MHB Finding Eigenvalues, Eigenvectors, [3]

[shamieh]

Consider the system:

$x' = x + y + z$
$y' = 0x + 2y + 3z$
$z' = 0x + 0y + 3z$

a)Find the eigenvalues for the systemSo after doing my $3 \times 3$ matrix I got: $\lambda_1 = -3$, $\lambda_2 = 1$, and $\lambda_3 = 2$ , is this correct?

b)Find an eigenvector for the smallest eigenvalue

So I am getting the eqn(s): $-4v_1 - 6v_2 - 10v_3 = 0$ but I'm stuck on how to solve now.. I am thinking the only way would be to do $v_1 = v_2 =1$ and $v_3 = -1$ so then wouldn't i have $(^1_{1_{-1}})$

 

[Euge]

Hi shamieh,

Your $\lambda_1$ should be $3$ instead of $-3$.

 

[shamieh]

How? I got det$(A) = 3\lambda^2 - 9\lambda + 6$

 

[Euge]

The matrix is 3 x 3, so how can your characteristic polynomial be second-degree? The characteristic polynomial should be $(\lambda -1)(\lambda -2)(\lambda -3) $.

 

[shamieh]

Euge, you are correct. I was doing the 3x3 matrix but forgetting to put my - $\lambda$ within the corresponding columns. I ended up with $\lambda_1 = 1$ , $\lambda_2 = 3$, $\lambda_3 = 2$ Does this seem correct?

 

[shamieh]

for part b) Find an eigenvector for the smallest eigenvalue I got this for my eigenvector \begin{bmatrix} 0 \\ 4 \\ -2 \end{bmatrix} , is this correct? or should it just be \begin{bmatrix} 4 \\ -2 \end{bmatrix} ? Thanks again in advance for your alls help you are seriously awesome

 

[shamieh]

Also for part c) find an eigenvector for the middle eigenvalue I got \begin{bmatrix} 6 \\ 1 \\ 1 \end{bmatrix}

 

[shamieh]

And also for part d) find an eigenvector for the largest eigenvalue I got \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} then for part e) write the general solution i got:

$X(t) = C_1 \begin{bmatrix} 0\\ 4 \\ 2 \end{bmatrix}e^{t} + C_2 \begin{bmatrix} 6 \\ 1 \\ 1 \end{bmatrix} e^{3t} + C_3 \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} e^{3t}$

Can anyone verify these solutions. Sorry for the triple post, and thanks again.

 

[Euge]

Euge, you are correct. I was doing the 3x3 matrix but forgetting to put my - $\lambda$ within the corresponding columns. I ended up with $\lambda_1 = 1$ , $\lambda_2 = 3$, $\lambda_3 = 2$ Does this seem correct?

Yes, it does -- the eigenvalues are indeed $1,2,3$.

for part b) Find an eigenvector for the smallest eigenvalue I got this for my eigenvector \begin{bmatrix} 0 \\ 4 \\ -2 \end{bmatrix} , is this correct? or should it just be \begin{bmatrix} 4 \\ -2 \end{bmatrix} ? Thanks again in advance for your alls help you are seriously awesome

Since the matrix is $3 \times 3$, conceptually it does not make sense for an eigenvector of this matrix to be $2\times 1$. So you can eliminate the latter option. The former option is incorrect. If say, your matrix is $A$, then any eigenvector $x$ corresponding to the smallest eigenvalue must satisfy $Ax = x$; you can check that your vector does not satisfy this. A suitable eigenvector is $\begin{bmatrix}1\\0\\0\end{bmatrix}$.

Also for part c) find an eigenvector for the middle eigenvalue I got \begin{bmatrix} 6 \\ 1 \\ 1 \end{bmatrix}

Incorrect. A suitable choice for an eigenvector is $\begin{bmatrix}1\\1\\0\end{bmatrix}$.

And also for part d) find an eigenvector for the largest eigenvalue I got \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} then for part e) write the general solution i got:

$X(t) = C_1 \begin{bmatrix} 0\\ 4 \\ 2 \end{bmatrix}e^{t} + C_2 \begin{bmatrix} 6 \\ 1 \\ 1 \end{bmatrix} e^{3t} + C_3 \begin{bmatrix} -1 \\ -3 \\ 1 \end{bmatrix} e^{3t}$

None of your eigenvectors are correct, so your final answer is incorrect. For the last eigenvector, $\begin{bmatrix}2\\3\\1\end{bmatrix}$ is a suitable choice.

 

[shamieh]

For the smallest eigenvector how are you getting \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} when you have \begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix} which turns to three equations...:

$(0) * v_1 -v_2 -v_3 = 0$
$(0) * v_1 - v_2 -3v_3 = 0$
$(0) * v_1 + (0) * v_2 -2v_3 = 0$

 

[shamieh]

So how are you plugging in a $1$ for $v_1$ when $v_1$ doesn't exist... We multiplied it by $0$...

 

[Euge]

For the smallest eigenvector how are you getting \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} when you have \begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix} which turns to three equations...:

$(0) * v_1 -v_2 -v_3 = 0$
$(0) * v_1 - v_2 -3v_3 = 0$
$(0) * v_1 + (0) * v_2 -2v_3 = 0$

What do you mean by the "smallest eigenvector"? The vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ is an eigenvector associated with the smallest eigenvalue $1$. You can see this from your own equations. Your third equation gives $v_3 = 0$. Plugging this back into the first equation, we get $v_2 = 0$. The variable $v_3$ is free to roam. Setting $v_3 = 1$, we obtain $\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$.

So how are you plugging in a $1$ for $v_1$ when $v_1$ doesn't exist... We multiplied it by $0$...

I do not understand the question.

 

[shamieh]

Nevermind , I get it now. So for part d) Find an eigenvector for the largest eigenvalue would this \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} also be correct for the largest eigenvalue vector? because 2 -1 - 1 = 0?

 

[Euge]

Nevermind , I get it now. So for part d) Find an eigenvector for the largest eigenvalue would this \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} also be correct for the largest eigenvalue vector? because 2 -1 - 1 = 0?

How did you get this vector? And what does it have to do with $2 - 1 - 1 = 0$? Please explain.

 

[shamieh]

Ignore my comment, Lol. I understand now. I was thinking way too hard. I am giong to post what I think the general solution is in a second.

 

[shamieh]

so the general solution is: $X(t) = C_1 \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix}e^{t} + C_2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} e^{2t} + C_3 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} e^{3t}$

 

[Euge]

Yes, this is correct. :D

 

[shamieh]

Sorry to be a nuisance Euge, it took me a good 10 minutes to realize that I was solving for what I had and then needed to recursively solve the larger equation. It also took me a moment to notice that you can not have a eigenvector that doesn't contain at least one non-zero. Thanks so much for your help.