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I have generated many random tridiagonal matrices and I considered that the null space of all of them is empty. Is this true for all tridiagonal matrices?

M a m a D
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2 Answers2

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The answer is no. For a nontrivial example consider $$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{bmatrix}$$

mechanodroid
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  • Thanks. I guess in general, for random matrices, the null space is empty. I have generated hundreds of matrices, the null space of all of them was empty – M a m a D Jul 07 '19 at 22:12
  • @MamaD What do you mean by that? Couldn't you have come across the above matrix if you randomly generated them? – Maximilian Janisch Jul 07 '19 at 22:13
  • @MaximilianJanisch you are right, this is an example which rejects the title. I have generated about 4000 symmetric tridiagonal random matrices and all of them had empty null space. I didn't come across the matrix given as solution. – M a m a D Jul 07 '19 at 22:17
  • @MamaD I see what you mean; I have added an explanation for this phenomenom below – Maximilian Janisch Jul 07 '19 at 22:27
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@mechanodroid has already provided a counter-example. I want to comment on why you didn't come accross any tri-diagonal matrices that have a non-trivial null space:

Note that the null space of a matrix $A$ equals $\{0\}$ if and only if the $A$ is not invertible). However, the Lebesgue measure (on $\Bbb R^{(n^2)}$) of the set of non-invertible $n\times n$ matrices is $0$.

A similar thing happens if you construct a Lebesgue measure (on $\Bbb R^{3n-2}$) only for tri-diagonal matrices. So, informally (since the Lebesgue measure is not at all a probability measure), the probability of randomly choosing a tri-diagonal matrix with non-trivial null-space is $0$.

Maximilian Janisch
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