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Compute the eigenvalues ​​and eigenvectors of the following matrix T:

$$T =\begin{pmatrix} 1&4\\ 0&1 \end{pmatrix}$$

I have $(T-\lambda I)=0$ and from this I found $(\lambda -1)^2$, so $\lambda = 1$

But when I found eigenvectors, $\left( \begin{smallmatrix} 0&4\\ 0&0 \end{smallmatrix} \right)$ $\left( \begin{smallmatrix} X1\\ X2 \end{smallmatrix} \right)$ = $\left( \begin{smallmatrix} 0\\ 0 \end{smallmatrix} \right)$

X2 = 0

X1 can take any value.

Using WA I have two eigenvectors, (0,0) and (1,0) and the same eigenvalue λ=1. What I'm doing wrong? Any help is welcomed :)

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  • $\begingroup$ $(0,0)$ is by definition not an eigenvector (eigenvectors have to be nonzero). $\endgroup$ – flawr Aug 1 '14 at 14:51
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For any triangular matrix the coefficients on the diagonal are the eigenvalues of the given matrix. In our case the only eigenvalue is $1$ and the vector $(1,0)^T$ is an eigenvector associated to $1$. There isn't other linearly independent eigenvector since the matrix isn't diagonalizable.

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  • $\begingroup$ I'm a little confused, lets say that $\lambda=8$ and the matrix is triangular too, all the conditions are the same. Is $(8,0)^T$ the eigenvector of this new matrix? $\endgroup$ – Luis Armando Aug 1 '14 at 15:05
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    $\begingroup$ For this matrix $$\begin{pmatrix}\lambda&\alpha\\0&\lambda\end{pmatrix}$$ $\lambda$ is the unique eigenvalue and $(\beta,0)^T$ for all $\beta\ne0$ is an eigenvector associated to this eigenvalue. $\endgroup$ – user63181 Aug 1 '14 at 15:11
  • $\begingroup$ I'm agree with you. Then, why $(1,0)^T$ is the response, could have be any value instead 1? $\endgroup$ – Luis Armando Aug 1 '14 at 15:21
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    $\begingroup$ Yes as I said $(\beta,0)^T$ is an eigenvector for all $\beta\ne0$; but notice that $$(\beta,0)^T=\beta (1,0)^T$$ which means that $(1,0)^T$ generates all other eigenvectors $\endgroup$ – user63181 Aug 1 '14 at 15:25
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As it ends up, this matrix has only the one eigenvalue ($\lambda = 1$) and the one dimensional eigenspace spanned by $\pmatrix{1&0}^T$. The $0$-vector is never considered an eigenvector.

Note that it is, in a sense, unusual for an $n \times n$ matrix to have fewer than $n$ linearly independent eigenvectors (which is why WA looks for a $2$nd eigenvector where there is none). However, it is possible, as this problem exemplifies. Matrices with fewer than $n$ linearly independent eigenvectors are non-diagonalizable.

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