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If $A \in \mathbb{M}_{n\times n}(\mathbb{R})$ with $n\ge 2$ has rank $1$, then the minimal polynomial of $A$ is of degree $2$.

I think it is true because i did not get any example which makes it false. So either give its proof or any counter example to disprove this. Thanks in advance.

Pizza or Pasta
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Deepak Gupta
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3 Answers3

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Of course, $0$ is an eigenvalue of $A$ with geometric multiplicity $n-1$. From there we have two possibilities:

One possibility is that $A$ has a second eigenvalue besides zero, which is true if and only if $\operatorname{rank}(A^2) = \operatorname{rank}(A) = 1$ (since the image of $A$ is a one-dimensional invariant subspace). In this case, $A$ is diagonalizable with minimal polynomial $p(x) = x(x- \lambda)$ for non-zero eigenvalue $\lambda$.

The other possibility is that $0$ is the only eigenvalue of $A$. In this case, we find that since $\operatorname{rank}(A^2)<\operatorname{rank}(A)$, we must have $A^2 = 0$. So, the minimal polynomial of $A$ is $p(x) = x^2$.

In either case, the minimal polynomial of $A$ has degree $2$, as we wanted to show.

Ben Grossmann
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  • How do you claim A is diagonalizable?please explain – Cloud JR K Nov 26 '18 at 07:03
  • @CloudJR The non-zero eigenvalue must have an eigenspace of dimension at least $1$. Thus, since $0$ has an $n-1$ dimensional eigenspace, we can select $n$ linearly independent eigenvectors of $A$, which is to say an eigenbasis. – Ben Grossmann Nov 26 '18 at 15:44
  • Got it thanks.. – Cloud JR K Nov 26 '18 at 18:00
  • How can we say that If zero is the only eigen value,then rank(A^2) < rank(A)? – Balasubramannyan S Nov 23 '19 at 10:40
  • If $A,A^2$ have the same rank, then the restriction of $A$ to the column-space of $A$ is invertible, which means that the column-space of $A$ Is a 1 dimensional invariant subspace corresponding to a non-zero eigenvalue – Ben Grossmann Nov 23 '19 at 20:05
  • Or if you prefer: Cayley-Hamilton says that if zero is the only eigenvalue then $A$ is nilpotent. However, if $A,A^2$ have the same rank then $A^k$ has the same rank for all $k$, which means that $A$ can’t be nilpotent – Ben Grossmann Nov 23 '19 at 20:07
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The degree of the minimal polynomial of $A$ cannot be one, otherwise $A$ would be a scalar multiple of $I$, meaning that the rank of $A$ is either zero or $n$ rather than one. Now, as $A$ has rank one, $A=uv^T$ for some vectors $u$ and $v$. Hence $A^2=(uv^T)(uv^T)=u(v^Tu)v^T=(v^Tu)A$, i.e. $A$ is annihilated by the quadratic polynomial $x^2-(v^Tu)x$. So, this must be the minimal polynomial of $A$.

user1551
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Hint (Up to permutation of blocks) the only Jordan normal form matrices of rank $1$ are $$\pmatrix{\lambda} \oplus {\bf 0}_{n - 1}, \quad \lambda \neq 0, \quad \qquad \textrm{and} \qquad \pmatrix{0&1\\0&0} \oplus {\bf 0}_{n - 2} .$$ What are the minimal polynomials of these matrices?

Travis Willse
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