# If $A \in \mathbb{M}_{n\times n}(\mathbb{R})$ with $n\ge 2$ has rank $1$ [duplicate]

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.

## marked as duplicate by Marc van Leeuwen linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 19 '17 at 8:22

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.

• How do you claim A is diagonalizable?please explain – Cloud JR Nov 26 '18 at 7:03
• @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. – Omnomnomnom Nov 26 '18 at 15:44
• Got it thanks.. – Cloud JR Nov 26 '18 at 18:00

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 $A$.

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?