Exercise 1.4.2

Suppose the rank-1 matrix $A$ is created by two vectors: $x{y}^T$, so we have for row $i$ and column $j$, $a_{\mathit{ij}}=x_i{y}_j$. Since we know $a_{11},\dots ,a_{1n}$ and $a_{11},\dots ,a_{m1}$, so if we assume $x_1$ is known, then it’s easy to see that $y_j=\frac{a_{1j}}{x_1}$ and $x_i=\frac{a_{i1}}{y_1}=\frac{a_{i1}{x}_1}{a_{11}}$. In the end, we have $a_{\mathit{ij}}=\frac{a_{i1}{a}_{1j}}{a_{11}}$.

For $a_{11}=2$, $a_{12}=3$, $a_{21}=4$, we have $a_{22}=\frac{a_{21}{a}_{12}}{a_{11}}=6$.

This formula breaks down when $x_1=0$, i.e. $a_{11},\dots ,a_{1n}$ are all zeros. The same happens when $y_1=0$. Then rank 1 is impossible (when both are zeros) or not unique.