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?
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2If $a=0$ then $A=0$ and $a'b=0$ and $A$ is diagonalizable. – Christoph Mark Dec 01 '16 at 07:09
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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 $J_2(0)$.
Christiaan Hattingh
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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.
Ben Grossmann
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