Real and Complex Solutions for Linear System with 2x2 Matrix

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  • Thread starter Dustinsfl
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In summary, a linear system with a 2x2 matrix can have real or complex solutions depending on the determinant of the matrix. If the determinant is non-zero, the system will have a unique solution with real values. If the determinant is zero, the system will have infinitely many solutions with complex values. The nature of the solutions can be determined through the use of the fundamental theorem of linear algebra and the properties of determinants.
  • #1
Dustinsfl
2,281
5
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
The characteristic equation is
$$
\lambda^2 -a^2 + 1 = 0\iff \lambda = \sqrt{a^2 - 1}.
$$
The eigenvalues are complex when $-1 < a < 1$ and real otherwise.
Let's consider the real case first.
We have that $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the real case is
$$
\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}.
$$
Next, let's look at the complex case.
We have $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the complex case is
$$
\mathbf{u} = A\exp\left[ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
i\sqrt{1 - a^2} - a
\end{pmatrix} +
B\exp\left[-ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
-i\sqrt{1 - a^2} - a
\end{pmatrix}.
$$

Correct?
 
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  • #2
dwsmith said:
$$
\mathbf{u}' = \begin{pmatrix}
a & 1\\
-1 & -a
\end{pmatrix}\mathbf{u}.
$$
The characteristic equation is
$$
\lambda^2 -a^2 + 1 = 0\iff \lambda = {\color{red}\pm}\sqrt{a^2 - 1}.
$$
The eigenvalues are complex when $-1 < a < 1$ and real otherwise.
Let's consider the real case first.
We have that $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the real case is
$$
\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}.
$$
Next, let's look at the complex case.
We have $(a - \lambda)u_1 + u_2 = 0\iff u_2 = (\lambda - a)u_1$.
Then
$$
u = \begin{pmatrix}
1\\
\lambda - a
\end{pmatrix}.
$$
The solution for the complex case is
$$
\mathbf{u} = A\exp\left[ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
i\sqrt{1 - a^2} - a
\end{pmatrix} +
B\exp\left[-ti\sqrt{1 - a^2}\right]\begin{pmatrix}
1\\
-i\sqrt{1 - a^2} - a
\end{pmatrix}.
$$

Correct?

Hi dwsmith, :)

I see no errors in your solution. But it is unnecessary to consider the real and complex cases separately. You can give the solution as,

\[\mathbf{u} = A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\mbox{ for }a\in\Re\]

Kind Regards,
Sudharaka.
 
  • #3
To expand on this now, referring to http://www.mathhelpboards.com/f10/change-variables-1871/#post8706.

How can I take u back to x with the u solution?
 
  • #4
dwsmith said:
To expand on this now, referring to http://www.mathhelpboards.com/f10/change-variables-1871/#post8706.

How can I take u back to x with the u solution?

Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.
 
  • #5
Ackbach said:
Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.

So it will be $xu$ where u is the Ay_1+By_2?
 
  • #6
Ackbach said:
Just multiply $\mathbf{u}$ by the necessary rotation matrix to get $\mathbf{x}$.

What Ackbach meant was,

\begin{eqnarray}

\mathbf{x} &=& \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\\

&=&\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\right]

\end{eqnarray}

Kind Regards,
Sudharaka.
 
  • #7
Sudharaka said:
What Ackbach meant was,

\begin{eqnarray}

\mathbf{x} &=& \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\\

&=&\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\ \sqrt{a^2 - 1} - a\end{pmatrix}+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}1\\-\sqrt{a^2 - 1} - a\end{pmatrix}\right]

\end{eqnarray}

Kind Regards,
Sudharaka.

So the solution is
\begin{alignat*}{3}
\mathbf{x} & = & \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}\right]\\
& = & A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t - \sin t\left(\sqrt{a^2 - 1} - a\right)\\
\sin t + \cos t\left(\sqrt{a^2 - 1} - a\right)
\end{pmatrix}\\
& & + B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t + \sin t\left(\sqrt{a^2 - 1} + a\right)\\
\sin t - \cos t\left(\sqrt{a^2 - 1} + a\right)
\end{pmatrix}
\end{alignat*}
 
  • #8
dwsmith said:
So the solution is
\begin{alignat*}{3}
\mathbf{x} & = & \begin{pmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{pmatrix}\left[A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
\sqrt{a^2 - 1} - a
\end{pmatrix}
+ B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
1\\
-\sqrt{a^2 - 1} - a
\end{pmatrix}\right]\\
& = & A\exp\left[t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t - \sin t\left(\sqrt{a^2 - 1} - a\right)\\
\sin t + \cos t\left(\sqrt{a^2 - 1} - a\right)
\end{pmatrix}\\
& & + B\exp\left[-t\sqrt{a^2 - 1}\right]\begin{pmatrix}
\cos t + \sin t\left(\sqrt{a^2 - 1} + a\right)\\
\sin t - \cos t\left(\sqrt{a^2 - 1} + a\right)
\end{pmatrix}
\end{alignat*}

Correct. (Yes)
 
  • #9
The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.
 
  • #10
Ackbach said:
The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.

What?
 
  • #11
Ackbach said:
The arguments of the trig functions are all $2t$, not $t$. But that's definitely the right idea.

Hi Ackbach, :)

dwsmith has mentioned(http://www.mathhelpboards.com/f10/change-variables-1871/#post8662) that he had made a typo and the transformation should be,

\[\mathbf{x} = \begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix}\mathbf{u}\]

I was referring to this in my previous post.

Kind Regards,
Sudharaka.
 
  • #12
All I know is this: assuming the original DE is
$$\dot{\mathbf{x}}=a\begin{bmatrix} \cos(2t) & \sin(2t)\\ \sin(2t) & -\cos(2t)\end{bmatrix}\mathbf{x},$$
the substitution
$$\mathbf{x}=\begin{bmatrix} \cos(2t) &-\sin(2t)\\ \sin(2t) & \cos(2t)\end{bmatrix}\mathbf{u}$$
reduces the DE to
$$\dot{\mathbf{u}}=\begin{bmatrix} a &2\\ -2 &-a\end{bmatrix}\mathbf{u}.$$
So you solve this new DE in $\mathbf{u}$ using the usual eigenvalue method, and then multiply by the rotation matrix
$$\begin{bmatrix} \cos(2t) &-\sin(2t)\\ \sin(2t) & \cos(2t)\end{bmatrix}$$
to get the final result.

Whatever substitution you make must have the same arguments for the trig functions as are in the original DE, or else you won't get the nice cancellations you need to get a constant-coefficients equivalent DE in $\mathbf{u}$. If the original DE had simply $t$ as the argument for the trig functions, then so must the substitution.
 

FAQ: Real and Complex Solutions for Linear System with 2x2 Matrix

1. What is the difference between real and complex cases?

Real cases involve only real numbers, while complex cases involve both real and imaginary numbers.

2. How are real and complex cases used in science?

Real and complex cases are used in various fields of science, such as physics, engineering, and mathematics, to model and solve real-world problems that involve both real and imaginary quantities.

3. What are some real-life applications of complex cases?

Complex cases are used in many practical applications, including signal processing, image processing, and control systems. They are also essential in understanding and analyzing phenomena such as electrical circuits, fluid dynamics, and quantum mechanics.

4. Can real and complex cases be solved using the same methods?

No, real and complex cases require different methods of solving. Real cases can be solved using algebraic methods, while complex cases often require the use of complex analysis techniques, such as contour integration and the Cauchy-Riemann equations.

5. Are there any drawbacks to using complex cases?

One potential drawback of using complex cases is that they can be more challenging to visualize and understand compared to real cases. Additionally, some complex problems may not have real solutions, making it harder to interpret the results. However, the benefits of using complex cases often outweigh these challenges in many scientific applications.

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