About this question:

Just want to check whether I understood it correctly.

Example of such matrix would be $A=\begin{pmatrix}3&1&0\\0&3&1\\0&0&3\end{pmatrix}$.

Unique eigenvector is $\begin{pmatrix}1\\0\\0\end{pmatrix}$.

We also have generalized eigenvectors $\begin{pmatrix}0\\1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\0\\1\end{pmatrix}$.

Fundamental solutions are $\begin{pmatrix}e^{3t}\\0\\0\end{pmatrix}$, $\begin{pmatrix}te^{3t}\\e^{3t}\\0\end{pmatrix}$, $\begin{pmatrix}\frac{t^2}2e^{3t}\\te^{3t}\\e^{3t}\end{pmatrix}$.

To me they seem to be linearly independent.

BTW is $e^{\lambda_{3}t}(\vec{x_{3}}+t\vec{x_{3}}+ \frac{t^2}{2!}\vec{x_{1}})$ in your post a typo? Should there be $\vec x_3+t\vec x_2+\frac{t^2}{2!}\vec x_1$?

LaTeX tip: You can write \lambda_1 or \vec x_1 instead of \lambda_{1} or \vec x_{1}. If the index consists only of one item, you do not need the curly brackets. That makes typing math expressions a bit faster.

@MartinSleziak, is indeed a correct interpretation of my question. I doubt that you can create 2 vectors like this with the eigen vector you showed. I mean, from a logical point of view... Those 3 eigen vectors are indeed linear independent meaning FM will be as well

indeed had a typo, thx will remember the tip

@privetDruzia So you are saying that the fundamental solutions I wrote above are different from what you wrote in your post? (Or maybe even incorrect.)

no no, I didn't link your example to my theoretical explanation. I just saw it as an different case

If I follow the notation from your post, I get $e^{3t}\vec x_1=e^{3t}\begin{pmatrix}1\\0\\0\end{pmatrix} =\begin{pmatrix}e^{3t}\\0\\0\end{pmatrix}$.

with which we tried to check my theory

hmm if this is what you get, than u might be right

(that s all new to me)

if you applied the theory I explained and got such results that would mean that the columns of FM are linear independent

but this is so strange:

$e^{3t}(t\vec x_1+\vec x_2) = e^{3t} \begin{pmatrix}t\\1\\0\end{pmatrix} = \begin{pmatrix}te^{3t}\\e^{3t}\\0\end{pmatrix}$

$e^{3t}(\frac{t^2}2\vec x_1+t\vec x_2+\vec x_3) = e^{3t} \begin{pmatrix}\frac{t^2}2\\t\\1\end{pmatrix} = \begin{pmatrix}\frac{t^2}2e^{3t}\\te^{3t}\\e^{3t}\end{pmatrix}$

Of course, I might have made a mistake. (And I have to admit, I did not do anythings with odes for a long time.)

two vectors are linear independent if you cant create the first one with the second one. But here we make the generalized eigenvectors with the first eigenvector => not linear independent. However the result you showed me shows two linear independent vecots. WTH?

The phrase can't create is a bit vague.

I would be more inclined to say: "Two vectors are linearly independent if one of them is not multiple of the other one."

yes

Three vectors are linearly independent if neither of them can be obtained as a linear combination of the other two vectors.

ooh

and the rowexpansion is not a linear combination?

Fact that $(A-\lambda I)\vec x_2=\vec x_1$ does not mean that $\vec x_2$ and $\vec x_1$ are linearly dependent, since you multiplied by a matrix, not by a constant.

@privetDruzia I do not know what rowexpansion is.

Or maybe I am used to a different terminology than you are.

sorry in english it s called: series expansion

so series expansion is not considered as a sort of linear combination?

You mean something like whether $f_3(t)=tf_2(t)+f_1(t)$ means that $f_3$ is linear combination of $f_2$ and $f_1$?

yes

that s an example of a linear conbination

Well, you can call is "a sort of linear combination".

But it is not linear combination in the sense it is defined in linear algebra.

ok that clarifies everything!

Here the vector space you are working with is space of complex-valued functions. It is a vector spaces over the field $\mathbb C$.

So if you are talking about linear dependence/independence in this vectors space, the coefficients of linear combinations should be complex numbers.

And I suppose this is the meaning in which the solution in the fundamental matrix are supposed to be independent.

series expansion is not a linear combination. So you can obtain linear independent vectors a and b with this method altough your calculation is based on some vector previously calculated vector a

mhm

Now I read on Wikipedia that columns of the fundamental matrix have to be independent for all t.

So you can simply fix t, and consider them n-tuples.

But still you are working with a vector space over $\mathbb C$, so the coefficients should be complex numbers.

so what?

what does this imply?

Well, that is the question for you.

Since I did not really understood your argument why you thought that the vectors $\vec y_1$, $\vec y_2$, $\vec y_3$ are linearly dependent.

I tried my best to show you on example that they are not.

mhm

And also I tried to explain what linear dependence/independence means.

I thought they were linear dependent because

you "make" $vec{y_2}$ with $vec{y_1}$

.

Well, but you don't.

yes in the row expansion you use $vec{y_1}$

You "make" $\vec y_2$ from $\vec x_1$ and $\vec x_2$.

right...

At least that seems that case in what you wrote there.

thats completel true

and a very good point!

Sorry, I will have to leave now.

But there was already an answer posted to your question.

As you see, a very experienced user had a look at your question. Which is a good sign.

But maybe I should mention one more thing - to explain why I edited the tags on your question.

When you add tags to a question, tag-exceprt is always shown to you. This is brief description saying what that tag is for.

If you read it, you will see that the tag independence is for questions about probability. And the tag fundamental groups is about fundamental group of a topological space.

So neither of those two tags was related to your question.

I have added linear-algebra, since your question is related to linear independence, linear combinations, eigenvalues, etc.

You might also consider whether some of the tags ode and systems-of-equations could be suitable.