# 28.2 Solve the system y'={}Y

#### karush

##### Well-known member
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)

View attachment 8922

#### topsquark

##### Well-known member
MHB Math Helper
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:
$$\displaystyle \left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right ) \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) = 2 \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )$$

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

-Dan

#### karush

##### Well-known member
Okay, so now find your eigenvectors:
$$\displaystyle \left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right ) \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) = 2 \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )$$

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

-Dan
So would $\lambda=4$ just be:
$$\displaystyle \left ( \begin{matrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 4 \end{matrix} \right ) \left ( \begin{matrix} a \\ b \\ c \end{matrix} \right ) =4\left ( \begin{matrix} a \\ b \\ c \end{matrix} \right )$$

#### topsquark

##### Well-known member
MHB Math Helper
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 $$\displaystyle \left ( \begin{matrix} 1 \\ 2 \\ 4 \end{matrix} \right )$$.

Can you get the other two?

-Dan

Addendum: Normally (pun intended) for Physics I would normalize this to $$\displaystyle \dfrac{1}{\sqrt{21}} \left ( \begin{matrix} 1 \\ 2 \\ 4 \end{matrix} \right )$$, but I don't know what conventions a Mathematician would ordinarily use.