$A$ be $n \times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of $A$

eigenvalues-eigenvectorslinear algebraminimal-polynomials

I am trying assignment questions in Linear Algebra and this question could not be solved by me. So, I thought of posting it here.

Let A be an $n\times n$ matrix over $\mathbb{C}$ such that every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A. Then which of the following are true:

  1. All eigenvalues of A are equal
  2. All eigenvalues of A are distinct
  3. $A=\lambda I $ for some $\lambda \in \mathbb{C}$ , where I is the n times n identity matrix
  4. The minimal and characterstic polynomial for A are equal.

I tried by taking eigenvectors to be distinct and the hypothesis but couldn't solve any options. Can you please give hints how to start the problem!

Thank you!

Best Answer

I assume our OP Avenger means "every non-zero vector of ${\mathbb{C}}^n$ is an eigenvector of A" instead of "every non-zero vector of ${\mathbb{C}}^n$ is an eigenvalue of A"; otherwise, the problem makes no sense.

If every non-zero vector $w$ is an eigenvector, each with eigenvalue $\mu_w$,

$Aw = \mu_w w \tag 1$

by choosing two linearly independent vectors $u$ and $v$ we have

$\mu_u u + \mu_v v = Au + Av = A(u + v)$ $= \mu_{u + v}(u + v) = \mu_{u + v}u + \mu_{u + v}v; \tag 2$

linear independence of $u$ and $v$ then forces

$\mu_u = \mu_{u + v} = \mu_v; \tag 3$

thus all eigenvalues are equal to some

$\mu \in \Bbb C, \tag 4$

that is, for any vector $w$,

$Aw = \mu w, \tag 5$

or

$A = \mu I; \tag 6$

$A$ is a scalar multiple of $I$.

The characteristic polynomial of $A$ is

$\det(A - xI) = \det (\mu I - xI)$ $= \det((\mu - x)I) = (\mu - x)^n; \tag 7$

however, $A = \mu I$ also satisfies its minimal polynomial

$m(x) = \mu - x, \tag 8$

and

$m(x) \ne (\mu - x)^n \tag 9$

unless $n = 1$.

So we have items (1) and (3) in the OP's question are true, but (2) and (4) are false.

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