Matrix Multiplication is Homogeneous of Degree 1

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.