# Eigenvalue of rank 2 matrix

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?

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.

• 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 3, 2019 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 $(y^Tx)x$. Apr 4, 2019 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 4, 2019 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. Apr 4, 2019 at 17:30

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.