MHB How to find the eigenvectors for a given matrix using the example in a textbook?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    System
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Solve the system
$$Y'=\begin{bmatrix}2 & 1 & 0 \\0 & 2 & 1 \\ 0 & 0 & 4 \end{bmatrix}Y$$
subtract $\lambda$ from the diagonal entries of the given matrix and take det:
$$\left|
\begin{array}{ccc}
- \lambda + 2 & 1 & 0 \\
0 & - \lambda + 2 & 1 \\
0 & 0 & - \lambda + 4
\end{array}\right|
=(-\lambda+2)^{2}(-\lambda+4)$$
the roots are:
$$\lambda_1=2,\quad\lambda_2=2, \quad\lambda_3=4$$this is the example in the book I am trying to follow but I don't see how they got these vectors or the rest of it
(my matrix is similiar)

https://www.physicsforums.com/attachments/8922
 
Physics news on Phys.org
karush said:
Solve the system
$$Y'=\begin{bmatrix}2 & 1 & 0 \\0 & 2 & 1 \\ 0 & 0 & 4 \end{bmatrix}Y$$
subtract $\lambda$ from the diagonal entries of the given matrix and take det:
$$\left|
\begin{array}{ccc}
- \lambda + 2 & 1 & 0 \\
0 & - \lambda + 2 & 1 \\
0 & 0 & - \lambda + 4
\end{array}\right|
=(-\lambda+2)^{2}(-\lambda+4)$$
the roots are:
$$\lambda_1=2,\quad\lambda_2=2, \quad\lambda_3=4$$this is the example in the book I am trying to follow but I don't see how they got these vectors or the rest of it
(my matrix is similiar)
Okay, so now find your eigenvectors:
[math]\left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right ) \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) = 2 \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )[/math]

etc. (Remember that you have that double eigenvalue, too!)

-Dan
 
topsquark said:
Okay, so now find your eigenvectors:
[math]\left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right ) \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) = 2 \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )[/math]

etc. (Remember that you have that double eigenvalue, too!)

-Dan

So would $\lambda=4$ just be:
$$\displaystyle \left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right )
\left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) =4\left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )$$
 
karush said:
So would $\lambda=4$ just be:
$$\displaystyle \left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right )
\left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) =4\left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )$$
Correct. So one choice of the eigenvector corresponding to the eigenvalue 4 is [math]\left ( \begin{matrix} 1 \\ 2 \\ 4 \end{matrix} \right )[/math].

Can you get the other two?

-Dan

Addendum: Normally (pun intended) for Physics I would normalize this to [math]\dfrac{1}{\sqrt{21}} \left ( \begin{matrix} 1 \\ 2 \\ 4 \end{matrix} \right )[/math], but I don't know what conventions a Mathematician would ordinarily use.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top