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I'm trying to understand division by zero cases in the eigenvector of a three-by-three real symmetric matrix, and how to avoid them.

I have the following matrix, where every value is real: \begin{equation} M_3 = \begin{bmatrix} a & d & f \\ d & b & e \\ f & e & c \end{bmatrix} \end{equation}

For this question, assume that the eigenvalues are already known: $\lambda_1, \lambda_2, \lambda_3$, with $\lambda$ standing in for any one of them. With that in mind, I can solve $\left(M_3 - \lambda \text{I} \right) v = \left(M_3 - \lambda \text{I} \right) \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = 0$ and get the following equations:

\begin{equation} \left( a - \lambda \right) v_1 + d v_2 + f v_3 = 0\\ d v_1 + \left( b - \lambda \right) v_2 + e v_3 = 0\\ f v_1 + e v_2 + \left( c - \lambda \right) v_3 = 0 \end{equation}

From there, I set $v_3 = 1$, use the first equation to solve for $v_1$, then use the middle equation to solve for $v_2$. I end up with:

\begin{equation} v = \begin{bmatrix} \dfrac{f^2 - \left(b - \lambda \right) - d e }{d^2 - \left(a - \lambda \right) \left(b - \lambda \right)} \\\phantom{1}\\ \dfrac{e \left( a - \lambda \right) - d f}{ d^2 - \left(a - \lambda \right) \left(b - \lambda \right)} \\\phantom{1}\\ 1\end{bmatrix} \end{equation}

I've already isolated a few zero cases. If $d=e=f=0$, then $M_3$ is already diagonal and the answer is trivial. And if $d = f = 0$ (or $d = e = 0$ or $e = f = 0$), then one of the eigenvalues is isolated and the other two are from the remaining 2x2 submatrix.

But if $d^2 - \left(a - \lambda \right) \left(b - \lambda \right) = 0$, then uh... I'm not sure what to do about it. This looks like it means the upper left $2\times2$ sub-matrix within the $3\times3$ matrix is singular, but I don't know what to make of that - it's been a long time since I dealt with eigen-things, and my old textbook is still un-enlightening on the subject. This is where I'm stuck. I'm hoping to find a closed-form or non-iterative method.

If having examples where this happens might help, here are two: \begin{equation} \begin{bmatrix}4&4&4\\4&4&4\\4&4&4\end{bmatrix}, \lambda_1 = 12, \lambda_2 = 0, \lambda_3 = 0 \\ \begin{bmatrix}10&6&4\\6&1&2\\4&2&7\end{bmatrix}, \lambda_1 \approx 15.385, \lambda_2 \approx 4.615, \lambda_3 = -2 \end{equation}

And there are so many ways to solve for the eigenvector with those three equations, all of which are going to have a possible division by zero. With just $v_3 = 1$, I'll have three equations with two unknowns and any two of those equations could be solved four ways meaning there's twelve ways to solve with just $v_3 = 1$, suggesting there's 36 ways to solve for the eigenvector if I try with $v_1 = 1$ or $v_2 = 1$ (I think, I might be wrong here). So that could be 36 (or more) possible division by zero cases, even though there's not that many unique numbers within the matrix. I'm not sure solving it another way would reduce or remove the number of possible division by zero cases.

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    $\begingroup$ Your solution method is interesting and will work whenever $v_3$ is a free variable. However it does not need to be. More general approaches would include, for example, using row operations to reduce $M-\lambda I$ to reduced row echelon form. $\endgroup$ Commented Oct 27 at 23:59
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    $\begingroup$ I should add that when I say "your method will work" I have not verified the algebra. I mean only that whenever $v_3$ is free, making the simplifying $v_3=1$ should lead to a valid solution. $\endgroup$ Commented Oct 28 at 0:04
  • $\begingroup$ Well, if $v_3 \neq 0$, we can rescale it to 1. So, we would only have to consider the $v_3 = 0$ as well. $\endgroup$ Commented Oct 28 at 2:14

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In general, there is not any simple closed-form solution formula, because your symmetric matrix can have a repeated eigenvalue, and there is no canonical way to pick a single-dimensional subspace from a multi-dimensional eigenspace.

However, if $\lambda$ is not a repeated eigenvalue, then its eigenspace is one-dimensional. Therefore $\operatorname{adj}(M-\lambda I)$ is rank-one and any nonzero column of $$ \operatorname{adj}(M-\lambda I) =\pmatrix{ (b-\lambda)(c-\lambda)-e^2 &ef-(c-\lambda)d &de-(b-\lambda)f\\ ef-(c-\lambda)d &(a-\lambda)(c-\lambda)-f^2 &df-(a-\lambda)e\\ de-(b-\lambda)f &df-(a-\lambda)e &(a-\lambda)(b-\lambda)-d^2} $$ can serve as an eigenvector that spans the eigenspace.

Since you don’t know which column is nonzero in advance, you should not pick a fixed column in the above and use it as a solution formula, because that column may turn out to be zero.

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  • $\begingroup$ I'm not familiar with this adj operator - I can kinda see how you got the last column, but not the others. Can you show details on how you calculated them? Also, what happens if every column is zero? $\endgroup$ Commented Nov 5 at 22:07
  • $\begingroup$ So nothing can be done when there's a repeated eigenvalue (when not orthogonal I think)? Even symbolic solvers can spit out something - not that they're consistent, but still... something. $\endgroup$ Commented Nov 5 at 22:29
  • $\begingroup$ @Status Something can be done. You just need more branching and need to make more arbitrary choices. Since your matrix is only 3x3, if $\lambda$ is a repeated eigenvalue, the rank of $M-\lambda I$ is either 0 or 1. If the rank is 0, you may choose any nonzero vector as an eigenvector. If the rank is $1$, then $M-\lambda I$ must have a nonzero row $u$. Hence you may pick any nonzero vector $v$ such that $u\cdot v=0$. E.g. if $u_1=0$, you may set $v=(1,0,0)$; if $u_1\ne0=u_2$, you may set $v=(0,1,0)$; if $u_1,u_2\ne0$, you may set $v=(u_1,-u_2,0)$. These choices are completely arbitrary. $\endgroup$ Commented Nov 6 at 3:16
  • $\begingroup$ If you add how the columns of $\operatorname{adj}(M-\lambda I)$ were found and what can be done with the eigenvectors when the matrix is rank 1 into your answer I'll accept it. $\endgroup$ Commented Nov 22 at 14:55
  • $\begingroup$ @Status The calculation of the adjugate matrix is covered in most introductory linear algebra texts. $\endgroup$ Commented Nov 26 at 8:55

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