix has some particular properties that could be exploited. In this question, all-1 matrix has an easily calculated powers.
– xbh
12 hours ago