Matrix Multiplication is Homogeneous of Degree 1
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proof literally carries over for any commutative ring in place of $\Bbb F$
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Theorem
Let $\mathbf A$ be an $m \times n$ matrix and $\mathbf B$ be an $n \times p$ matrix such that the columns of $\mathbf A$ and $\mathbf B$ are members of $\R^m$ and $\R^n$, respectively.
Let $\lambda \in \mathbb F \in \set {\R, \C}$ be a scalar.
Then:
- $\mathbf A \paren {\lambda \mathbf B} = \lambda \paren {\mathbf A \mathbf B}$
Proof
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$
| \(\, \displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 p: \, \) | \(\displaystyle \mathbf A \paren {\lambda \mathbf B}\) | \(=\) | \(\displaystyle \lambda \sum_{k \mathop = 1}^n a_{i k} b_{k j}\) | Definition of Matrix Product (Conventional) and Definition of Matrix Scalar Product | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n a_{i k} \paren {\lambda b_{k j} }\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \mathbf A \paren {\lambda \mathbf B}\) | Definition of Matrix Product (Conventional) and Definition of Matrix Scalar Product |
$\blacksquare$
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proof literally carries over for any commutative ring in place of $\Bbb F$
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