Rank-1 matrices¶
Outer product¶
Consider the following matrix product:
If we notice the matrix on the right, each row is a multiple of the row vector \(\begin{bmatrix}x\\ y \\z\end{bmatrix}^T\) and each column is a multiple of the column vector \(\begin{bmatrix}a\\b\\c\end{bmatrix}\). From this we can deduce that this matrix has rank \(1\). The product of the two vectors on the left is called the outer product. We can go the other way and claim that every matrix of unit rank can be expressed as the outer product of two vectors:
To see why this is true, start with any \(m \times n\) matrix \(A\) of unit rank. If the rank is \(1\), then every vector in the column space of \(A\) is of the form \(ku\), where \(u = \begin{bmatrix}u_1 & \cdots & u_m\end{bmatrix}^T\) . Then, for each vector in the standard basis, we have scalars \(v_1, \cdots, v_n\) such that \(Ae_i = v_i u\). Note that \(Ae_i\) is nothing but the \(i^{th}\) column of the matrix \(A\). Therefore, we have:
Note
Every \(m \times n\) matrix of unit rank is of the form \(uv^T\) where \(u \in \mathbb{R}^m\) and \(v \in \mathbb{R}^n\).