Given the constant matrix

A = \begin{bmatrix}4&-1&0\\3&1&-1\\1&0&1\end{bmatrix}

Find the fundamental matrix.

After finding the eigenvalue $\lambda$=2 with multiplicity 3, then trying to find the eigenvector corresponding to the eigenvalue, I end up with matrix \begin{bmatrix}2&-1&0\\3&-1&-1\\1&0&-1\end{bmatrix}, then after trying to find the corresponding eigenvector I end up with u=(1,2z,z), which I obtained from letting x=1 and subtracting the first and second rows with appropriate multiplication of 3$R_1$ and -2$R_2$ and adding the rows. Also, the reduced matrix has the form


Then the first eigenvector u is u=(1,2,1). Nonetheless, since the eigenvalue has muliplicity 2 shouldn't there be two more eigenvectors in terms of some variable t?

  • $\begingroup$ Note that $$A-2I=\begin{bmatrix}2&-1&0\\3&-1&-1\\1&0&\color{#f00}{-3}\end{bmatrix}.$$ $\endgroup$ – awllower Oct 20 '18 at 8:07
  • $\begingroup$ Are you sure that you wrote the right matrix. I ask this because $2$ is not an eigenvalue of the one that you mentioned (in fact, it has no rational eigenvalues). Besides, what is a “fundamental matrix”? $\endgroup$ – José Carlos Santos Oct 20 '18 at 8:10
  • $\begingroup$ I apologize I have fixed the matrix, it seems I accidentally added a negative to the element (3,3) it should be just 1, NOT -1 $\endgroup$ – lastgunslinger Oct 20 '18 at 8:20
  • $\begingroup$ Eigenvector is $u=(z,2z,z)$ or: any non-zero multiple of $(1,2,1).$ $\endgroup$ – user376343 Oct 20 '18 at 9:03
  • 1
    $\begingroup$ The algebraic multiplicity of the eigenvalue $2$ is $3$, but the geometric multiplicity is $1$. $\endgroup$ – Math1000 Oct 20 '18 at 10:37

Your Answer

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

Browse other questions tagged or ask your own question.