Mixing
General solution to linear system with general form
$begingroup$
I am trying to find the general solution of the system
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X$$
where $a+d not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong?
ordinary-differential-equations
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
$endgroup$
add a comment |
$begingroup$
I am trying to find the general solution of the system
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X$$
where $a+d not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong?
ordinary-differential-equations
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
$endgroup$
$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
2
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55
add a comment |
$begingroup$
I am trying to find the general solution of the system
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X$$
where $a+d not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong?
ordinary-differential-equations
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
$endgroup$
I am trying to find the general solution of the system
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X$$
where $a+d not= 0 $ and $ad-bc=0$
I find that the eigenvalues are 0 and $a+d$ with corresponding eigenvectors both $[0,0]$ which implies that the general solution is simply $X(t)=0$ but this solution does not seem right to me. Is there something I have done wrong?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
asked Jan 31 at 2:24
Richard VillalobosRichard Villalobos
1807
1807
$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
2
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55
add a comment |
$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
2
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55
$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
2
2
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We are given
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X\text{where}~ ~~a+d ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - lambda I| = 0$, yielding
$$lambda_{1,2} = frac{1}{2} left(-sqrt{a^2-2 a d+4 b c+d^2}+a+dright),frac{1}{2} left(sqrt{a^2-2 a d+4 b c+d^2}+a+dright)$$
We can then find the associated eigenvectors by solving $[A-lambda_i]v_i = 0$, yielding
$$v_1 = begin{pmatrix}
-dfrac{sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1
end{pmatrix}, v_2 = begin{pmatrix} -dfrac{-sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$lambda_1 = 0, v_1 = begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} \ lambda_2 = a + d , v_2 = begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{lambda_1 t} ~v_1 + c_2 ~e^{lambda_2 t}~ v_2 = c_1 begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} + c_2~ e^{(a+d)t} begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
answered Jan 31 at 5:00
MooMoo
5,64031020
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
We are given
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X\text{where}~ ~~a+d ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - lambda I| = 0$, yielding
$$lambda_{1,2} = frac{1}{2} left(-sqrt{a^2-2 a d+4 b c+d^2}+a+dright),frac{1}{2} left(sqrt{a^2-2 a d+4 b c+d^2}+a+dright)$$
We can then find the associated eigenvectors by solving $[A-lambda_i]v_i = 0$, yielding
$$v_1 = begin{pmatrix}
-dfrac{sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1
end{pmatrix}, v_2 = begin{pmatrix} -dfrac{-sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$lambda_1 = 0, v_1 = begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} \ lambda_2 = a + d , v_2 = begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{lambda_1 t} ~v_1 + c_2 ~e^{lambda_2 t}~ v_2 = c_1 begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} + c_2~ e^{(a+d)t} begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
answered Jan 31 at 5:00
MooMoo
5,64031020
$endgroup$
add a comment |
$begingroup$
We are given
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X\text{where}~ ~~a+d ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - lambda I| = 0$, yielding
$$lambda_{1,2} = frac{1}{2} left(-sqrt{a^2-2 a d+4 b c+d^2}+a+dright),frac{1}{2} left(sqrt{a^2-2 a d+4 b c+d^2}+a+dright)$$
We can then find the associated eigenvectors by solving $[A-lambda_i]v_i = 0$, yielding
$$v_1 = begin{pmatrix}
-dfrac{sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1
end{pmatrix}, v_2 = begin{pmatrix} -dfrac{-sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$lambda_1 = 0, v_1 = begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} \ lambda_2 = a + d , v_2 = begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{lambda_1 t} ~v_1 + c_2 ~e^{lambda_2 t}~ v_2 = c_1 begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} + c_2~ e^{(a+d)t} begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
answered Jan 31 at 5:00
MooMoo
5,64031020
$endgroup$
add a comment |
$begingroup$
We are given
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X\text{where}~ ~~a+d ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - lambda I| = 0$, yielding
$$lambda_{1,2} = frac{1}{2} left(-sqrt{a^2-2 a d+4 b c+d^2}+a+dright),frac{1}{2} left(sqrt{a^2-2 a d+4 b c+d^2}+a+dright)$$
We can then find the associated eigenvectors by solving $[A-lambda_i]v_i = 0$, yielding
$$v_1 = begin{pmatrix}
-dfrac{sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1
end{pmatrix}, v_2 = begin{pmatrix} -dfrac{-sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$lambda_1 = 0, v_1 = begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} \ lambda_2 = a + d , v_2 = begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{lambda_1 t} ~v_1 + c_2 ~e^{lambda_2 t}~ v_2 = c_1 begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} + c_2~ e^{(a+d)t} begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
answered Jan 31 at 5:00
MooMoo
5,64031020
$endgroup$
We are given
$$X' = begin{bmatrix}
a & b \
c & d
end{bmatrix}X\text{where}~ ~~a+d ne 0 , ~~ad-bc=0$$
We can find the eigenvalues using the characteristic polynomial by solving $|A - lambda I| = 0$, yielding
$$lambda_{1,2} = frac{1}{2} left(-sqrt{a^2-2 a d+4 b c+d^2}+a+dright),frac{1}{2} left(sqrt{a^2-2 a d+4 b c+d^2}+a+dright)$$
We can then find the associated eigenvectors by solving $[A-lambda_i]v_i = 0$, yielding
$$v_1 = begin{pmatrix}
-dfrac{sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1
end{pmatrix}, v_2 = begin{pmatrix} -dfrac{-sqrt{a^2-2 a d+4 b c+d^2}-a+d}{2 c}\1end{pmatrix}$$
Now, we can use the conditions we are given, namely $a+d not= 0 ,ad-bc=0 $, by substituting $bc = ad$ in each eigenvalue/eigenvector pair yielding
$$lambda_1 = 0, v_1 = begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} \ lambda_2 = a + d , v_2 = begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
We can now write
$$X(t) = c_1~ e^{lambda_1 t} ~v_1 + c_2 ~e^{lambda_2 t}~ v_2 = c_1 begin{pmatrix} -dfrac{d}{c} \ 1 end{pmatrix} + c_2~ e^{(a+d)t} begin{pmatrix} dfrac{a}{c} \ 1 end{pmatrix}$$
answered Jan 31 at 5:00
MooMoo
5,64031020
edited Jan 31 at 12:10
answered Jan 31 at 5:00
MooMoo
5,64031020
answered Jan 31 at 5:00
MooMoo
5,64031020
answered Jan 31 at 5:00
MooMoo
5,64031020
5,64031020
add a comment |
add a comment |
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$begingroup$
If the matrix $A$ is singular, then $0$ is guaranteed to be (at least) one of the eigenvalues.
$endgroup$
– Decaf-Math
Jan 31 at 2:39
$begingroup$
I found that 0 is one of its eigenvalues but also that a+d is also an eigenvalue
$endgroup$
– Richard Villalobos
Jan 31 at 2:48
2
$begingroup$
"with corresponding eigenvectors both $[0,0]$" <-- That can't be right, because by definition eigenvectors have to be nonzero vectors.
$endgroup$
– zipirovich
Jan 31 at 5:55