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I have a matrix $$A = \begin{bmatrix}1 & 1 \\ -1 & 3\end{bmatrix}$$ I want to find out the generalised Eigenvectors. The Eigen values corresponding to the characteristic equation is $\lambda = 2$ and the Eigenvector correspondig to the eigenvalue is found to be $\begin{bmatrix}1 \\ 1\end{bmatrix}$.

So how to calculate the generalised Eigen vector for this matrix.


What I did is ,

I took $(A-\lambda\cdot I)^2 \nu = 0$.

then solving the $(A-\lambda I)^2 = \begin{bmatrix}1-\lambda & 1 \\ -1 & 3-\lambda \end{bmatrix}^2 = \begin{bmatrix}\lambda^2-2\lambda & 4 - 2\lambda \\ 2\lambda-4 & \lambda^2 - 6\lambda+8 \end{bmatrix}$

At this point I don't know whether I am doing the things correct . as finding the determinant will take this to $\lambda^4$.

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2 Answers 2

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Since $\lambda=2$ has algebraic multiplicity equal to 2, the Jordan form is

$$J = \begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}$$

then by $P=[v_1\,v_2]$

$$P^{-1}AP=J \implies AP=PJ$$

that is

  • $Av_1=2v_1$ (already found)
  • $Av_2=v_1+2v_2 \implies (A-2I)v_2=v_1$
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    $\begingroup$ I am sorry , but I didn't understand it . So no need to find , $(A-\lambda I)^2$ to find the generalised Eigen vectors.? $\endgroup$
    – Lawliet
    Sep 23, 2020 at 13:05
  • $\begingroup$ I'm not sure you need to use Jordan forms here, I mean it works but there are simpler ways. $\endgroup$
    – o's1234
    Sep 23, 2020 at 13:06
  • $\begingroup$ In your solutions, we know J, we know a, but P is partially unknown as we only know $v_1$, then how are you finding the $v_2$. $\endgroup$
    – Lawliet
    Sep 23, 2020 at 13:09
  • $\begingroup$ @Lawliet From the last system $(A-2I)v_2=v_1$. $\endgroup$
    – user
    Sep 23, 2020 at 13:10
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    $\begingroup$ @Lawliet When we can guess Jordan form, this is a way to proceed. User o's 1234 is suggesting another way, maybe preferable in this case. $\endgroup$
    – user
    Sep 23, 2020 at 13:35
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$$ \begin{bmatrix} \lambda^2-2\lambda & 4-2\lambda \\ 2\lambda-4 & \lambda^2-6\lambda+8 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix} $$

Therefore, you can pick any eigenvector you want, so long as it does not satisfy $(A-\lambda I)$.

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  • $\begingroup$ How to find such a vector .? $\endgroup$
    – Lawliet
    Sep 23, 2020 at 13:07
  • $\begingroup$ Because the vector multiplied by the zero vector need be zero, you can pick basically any vector you want, so long as it is linearly independent from the vector you already chose. $\endgroup$
    – o's1234
    Sep 23, 2020 at 13:08
  • $\begingroup$ Ok, can you just elaborate your answer with one of such vector as an example .? $\endgroup$
    – Lawliet
    Sep 23, 2020 at 13:10
  • $\begingroup$ How about $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ $\endgroup$
    – o's1234
    Sep 23, 2020 at 13:36

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