Given vector \(a\) with length \(m\) and \(b\) with length \(n\), the outer product of them defined as:
\[ a b^\top = C \tag{1-1} \]While matrix \(C\) has dimension \(m \times n\).
Here is a brief discussion of the property of matric \(C\)
Special Case: \(a = b\)
Then we have:
\[ A = a a^\top \tag{1-2} \] \[ A = \left[ \begin{matrix} a_1 a & a_2 a & \cdots & a_n a \end{matrix} \right] = \left[ \begin{matrix} a_1 a^\top \\ a_2 a^\top \\ \vdots \\ a_n a^\top \end{matrix} \right] \tag{1-3} \] \[ A = \left[ \begin{matrix} a_1 a_1 & a_1 a_2 & \cdots & a_1 a_n \\ a_2 a_1 & a_2 a_2 & \cdots & a_2 a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_n a_1 & a_n a_2 & \cdots & a_n a_n \\ \end{matrix} \right] \tag{1-4} \]Here are several properties of matrix A:
- \(det(A) = 0\)
- \(trace(A) = \|a\|^2\)
- \(rank(A) = 1\)
- Eigenvalues of \(A\) are: \([\|a\|^2, 0, \cdots, 0]\)
According to equation (1-3), all column or row vectors of matrix \(A\) are linear correlated, so its rank is 1. Meanwhile, \(A\) is a symmetrical semi positive definite matrix, \(trace(A) = \Sigma_{i = 1}^n \lambda_i\), \(\lambda\) is the eigenvalue of \(A\). Because \(rank(A) = 1\), only the first eigenvalue of \(A\) is not zero. Therefore, \(trace(A) = \lambda_1 = \|a\|^2\).
General Case
\[ a b^\top = C \tag{2-1} \]Properties of matrix C:
- Denote \(k = \text{min}(m,n)\)
- \(\text{det}(C) = 0\)
- \(\text{trace}(C) = a[0:k]^\top b[0:k] \)
- \(\text{rank}(C) = 1\)
- Singular values of \(C\) are: \([\|a\|\|b\|, 0, \cdots, 0]\)
Decomposition
Given a 1-rank matrix \(A \in \mathbb{R}^{m \times n}\), it can be decomposed to:
\[ A = u v ^\top \] \[ u = k \left[ \begin{matrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \\ \end{matrix} \right] \qquad v ^\top = \frac{1}{k a _{11}} \left[ \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \end{matrix} \right] \quad (k \neq 0) \tag{2-2} \]Because of the existence of \(k\), the decomposition is not unique.
Specially, if \(u = v\), \(u = \frac{1}{\sqrt{a_{11}}}[ \begin{matrix}a_{11} & a_{12}& \cdots& a_{1n} \end{matrix} ] ^\top\).