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Given the matrix $A=\begin{pmatrix} 5 & 1 \\ -4 & 1 \end{pmatrix}$. I am supposed to find the general solution of the form $y'=Ay$. Since it has $\lambda=3$ with an algebraic multiplicity of 2, and I find $v_1 =(1,-2)^T$.

How am I supposed to find another eigenvector?

I was given the equation $(A- \lambda_1 I) v_2=v_1$ to find another eigenvector $v_2$. How does this idea come about?

xyz12354
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    You can't. The dimension of $\ker (A-3I) $ is one. You can find a generalised eigenvector. – copper.hat Mar 04 '20 at 06:12
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    Another eigenvector: $(2, -4)^\top$. Another linearly independent eigenvector? It can't be done. – user754697 Mar 04 '20 at 06:18
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    Well, that is a good point. Anything in $\ker (A-3I)$ is an eigenvector, but the dimension of the space in this problem in one, so you can't diagonalise the matrix with eigenvectors. – copper.hat Mar 04 '20 at 06:21
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    What do you mean by "find the general solution for this matrix"? What is the goal? –  Mar 04 '20 at 06:27
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    You’re clearly meant to find a generalized eigenvector and compute the Jordan normal form per the conventional method that is taught. However, there’s no need to find any eigenvectors, generalized or otherwise when working with a real $2\times2$ matrix. See this answer for all three cases of the exponential. – amd Mar 04 '20 at 07:05
  • @amd that is helpful – xyz12354 Mar 04 '20 at 15:02
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    The addition of the $y'=Ay$ really changes the question in a significant way. – copper.hat Mar 04 '20 at 17:28

2 Answers2

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You can find $v_1 = (1,-2)^T$ and one solution of $(A-3I)v_2 = v_1$ is $v_2 = (0, 1)^T$, so we see that $v_1,v_2$ is a basis. Since $(A-3I)v_1 = 0$, we see that $(A-3I)^2 = 0$ (this could have been figured out from the characteristic polynomial as well, so in this case there is no need to find eigenvectors).

We see that $e^{(A-3I)t} = \sum_{k=0}^\infty {t^k \over k!} (A-3I)^k = I+(A-3I)t$.

Hence $e^{At} = e^{3t} e^{(A-3I)t} = e^{3t}(I + (A-3I)t) = e^{3t} \begin{bmatrix} 1+2t & t \\ -4t & 1-2t\end{bmatrix}$.

copper.hat
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Here the first eigen vector corresponding to the eigen value $~\lambda = 3~$ is $~v_1 =\begin{pmatrix} 1 \\ -2 \end{pmatrix}~$. We have to find the second eigen vector corresponding to the repeated eigen value $~\lambda = 3~$.

For these we have to find the vector $~v_2=\begin{pmatrix} \alpha \\ \beta \end{pmatrix}~$ such that $~(A-3 I)v_2=v_1~$. Now $$~(A-3 I)v_2=v_1$$ $$\implies \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix}\cdot \begin{pmatrix} \alpha \\ \beta \end{pmatrix}=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$$ $$\implies 2\alpha~+~\beta~=~1$$Putting $~\alpha=1~,\beta=-1~$.

$$\therefore v_2=\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

So the general solution is $$y(x)=c_1~v_1~e^{3x}~+~c_2\left(v_1~x~+~v_2\right)~e^{3x}$$ $$y(x)=c_1~e^{3x}~\begin{pmatrix} 1 \\ -2 \end{pmatrix}~+~c_2~e^{3x}~\left[\begin{pmatrix} 1 \\ -2 \end{pmatrix}~x~+~\begin{pmatrix} 1 \\ -1 \end{pmatrix}\right]$$where $~c_1,~c_2~$ are constants.

Note: For the question How does this idea come about ?, please find the reference given below :

"Differential Equations" by Shepley L. Ross

nmasanta
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