3

This seems like a rather trivial question, but I just want to know first if I am correct that this step is valid and, if so, if there's a name to it. It surely isn't linearity, unless I am mistaken.

Suppose we have matrices A and B. Further, B is a scalar multiple of some other matrix, so perhaps we have $B = 3C$. My question is, would it be valid in multiplying $A$ and $B$ to pull out the constant $3$ and then multiply it by the product of $A$ and $B$? In other words: $$AB = A(3C) = 3(AC)$$ In other words, it is necessarily the case that the above must hold? Is there a name for such a property?

Thanks. I apologize for how elementary this surely is.

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    Yes, it is valid since 3 is a $\color{grey}{\textrm{scalar}}$. See here. Don´t forget to mark an answer as accepted if you appreciate the effort of the respondents. – callculus42 May 17 '19 at 19:43

3 Answers3

2

Yes, constants commute in and out of the matrix multiplication order. In other words, if $x$ is a constant, $A$ is an $n \times k$ matrix and $B$ is $k \times m$ matrix for positive integers $k,m,n$, then $$ x(AB) = (Ax)B = A(xB) = (AB)x. $$

gt6989b
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2

Yes, it does. In fact, this is actually essential by definition of the matrix as a linear operation. A linear operation adheres to two conditions given as follows.

\begin{equation} f(x + y) = f(x) + f(y) \\ f(c \cdot x) = c \cdot f(x) \end{equation}

Every matrix is a linear operation, but not every linear operation is necessarily a matrix.

WafflesTasty
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The answer is yes.

When a matrix is multiplied by a constant each row is multiplied by the same constant. When you multiply matrices you find dot product of the rows of the first one and columns of the second one so the constant could be factored out.

Mohammad Riazi-Kermani
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