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

  1. All eigen values of $A$ are equal.
  2. All eigen values of $A$ are distinct.
  3. $A=\lambda I$ for $\lambda \in \mathbb{C}$ where $I$ is an $n\times n$ identity matrix.
  4. If $\chi_A$ and $m_A$ denote the characteristic polynomial and minimal polynomial respectively, then $\chi_A=m_A$.

Now for $4$, I know that $\chi_A=m_A$ implies the eigen space has dimension $1$, that is all the eigen-values are different. But how can I conclude if every non-zero vectors are eigen-vectors or not if this happens.

Again if $A=\lambda I$ then all the eigen values are same and every non-zero vector is an eigen vector. But I am confused to analyze it properly. Any help please.


3 Answers 3


HINT: It’s easy to see that if $A=\lambda I$, then every non-zero vector in $\Bbb C^n$ is an eigenvector of $A$ for the eigenvalue $\lambda$, so it’s certainly possible for all of the eigenvalues of $A$ to be equal; this shows you that (2) and (4) are not necessarily true, while (1) and (3) can be true. All that remains is to decide whether (3) must be true (in which case (1) will also be true).

Suppose that $x,y\in\Bbb C^n$, $Ax=\lambda_1x$, and $Ay=\lambda_2y$, where $\lambda_1\ne\lambda_2$. Then


and $A(x+y)=\lambda(x+y)$ for some $\lambda\in\Bbb C$, so $\lambda x+\lambda y=\lambda_1x+\lambda_2y$, and therefore


Observe that at least one of $\lambda-\lambda_1$ and $\lambda_2-\lambda$ is non-zero and derive a contradiction, thereby showing that all eigenvalues of $A$ must be equal and hence that (1) must be true.

Now if all of the eigenvalues are equal to $\lambda$, say, so that $Ax=\lambda x$ for each $x\in\Bbb C^n$, then $A-\lambda I=0$ for all $x\in\Bbb C^n$, so what must $A$ be?

  • $\begingroup$ One question: why one of $\lambda-\lambda_1$ and $\lambda_2-\lambda$ is non-zero? If one of them is non-zero then they are different and having same eigenvectors to different eigen-values will lead to a contradiction. But why they should be non-zero in the first place? $\endgroup$ Commented May 16, 2016 at 2:33
  • $\begingroup$ @Kushal: We're assuming that $\lambda_1\ne\lambda_2$, so $\lambda_1-\lambda\ne\lambda_2-\lambda$. Thus, if one of those differences is $0$, the other one is not, and neither is its negative. $\endgroup$ Commented May 16, 2016 at 3:00
  • $\begingroup$ Oh missed that point. Got it now. $\endgroup$ Commented May 16, 2016 at 3:03

Hint Suppose that every nonzero vector is an eigenvector of $A$. We can recover algebraically the eigenvalue $\lambda$ of an eigenvector $\bf x$ using the fact that $$\frac{\bar{\bf x}^T A {\bf x}}{\bar{\bf x}^T {\bf x}} = \frac{\bar{\bf x}^T \lambda {\bf x}}{\bar{\bf x}^T {\bf x}} = \lambda .$$ This suggests considering the (continuous) function $${\bf y} \mapsto \frac{\bar{\bf y}^T A {\bf y}}{\bar{\bf y}^T {\bf y}} ,$$ which, since every nonzero vector is an eigenvector, returns the eigenvalue of $\bf y$. What does continuity of this function imply if $A$ has two different eigenvalues?


Suppose, for the sake of illustration, that $n = 3$.

Let $e_1, e_2, e_3$ be the standard basis, so that $e_1 = (1, 0, 0)$, etc.

Remember (or compute) that the columns of $A$ are precisely its action on the basis vectors. But the basis vectors are also eigenvectors of $A$! So (e.g.) $Ae_1 = (\lambda_1, 0, 0)$ for some $\lambda_1$. Deduce that all the off-diagonal entries of $A$ are zero.

Now consider $A$ applied to the vector $e_1 + e_2 = (1, 1, 0)$. You'll get $(\lambda_1, \lambda_2, 0)$. But this is also an eigenvector of $A$, so it is a multiple of $(1, 1, 0)$. What does that tell you about $\lambda_1$ and $\lambda_2$?


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