# Eigenvalues of $ab^T$

In deriving a Newton scheme, I end up with a Jacobian matrix of the form $J=I+ab^T$ where $a,b$ are vectors. For practical reasons, I want to approximate it by a symmetric positive definite matrix. Symmetrizing is easy: replace it by $J \approx I+\frac 12(ab^T+ba^T)$. But the (now real) eigenvalues may still be negative, so I further approximate $J \approx \tilde J := I+\frac 12\alpha (ab^T+ba^T)$ where I want to choose $\alpha$ so that $\tilde J$ is s.p.d.

To choose $\alpha$, I need the eigenvalues of the matrix $ab^T$. This ought to be simple enough, but I must be missing the key step to find a closed-form expression for the eigenvalues. Clearly, there will be at most two non-zero eigenvalues, and the corresponding eigenvectors must lie in the plane spanned by the vectors $a,b$. The eigenvalues are then the minimal and maximal value of the Rayleigh quotient $$R(x) := \frac{x^T (ab^T) x}{x^T x} = \frac{(a^Tx) (b^Tx)}{x^T x}$$ over all $x \in \text{span}(a,b)$.

I suspect that there is a closed form expression for the minimal and maximal value of $R(x)$, but its form eludes me.

• Isn't that simply $\lambda_{1,2} = a^Tb$? Commented Jul 6, 2016 at 13:08
• Sorry, should be $\lambda = a^Tb$, because $ab^T$ is rank 1 and therefore only has one non-zero eigenvalue (with eigenvector $a$). Commented Jul 6, 2016 at 13:15
• $(ab^T)a=(b^Ta)a$ implies that the right-eigenvector is $a$ and the eigenvalue is $b^Ta$ Commented Jul 6, 2016 at 13:56
• Also, since $ab^T$ is nonsymmetric in general, the maximal absolute value of its Rayleigh quotient (i.e. its numerical radius) is in general strictly larger than its maximum absolute eigenvalue (i.e. its spectral radius). Commented Jul 6, 2016 at 14:00
• Ah, interesting question. Maybe I have managed to mislead myself then. What I want is that $\tilde J$ is positive definite. For this, I actually need the minimal value of $R(x)$ (i.e., the most negative value, if $R(x)$ can actually be negative). As @RichardZhang points out, this may be a question unconnected to the eigenvalue(s) of $ab^T$. Commented Jul 6, 2016 at 20:51

This is sometimes known as Buzano's inequality (http://www.jstor.org/stable/2159168). In general, if $\|x\|=1$, $P=xx^{\top}$ is a projection operator, so a simple application of Cauchy-Schwarz leads to $$2|b^{\top}xx^{\top}a|-|b^{\top}a| \leq |b^{\top}(2P-I)a| \leq \|a\|\,\|b\|,$$ which gives Buzano's inequality $$|b^{\top}xx^{\top}a| \leq \frac{\|a\|\,\|b\|+|a^{\top}b|}{2}.$$
Without loss of generality let $a=(1,0)$, $b=(p,q)$ ($p^2+q^2=1$, $q\geq0$) and $x = (\alpha, \beta)$ ($\alpha^2+\beta^2=1$, $\alpha\geq0$). Finding the extrema of $b^{\top}xx^{\top}a = \alpha(\alpha p+\beta q)$ is easily done with Lagrange multipliers, giving the values $$\frac{p\pm1}{2}.$$ Here $p = \frac{a^{\top}b}{\|a\|\|b\|} = \cos\angle ab$.
From this $$\frac{a^{\top} b-\|a\|\|b\|}{2} \|x\|^2 \leq a^{\top} xx^{\top}b \leq \frac{ a^{\top}b+\|a\|\|b\|}{2} \|x\|^2$$ (with both inequalities being tight) so the admissible values of $\alpha$ are given by $$1 + \tfrac12(p\pm1)\alpha \|a\|\|b\| > 0, \qquad \frac{-2}{\|a\|\|b\| + a^{\top}b} < \alpha < \frac{2}{\|a\|\|b\|-a^{\top}b}.$$