- #1
castrodisastro
- 82
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Homework Statement
Find the general solution of y'=Ay. Your answer must be a real-valued function.
[tex]A=
\begin{pmatrix}
1 & 1\\
0 & 1\\
\end{pmatrix}
[/tex]
Homework Equations
The Attempt at a Solution
The first step would be to find the eigenvalues. I forgot the name of the term but if it is "triangular"(?) then the eigenvalues are the numbers along the diagonal (I took the determinant of the matrix and it also gave me λ=1(multiplicity 2))
So the next step would be to find the eigenvector corresponding to λ=1.
Plugging 1 for λ yields...
[tex]
\begin{pmatrix}
0 & 1\\
0 & 0\\
\end{pmatrix}
\begin{pmatrix}
a\\
b\\
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
\end{pmatrix}
[/tex]
[itex]\vec{v}[/itex]=[tex]
\begin{pmatrix}
a\\
b\\
\end{pmatrix}
[/tex]
This gives me b=0
So now what?
I know what I am supposed to do. I need to solve for the vector [itex]\vec{v}[/itex] in A[itex]\vec{v}[/itex]=0, but if b=0, then would [itex]\vec{v}[/itex]=
[tex]
\begin{pmatrix}
0\\
0\\
\end{pmatrix}?
[/tex]
That would make it a trivial solution, in other words, not the eigenvector.
This seems like it is the most simple problem. The professor worked another problem almost identical to this one, but he goes so fast I can only either copy the notes or follow his process, not both. In his example, he took b=0 and somehow it became the vector
[tex]
\begin{pmatrix}
1\\
0\\
\end{pmatrix}
[/tex]
I don't understand why...or maybe I wrote down the notes incorrectly.
I also know that once I solve for the eigenvector, then I can express the solution as
x=eλt[itex]\vec{v}[/itex]
Which I then have to break up into a real solution and an imaginary solution. I don't see how I would even get imaginary solutions since the eigenvalues are real numbers.
The professor and the book do not teach concepts, or attempt to help us understand what is happening with each of these processes, he just recites the process. Which doesn't give us the tools to figure things out in cases like these.
Some guidance would be much appreciated.