# When is an outer product $ab^\top$ diagonalisable? [duplicate]

If $a$ and $b$ are $n\times 1$ vectors and $A=ab^\top$, why is A diagonalizable if and only if $a^\top b \ne 0$ ?

From my understanding, $a^\top b$ is the sum of the diagonal of A. How does this relate to diagonalizability?

## marked as duplicate by Jack, user1551 matrices 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 1 '16 at 17:35

• If $a=0$ then $A=0$ and $a'b=0$ and $A$ is diagonalizable. – user213008 Dec 1 '16 at 7:09
As pointed out in the comments we need the additional condition that $a,b$ must be non-zero, else $ab^T$ is just the zero matrix, which is of course diagonal. Suppose $a,b$ are both nonzero (so then we must have $ab^T$ is nonzero) and we also have $a^Tb=0$: then notice that $b^Ta=0$ and $$(ab^T)^2 = ab^Tab^T = a(0)b^T = 0$$ so that $ab^T$ is square-zero, and a square-zero matrix which is not the zero matrix itself is not diagonalizable, since its minimal polynomial is $x^2$ and therefore its Jordan form contains a Jordan block $J_2(0)$.
I assume that $a$ and $b$ are not zero.
If $a^Tb = 0$, then $(ab^T)^2 = a(b^Ta)b^T = (a^Tb)ab^T = 0$. So, $ab^T$ is nilpotent but non-zero. So, $ab^T$ is not diagonalizable.
On the other had, if $a^Tb = \mu \neq 0$, then $(ab^T)^2 = (a^Tb) ab^T = \mu ab^T$, which means that if we take $p(x) = x^2 - \mu x$, then $p(ab^T) = 0$. Since $p$ is a polynomial with no repeated factors and $p(ab^T) = 0$, $ab^T$ must be diagonalizable.