MHB Solve Eigenvalues, Eigenvectors & General Solution for X'=AX

[shamieh]

Consider the system $x'_1 = x_1 + 2x_2$ and $x'_2 = 3x_1 + 2x_2$

If we write in matrix from as $X' = AX$, then

a) $X =$

b) $X' =$

c) $A =$

d) Find the eigenvalues of **A**.

e) Find eigenvectors associated with each eigenvalue. Indicate which eigenvector goes with which eigenvalue.

f) Write the general solution to the system.

g) Find the specific solution that satisfies the initial conditions $x_1(0) = 0$ and $x_2(0) = -4$


**Ok so here are my solutions so far**

a) $X = \vec{X} = (^{x_1}_{x_2})$

b) $X' =$ \begin{bmatrix} (1-\lambda) & 2 \\ 3 & (2-\lambda) \end{bmatrix}

c) $A =$ \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix}

d) $\lambda_1 = -1$ and $\lambda_2 = 4$

e,f,g) **Please Help!** Not sure what to do.

Thanks in advance. ( I also need someone to verify that my answers are correct).

 

[I like Serena]

Hi shamieh,

For b) you should have $X'=(^{x_1'}_{x_2'})$.
Your other answers are correct.

For e) you need to solve $(-I - A)X = 0$ respectively $(4I -A)X = 0$.
These equations should lead to infinitely many solutions, which are actually lines.
You need to find a vector on each line, which are the eigenvectors.

 

[shamieh]

Serena, I am kind of uncertain on how to do that. Do you mean that I should do this? $(-I - A)X =$ \begin{bmatrix} -2 & -3 \\ -4 & -3 \end{bmatrix} ?

I am a little uncertain on how to apply $(-I - A)X$

 

[I like Serena]

Serena, I am kind of uncertain on how to do that. Do you mean that I should do this? $(-I - A)X =$ \begin{bmatrix} -2 & -3 \\ -4 & -3 \end{bmatrix} ?

I am a little uncertain on how to apply $(-I - A)X$

What we have is:
$$(-I-A)X = \left(\begin{bmatrix}-1&0\\0&-1\end{bmatrix} - \begin{bmatrix}1&2\\3&2\end{bmatrix}\right)X
=\begin{bmatrix}-2&-2\\-3&-3\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}
= 0$$

The standard way to solve such an equation is with Gausssian elimination, also known as row reduction.
Are you familiar with that process?

 

[shamieh]

So now would I get
$-2x_1 - 2x_2 = 0$
$-3x_1 - 3x_2 = 0$

 

[I like Serena]

So now would I get
$-2x_1 - 2x_2 = 0$
$-3x_1 - 3x_2 = 0$

Yes. Can you solve it?
That is, find at least one non-zero solution, considering that any multiple will also be a solution?

 

[shamieh]

So I'm thinking that $x_1 = 1$ & $x_2 = -1$ because $3x_1 + 3x_2 = 0$ ?

 

[I like Serena]

So I'm thinking that $x_1 = 1$ & $x_2 = -1$ because $3x_1 + 3x_2 = 0$ ?

Yep. You've found the eigenvector corresponding to eigenvalue -1.

 

[shamieh]

So my first vector will be $v_1 = e^{-t} (^1_{-1}) = (^{e^{-t}}_{-e^{-t}})$ ?

 

[shamieh]

Also does \begin{bmatrix} stuff && stuf \\ morestuf && mrstuf \end{bmatrix} and $(^s_x)$ both mean the same thing ? Are they both implying Matrices or is the parenthesis implying something different? I've noticed some people use them interchangeably and some people not. I noticed that parenthesis are used in sets $(^n_k)$

 

[I like Serena]

So my first vector will be $v_1 = e^{-t} (^1_{-1}) = (^{e^{-t}}_{-e^{-t}})$

You seem to be jumping ahead, and it's not quite correct yet.

The eigenvector for $\lambda=-1$ is $(^{1}_{-1})$.
This is the first half of the answer to e).

 

[shamieh]

So finally I have: $v_1 = 1$ , $v_2 = -1$ so $eigenvector_1 =$ \begin{bmatrix} e^{-t} \\ -e^{-t} \end{bmatrix} and $w_1 = 2$, $w_2 = 3$ so $eigenvector_2 =$ \begin{bmatrix} 2e^{4t} \\ 3e^{4t} \end{bmatrix}

- - - Updated - - -

Oh I get what your saying. The eigenvector corresponding to $\lambda_1$ is $(^1_{-1})$

 

[I like Serena]

Also does \begin{bmatrix} stuff && stuf \\ morestuf && mrstuf \end{bmatrix} and $(^s_x)$ both mean the same thing ? Are they both implying Matrices or is the parenthesis implying something different? I've noticed some people use them interchangeably and some people not. I noticed that parenthesis are used in sets $(^n_k)$

More or less. The first two both signify matrices, although the second is really a vector that just might be interpreted as a matrix. However, the notation $\binom n k$ usuaĺly denotes binomium or combination, which is something completely different.

 

[shamieh]

so that means that the second eigenvector corresponding with $\lambda_2$ is going to be ($ ^{2e{4t}}_{3e^{4t}}$)

 

[I like Serena]

so that means that the second eigenvector corresponding with $\lambda_2$ is going to be ($ ^{2e{4t}}_{3e^{4t}}$)

No. The eigenvector is not a function of t.
Those functions only come in the later questions.

 

[shamieh]

so for part f) finding the general solution would it just be $X(t) = C_1 (^2_3)e^{4t} + C_2 (^1_{-1})e^{-t}$ ?

- - - Updated - - -

No. The eigenvector is not a function of t.
Those functions only come in the later questions.

So then how do I write the final solution for part e) ?

 

[shamieh]

and for g) I got $X(t) = -4/5 (^2_3) e^{4t} - 8/5(^1_{-1})e^{-t}$

 

[I like Serena]

So then how do I write the final solution for part e) ?

The second part is that the eigenvector for $\lambda=4$ is $\binom 2 3$.

so for part f) finding the general solution would it just be $X(t) = C_1 (^2_3)e^{4t} + C_2 (^1_{-1})e^{-t}$ ?

and for g) I got $X(t) = -4/5 (^2_3) e^{4t} - 8/5(^1_{-1})e^{-t}$

Looks good! (Nod)