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Let $A$ be an $n\times n$ matrix over $\mathbb{C}$ such that every nonzero vector of $\mathbb{C}^n$ is an eigenvector of $A$. Then which of the following hold?

  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. If $\chi_A$ and $m_A$ denote the characteristic polynomial and the minimal polynomial, respectively, then $\chi_{A}=m_{A}$.

I tried the problem, but I can not reach any conclusion.

Solution: I know that for every eigenvector, there corresponds unique eigenvalue, so the matrix has distinct eigenvalues, and hence its characteristic and minimal polynomials are the same. I have some confusion about my answer, please help me.

Jon Warneke
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Warrior
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    If every nonzero vector is an eigenvector, then there is certainly a basis of $\mathbb{C}^n$ consisting only of eigenvectors. Choose such a basis, and denote the eigenvalue corresponding to the first vector $x_1$ in the basis by $\lambda$. What must be the eigenvalue(s) of the other vectors in the basis? Hint: $x_1,x_2$, and $x_1+x_2$ are all eigenvectors; what must their eigenvalues be? – Ian Dec 23 '15 at 15:52
  • Sorry: what is the problem asking you to do? Multiple choice? Prove something? – Alekos Robotis Dec 23 '15 at 15:53
  • yes this is a multi correct answer, – Warrior Dec 23 '15 at 15:54
  • What if $A = I$? Then not all eigenvalues of $A$ are distinct. – Jon Warneke Dec 23 '15 at 15:57
  • @JonWarneke according to you 1 and 3 are correct. am I correct? but i have a little confusion about the option 2 and 4. are these wrong? please clarify. – Warrior Dec 23 '15 at 16:09
  • If the eigenvalues are distinct then not every vector can be an eigenvector as Ian's hint helps you to show. – mlu Dec 23 '15 at 16:14

2 Answers2

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We can see that all the eigenvalues must be equal. Suppose they were not, that is for some vectors $v,w$ we had $$ Av=\lambda_1 v, Aw=\lambda_2w$$ where $ \lambda_1\ne\lambda_2$ and $\lambda_1\ne0,\lambda_2\ne 0$. Then $$ A(v+w)=\lambda_1v+\lambda_2w$$ and $v+w$ is not an eigenvector for $A$, because there is no $\lambda_3$ such that $$ A(v+w)=\lambda_3(v+w).$$ So, the eigenvalues must be equal and not distinct. If this matrix scales every vector by the same value $\lambda$, we know that it is a scalar multiple of the identity based on the matrix representation. $$ A= \begin{pmatrix} \lambda&0&0&0\\ 0&\lambda&0&0\\ 0&0&\ddots&0\\ 0&0&0&\lambda \end{pmatrix} =\lambda \begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\ddots&0\\ 0&0&0&1 \end{pmatrix}=\mathbb{I}.$$ Finally, we must determine whether $\chi_A=m_A.$ Because the Jordan blocks are all of size one, we can see that the minimal and characteristic polynomials are not equal, but maybe you haven't learned this yet.

We know the minimal polynomial is the monic polynomial of least degree such that $$ m_A(A)=0.$$ Whereas, the characteristic polynomial $\chi_A$ has the eigenvalues of $A$ as roots, repeated according to multiplicity. So: $$ \chi_A(z)=(z-\lambda)^n$$ where $n$ is the dimension of the vector space. We can see that to be the minimal monic polynomial such that $m_A(A)=0$, $$ m_A(z)=(z-\lambda).$$ So, the two polynomials are not equal.

In conclusion, options $1$ and $3$ hold.

Alekos Robotis
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  • thank you, +1 for you @Antonios-Alexanders Robotis. – Warrior Dec 23 '15 at 16:15
  • You do not need the assumption than $\lambda_1$ and $\lambda_2$ are nonzero, only distinct, but the corresponding eigenvectors must be nonzero. – mlu Dec 23 '15 at 16:19
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As explained in the answers of the question Demonstration: If all vectors of $V$ are eigenvectors of $T$, then there is one $\lambda$ such that $T(v) = \lambda v$ for all $v \in V$., the hypothesis implies that you are dealing with a scalar matrix (a scalar multiple of the identity matrix). This implies that points 1 and 3 are true, while points 2* and 4 are false unless $n\leq 1$.

*Provided that one interprets "all eigenvalues" as "all roots of the characteristic polynomial", so that the eigenvalues being distinct is a non-trivial property. With the more natural definition "all scalars $\lambda$ such that $\dim(\ker(T-\lambda I))>0$", saying that all eigenvalues are distinct would be trivially true.

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Marc van Leeuwen
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