4

How to show that the matrix

$$B= \left(\begin{array}(\lambda_1 & a & b \\ 0 & \lambda_1 & c\\ 0 & 0 & \lambda_2\end{array}\right)$$

is diagonalizable when $a\neq0$, when $\lambda_1\neq \lambda_2$. How should I work this out? I tried comparing the algebraic and geometric multiplicities of the eigenvalues of $B$ but I didn't succeed in this approach yet.

Daniel Fischer
  • 206,697
jjepsuomi
  • 8,619

2 Answers2

4

It suffices to note that $B - \lambda_1 I$ has nullity (null-space dimension) $1$ when $a \neq 0$ and $2$ when $a = 0$. So, the geometric multiplicity of $\lambda_1$ is $1$ when $a \neq 0$ and $2$ when $a = 0$.

Ben Grossmann
  • 225,327
  • Thank you so much! I was on the right track then. Sry for my simple questions, I'm still learning the subject ;) – jjepsuomi Feb 09 '16 at 14:59
  • 1
    The fact that your questions admit a satisfactory, quick answer is not a sign that your questions are inherently simple; the way I see it, it means that you're asking exactly the right questions and doing the work in between on your own. Good luck! – Ben Grossmann Feb 09 '16 at 15:05
  • Nice to hear that. Appreciate it and thank you :) – jjepsuomi Feb 09 '16 at 15:12
1

Clearly, when the dimension is $3$ and the repeated diagonal entries appear only in contiguous blocks, the problem is not really destabilizing.

The case when the dimension is $n$ is studied in If a matrix is triangular, is there a quicker way to tell if it is can be diagonalized?

When the repeated diagonal entries appear only in contiguous blocks, the matrix $B$ is similar to the matrix of diagonal blocks; here $B$ is similar to $diag(\begin{pmatrix}\lambda_1&a\\0&\lambda_1\end{pmatrix},\lambda_2)$. Thus $B$ is diagonalizable iff each block is diagonal; consequently, the complexity of this algorithm is $0$ algebraic operation.

The case when the equal eigenvalues are not grouped is much more difficult. We have not been able to find an algorithm that does the job in $O(n^2)$ and even in $o(n^3)$.

Who can do better?

Community
  • 1