Matrix Multiplication is Homogeneous of Degree 1
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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:
- $\mathbf A \paren {\lambda \mathbf B} = \lambda \paren {\mathbf A \mathbf B}$
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Proof
Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$.
\(\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: \, \) | \(\ds \mathbf A \paren {\lambda \mathbf B}\) | \(=\) | \(\ds \lambda \sum_{k \mathop = 1}^n a_{i k} b_{k j}\) | Definition of Matrix Product (Conventional) and Definition of Matrix Scalar Product | ||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n a_{i k} \paren {\lambda b_{k j} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf A \paren {\lambda \mathbf B}\) | Definition of Matrix Product (Conventional) and Definition of Matrix Scalar Product |
$\blacksquare$
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This article is complete as far as it goes, but it could do with expansion. In particular: proof literally carries over for any commutative ring in place of $\Bbb F$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding this information. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Expand}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |