Let $v_1, \dots, v_4 \in \mathbb{R}^4$ and $$ M = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ v_4 \end{bmatrix} $$ be such that $M$ has four distinct, strictly positive eigenvalues $\lambda_1, \dots, \lambda_4 \in \mathbb{R}$.

For $w_{11}, w_{12}, w_{21}, w_{22} \in \mathbb{R}$, let $$ \tilde M = \begin{bmatrix} v_1 \\ v_2 \\ w_{11} v_1 + w_{12} v_2 \\ w_{21} v_1 + w_{22} v_2 \end{bmatrix}. $$

Clearly, $\tilde M$ is singular and has at most (or exactly?) two non-zero eigenvalues $\tilde \lambda_1, \tilde \lambda_2 \in \mathbb{C}.$ I am wondering: Is there any relationship between $\tilde\lambda_1, \tilde\lambda_2$ and $\lambda_1, \dots, \lambda_4$?

I know that all must come from the set $$ \Big\{ l \in \mathbb{R} \mid \exists x \in \mathbb{R}^4 : \Big(\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} - l \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix}\Big) x = 0 \Big\}, $$ but can one say anything more?


1 Answer 1


To answer the part about the number of zero eigenvalues. Clearly, $v_1$ is linearly independent with $v_2$ (otherwise $M$ would have zero determinant and thus a zero eigenvalue, which is forbidden).

The kernel of $\tilde M$ is$$\ker \tilde M=(span\{v_1,v_2\})^\bot,$$ and the dimension of this space is $4-\dim span\{v_1,v_2\}=4-2=2$, therefore the zero eigenvalue has geometric multiplicity $2$.


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