91Ó°ÊÓ

We are given an m × n matrix A with rank m. Let's denote the matrix as follows: \(A =\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \)

Our goal is to find an n × m matrix B such that AB = I_m. Let's denote matrix B as: \(B =\begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1m} \\ b_{21} & b_{22} & \cdots & b_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nm} \end{bmatrix} \)

Since A has rank m, we know that it has m linearly independent columns. As a consequence, we know that the null space of A contains only the zero vector: \(A\mathbf{x} = \mathbf{0} \implies \mathbf{x} = \mathbf{0}\)

We will now construct the matrix B that satisfies AB = I_m. For this purpose, we will use the linearly independent columns of A to form a basis for the column space of A. Let the vectors \( \mathbf{c}_1, \mathbf{c}_2, ..., \mathbf{c}_m \) denote the column basis of A. We will now define the matrix B such that its ith row is the transpose of the vector \( \mathbf{c}_i \). \(B =\begin{bmatrix} \mathbf{c}_1^T \\ \mathbf{c}_2^T \\ \vdots \\ \mathbf{c}_m^T \end{bmatrix}\)

We will now verify that AB = I_m. To do this, we will compute the product AB and check if it results in the identity matrix of order m. When we multiply A with B, the ij-th component of the product AB is given by the dot product of the ith row of A and the jth column of B: \((AB)_{ij} = \mathbf{r}_i \cdot \mathbf{c}_j\) Since A and B share the same column basis, their columns are linearly independent, so the ith row vectors of A will only be orthogonal to the non-corresponding jth row vectors of B. Hence, for i ≠ j, we have: \((AB)_{ij} = \mathbf{r}_i \cdot \mathbf{c}_j = 0\) On the other hand, for i = j, we have: \((AB)_{ij} = \mathbf{r}_i \cdot \mathbf{c}_i = 1\) Thus, the product AB is given by: \(AB =\begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} = I_m\) This confirms that there exists an n × m matrix B such that AB = I_m, as required.