# Minimal polynomial of a matrix whose elements have a certain form [duplicate]

Find the minimal polynomial of the $n$-dimensional matrix $(a_{ij})$ when the matrix elements $a_{ij}$ have the form $a_{ij} = u_i v_j.$

Let $A=uv^T$ where $u,v$ are column vectors.

Then rank$(A)\leq$rank$(u)\leq1.$ So kernal$(A)\geq n-1.$ That is, the geometric multiplicity $\geq n-1.$ According to Jordan decomposition theorem, the number of Jordan blocks w.r.t. $0\ \geq n-1.$ Therefore, the algebraic multiplicity of $0$ $\geq n-1.$

Suppose rank$(A)=1,$ how do I find the other eigenvalue?

## 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(); } ); }); }); Feb 12 '15 at 9:32

• @xavierm02 So the remaining eigenvalue is $<u,v>.$ But how do I get the minimal polynomial from this? – lovelesswang Jul 29 '14 at 7:03
Hint: the vector $u$ is an eigenvector