# Diagonalization of $M=ab^t+ba^t$

Given $a=(a_i)_{i=1}^n$ and $b=(b_i)_{i=1}^n$ vectors in $\mathbb{R}^n$, we define the matrix $M=(m_{ij})_{i,j=1}^n$ as: $$m_{ij}=a_ib_j + a_jb_i,$$ or equivalently $$M=ab^t+ba^t.$$ Note that $a,b$ are also $n\times 1$ matrices, $a^t,b^t$ are $1\times n$ matrices.

What are the eigenvalues and eigenvectors of $M$?

-
Since rank$(M)\le 2$, at least $n-2$ of the eigenvalues are zero, and the corresponding eigenvectors are orthogonal to $\vec{a}$ and $\vec{b}$. –  hardmath Jul 29 at 16:25

We are considering the following two cases:

Case I. $a$, $b$ linearly dependent. Let for example $b=\mu a$, then $M=2\mu\, a a^t$. But the rank of $aa^t$ is 1, as all its columns are multiple of the first column (if the first is nonzero.), and hence its eigenvectors are $a$, with eigenvalue $\lambda= 2\mu\, a^ta$, and $n-1$ eigenvectors corresponding to $\lambda=0$, which constitute a basis of the kernel of $aa^t$.

Case II. If $a$, $b$ linearly independent (and hence ∥a∥,∥b∥≠0). Then $$Mx=\lambda x \,\,\Longleftrightarrow\,\, a(b^tx)+b(a^tx)=\lambda x,\tag{1}$$ thus, if $\lambda\ne0$, then $x$ is a linear combination of $a$ and $b$. Hence $M$ has as eigenvalues $\lambda=0$ with multiplicity $n-2$, and eigenspace $$\{x:a^tx=b^tx=0\}.$$ Also, setting $x=c_1a+c_2b$ in $(1)$ we obtain the system $$a^ta c_1+a^tb c_2=\lambda c_1, \quad a^tb c_1+b^tb c_2=\lambda c_2,$$ which provides another two eigenvalues/eigenvectors. The eigenvalues satisfy the equation $$0=\left| \begin{matrix}a^ta-\lambda & a^tb \\ a^tb & b^tb-\lambda \end{matrix} \right|=\lambda^2-(|a|^2+|b|^2)\lambda+(|a|^2|b|^2-(a^tb)^2),$$ and thus \begin{align} \lambda_{1,2}&=\frac{|a|^2+|b|^2\pm\sqrt{(|a|^2+|b|^2)^2-4(|a|^2|b|^2-(a^tb)^2)}}{2} \\ &=\frac{|a|^2+|b|^2\pm\sqrt{(|a|^2-|b|^2)^2+4(a^tb)^2}}{2}. \end{align}

-
i just use matlab for showing result –  dato datuashvili Jan 13 at 12:05
@yoirgos: thanks ! Can you conclude and give the values of $\lambda_1$ and $\lambda_2$ and the formula for the associated eigenvectors ? –  mellow Jan 13 at 17:00
@Yiorgos I believe you made a mistake somewhere; by my calculations your system of equations should have been $$a^t b c_1 + a^tb c_1 = \lambda c_2,\\ a^t bc_1 + b^t b c_2 = \lambda c_1$$ I suggest you recheck your algebra –  Omnomnomnom Jan 13 at 18:11

Some small notes, since Yiorgos' has done a fairly thorough job:

The key insight here is that $M$ can be written very nicely as the sum of rank-1 matrices.

That is: if we take $a,b$ as vectors in $\mathbb{R}^n$, $ab^T$ gives us the matrix whose entries are $M'_{ij} = a_i b_j$, and $b a^T$ gives us the matrix whose entries are $M''_{ij} = a_j b_i$. Noting that $M_j = M'_j + M''_j$, it follows that $M = ab^T + ba^T$, and Yiorgos' analysis follows.

Note also that for vectors $u,v \in \mathbb{R}^n$, we can express the dot-product as $$u \cdot v = v^T u = u^T v$$ Hence, the set $$\{x:a^Tx=b^Tx=0\}.$$ is the set of all $x \in \mathbb{R}^n$ that are orthogonal to both $a$ and $b$.

So, suppose that $a,b$ are linearly independent. We note that we may select $n-2$ eigenvectors corresponding to $\lambda = 0$ by selecting an orthogonal basis for the subspace $$\{x:a^Tx = b^Tx = 0\} = [\operatorname{span}(a,b)]^T$$ The remaining 2 eigenvalues and eigenvectors, we note that because $M$ is symmetric, we may select a set of mutually orthogonal eigenvectors, so that two mutually orthogonal eigenvectors can necessarily be selected from the set $$\operatorname{span}(a,b) = \{c_1 a + c_2 b : c_1,c_2 \in \mathbb{R} \}$$ So, consider any $x$ that can be written as $x = c_1a + c_2 b$, and suppose that $x$ is an eigenvector for some $\lambda$. We have

$$a b^T (c_1 a + c_2 b) + b a^T(c_1 a + c_2 b) = \lambda(c_1 a + c_2 b)$$ We may rearrange the equation to get $$[(b^T a - \lambda) c_1 + b^T b c_2] a + [a^T a c_1 + (a^T b - \lambda) c_2] b = 0$$ Since $a$ and $b$ are linearly independent, this gives us a system of equations for $c_1$ and $c_2$: $$(b^T a - \lambda) c_1 + (b^T b) c_2 = 0\\ (a^T a) c_1 + (a^T b - \lambda) c_2 = 0$$ Or in other words, the set of $(c_1,c_2)$ corresponding to eigenvectors of the original matrix are precisely the eigenvectors of the new matrix $$\pmatrix{b^T a & b^Tb\\ a^T a & a^T b}$$

-