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Suppose $x$ and $y$ are two linearly independent nonzero vectors in $\mathbb{R}^n$. Then we know that the matrix $M = xy^T + yx^T$ is a rank 2 matrix. I seem to have made the observation that the two nontrivial eigenvalues of $M$ are given by $\lambda_{1,2} = x^Ty \pm \|x\|\|y\|$. For example if $x = \begin{bmatrix}a & b\end{bmatrix}^T$ and $x = \begin{bmatrix}c & d\end{bmatrix}^T$, then the characteristic polynomial is given by $${{s}^{2}}-\left( 2{{x}^{T}}y \right)\cdot s-{{\left( bc-ad \right)}^{2}}$$ and application of the quadratic formula verifies that discriminant part (under the square root) is given by: $4\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)$ so the observation holds. I have tried some numeric computation for $n=3,4,5$ and the formula seems to work. But I haven't been able to prove it.

Approach one:

Write,

\begin{align} \det \left( M-\lambda I \right) &=\det \left( x{{y}^{T}}+\underbrace{\left( -y{{x}^{T}}-\lambda I \right)}_{:=A} \right) \\ & =-\left( 1+{{y}^{T}}{{\left( -y{{x}^{T}}-\lambda I \right)}^{-1}}x \right)\det \left( y{{x}^{T}}+\lambda I \right) \\ \end{align}

and try to use the lemma but that didn't take me anywhere.

Approach two:

\begin{align} & Mv=x{{y}^{T}}v+y{{x}^{T}}v=\left( {{x}^{T}}y+\left\| x \right\|\left\| y \right\| \right)v \\ & \Rightarrow x{{y}^{T}}v+y{{x}^{T}}v={{x}^{T}}yv+v\sqrt{{{x}^{T}}x{{y}^{T}}y} \\ \end{align}

and try to match the left and right hand side, but I couldn't get that to work either.

Can someone provide a hint or proof? Also, how would one go about deriving what the eigenvectors corresponding to the two eigenvalues look like?

ITA
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2 Answers2

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An eigenvector $v$ of $M$, not belonging to the $0$ eigenvalue, should be in the subspace generated by $x$ and $y$. Therefore, $v=rx+sy$. If its eigenvalue is $\lambda$, then $\lambda (rx+sy)=Mv=x(y^T(rx+sy))+y(x^T(rx+sy))$. In other words,

$$(r(y^Tx)+s(y^Ty)-\lambda r)x+(r(x^Tx)+s(x^Ty)-\lambda s)y=0$$

Since $x,y$ are linearly independent, then each coefficient must be zero.

This is a homogeneous linear system in $r$ and $s$ with matrix $$\begin{pmatrix}y^Tx-\lambda&y^Ty\\x^Tx&x^Ty-\lambda\end{pmatrix}$$

Therefore, the eigenvalues are the roots of $$\lambda^2-(y^Tx+x^Ty)\lambda+(y^Txx^Ty-x^Txy^Ty)=0$$

Or what is the same

$$\lambda^2-\left[(x^Ty+\|x\|\|y\|)+(x^Ty-\|x\|\|y\|)\right]\lambda +\left((x^Ty)^2-(\|x\|\|y\|)^2\right)=0$$

from where you see that those are the roots.

Alternatively, try the vectors $v_\pm=\|y\|x\pm\|x\|y$. For these

$$Mv=x(\|y\|y^Tx\pm\|x\|\|y\|^2)+y(\|y\|\|x\|^2\pm\|y\|x^Ty)=(x^Ty\pm\|x\|\|y\|)v_\pm$$

Observe that, by Cauchy's inequality, the eigenvalues $x^Ty\pm\|x\|\|y\|$ are either equal or zero if and only if $x$ and $y$ are linearly dependent.

user647486
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  • Its your first equation correct? Expanding the first term for example gives: $r{{y}^{T}}xx+s{{y}^{T}}yx-\lambda rx$. But the original expansion has no $xx$ term, what is $xx$ when $x$ is a vector? The $x,y$ are vectors so the order matters. – ITA Apr 03 '19 at 23:04
  • @ITA The factors in parentheses are scalars. The equation is correct and the $x$ on the right is what is customary of writing the scalar on the left and the vector on the right. Now, when you 'expand' it you cannot write it as $y^Txx$ as you wrote. It must be kept as $(y^Tx)x$. – user647486 Apr 04 '19 at 13:30
  • To each his own. When you deal with matrices and vectors a lot, the non-commutativity matters quite a bit and the convention is to preserve order ... not inspect manually what quantity constitutes a scalar and what doesn't and then re-arrange as per "custom". – ITA Apr 04 '19 at 16:47
  • @ITA That is until you stop working with matrices and start working with linear operators, and move from having scalars in a field to having scalars in a, not necessarily commutative, ring. In that case, precisely due to the non-commutativity, you would like to write the scalars in the right place. – user647486 Apr 04 '19 at 17:30
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The rank-two matrix $M=xy^T+y x^T$ in matrix form can be written as $AB$, where $A=(x,y)$ and $B=(y^T;x^T)$. (A has two columns, and B has two rows.) It can be shown (Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$?) that the nontrivial eigenvalues of $AB$ equal those of $BA$. As a result, the nontrivial eigenvalues of $M$ are given by the two eigenvalues of the 2-by-2 matrix $BA$ \begin{pmatrix} yx^T & yy^T \\ xx^T & xy^T \\ \end{pmatrix} which is the same as the other answer.

Hanwen
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