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 :)

  • $\begingroup$ $(0,0)$ is by definition not an eigenvector (eigenvectors have to be nonzero). $\endgroup$ – flawr Aug 1 '14 at 14:51

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.