# 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. |