# Problem about real square matrix with rank 1 [duplicate]

Given $A \in \mathbb{R}^{n \times n}$ and $\text{rank}(A) = 1$. By working only on real field, show that $A$ is diagonalizable if and only if $\text{tr}(A) \neq 0$. Here, $\text{tr}(A)$ is the sum of all eigenvalues of $A$.
## 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(); } ); }); }); Apr 25 '15 at 11:55
• @Servaes Prove that 0 is an eigenvalue of $A$ and it has geometric multiplicity of $n-1$. If $A$ is diagonalizable, $0$ also has algebraic multiplicity of $n-1$, so there is another nonzero eigenvalue, so tr(A) is not zero. On the other hand, if $A$ is not diagonalizable, then $0$ must have algebraic multiplicity of $n$, so that tr(A) must be zero. Is it true? – Kevin Limanta May 23 '14 at 15:19