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Let $A$ be an $n\times n $ matrix over $\mathbb C$ such that every non-zero vector of $\Bbb C^n$ is an eigen vector of $A$. Then which are correct?

  1. All eigen values of $A$ are equal.
  2. All eigen values of $A$ are distinct
  3. $A=\lambda I$.
  4. If $\lambda_A $and $m_A $ are characteristic and minimal polynomial of $A$ then $\lambda_A =m_A $

I got that if every non-zero vector of $\Bbb C^n$ is an eigen vector of $A$ then $A$ is a scalar multiple of identity i.e. $A=\lambda I$.

Thus all eigen values of $A$ are equal and hence we have $1,3$ correct and $4$ is false.Am I right?

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1 and 3 are indeed correct: If $v$ and $w$ are eigenvectors of $A$ to different eigenvalues, then $v + w \neq 0$ is no eigenvector. Thus all eigenvalues of $A$ must be equal; call this eigenvalue $\mu$. Because $Ax = \mu x$ for every $x \in \mathbb{C}^n$ we have $A = \mu I$.

2 and 4 hold for $n=1$ and are wrong for $n \geq 2$: If $n = 1$ then all eigenvalues of $A$ are distinct, for $n \geq 2$ this does not hold because $\mu$ has multiplicity $n$. For $n = 1$ we have $\lambda_A(t) = t-\mu = m_A(t)$, but for $n \geq 2$ we have $\lambda_A(t) = (t-\mu)^n \neq t-\mu = m_A(t)$.

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