- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.21 Miscellaneous review exercises on matrices
1. $\quad$ If a square matrix has a row of zeros or a column of zeros, prove that it is singular.
$\quad$ Proof. $\quad$ Let $A$ be an $n \times n$ matrix whose $k^{th}$ column is zeros. Then, for any $n \times n$ matrix $B = (b)_{i, j = 1}^{n, n},$ the $k^{th}$ diagonal element of the product $BA$ is \begin{align*} \sum_{i=1}^n b_{ki}a_{ik} &= \sum_{i=1}^n 0(b_{ki}) \\ &= 0 \end{align*} But if this is true for any $n \times n$ matrix $B,$ then there is no $B$ such that $BA = I.$ Hence, if $A$ has a column of zeros, it is singular.
$\quad$ Now, assume $A$ is a nonsingular matrix with its $k^{th}$ row being zeros. Then, its left-inverse $B$ is also its right-inverse. In other words, $AB = I.$ But since the $k^{th}$ row of $A$ is zeros, the $k^{th}$ diagonal element of $AB$ is given by \begin{align*} \sum_{i=1}^n a_{ki}b_{ik} &= \sum_{i=1}^n 0(b_{ik}) \\ &= 0 \end{align*} But if the $k^{th}$ diagonal element of $AB$ is zero, then $AB \neq I,$ giving a contradiction. Hence, if $A$ has a row of zeros, it is singular. This completes the proof.