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I didn't know how to evaluate $$4A-\begin{bmatrix} 2 &-2 \\ 0 & 2 \end{bmatrix}A$$

and so I looked in the solutions, and what they did was they rewrote $4A$ as

$$\begin{bmatrix} 4 &0 \\ 0 & 4 \end{bmatrix}A-\begin{bmatrix} 2 &-2 \\ 0 & 2 \end{bmatrix}A$$

Is this allowed? Why does this work?

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    Have you tired to do it in components? – 2017-02-23
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    It's 4 multiplied by the identity matrix x A – 2017-02-23
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    It works by the **definition** of [scalar multiplication of matrices](https://en.wikipedia.org/wiki/Scalar_multiplication#Scalar_multiplication_of_matrices). – 2017-02-23
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    Assuming A is a 2x2 matrix (which it must be for the expression to make sense) then 4A = 4(I$\times$ A)= (4I)A$ and $4I $ is the the matrix wit 4 in its diagonal. – 2017-02-23

2 Answers 2

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Note that $A=IA$ and $kI = \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix}$ for all $k$, so it follows that $$4A = 4IA = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}A$$

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    The identity $4(IA)=(4I)A$ is implicitly used, which is not obvious and one might eventually end up with proving $4A=4(IA)$ by definition. – 2017-02-23
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    @Jack: I don't see the problem. Obvious or not, when doing matrix algebra (like in this question), it is fine to assume basic rules of matrix multiplication and scalar multiplication without re-proving them every time. – 2017-02-23
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    Surely my comment was not an objection to your answer. "it is fine to assume basic rules of matrix multiplication and scalar multiplication without re-proving them every time" I would assume that this is not a suggestion for one who does not know the "basic rules" or does not know the necessity of these rules (yet). – 2017-02-23
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It follows from the definition of scalar multiplication of matrices:

$$ k\begin{pmatrix} a&b\\ c&d \end{pmatrix}= \begin{pmatrix} ka&kb\\ kc&kd \end{pmatrix}. $$ By this definition, for any $2\times 2$ matrix $A$ and any scalar $k$, one can check by direct calculation that $$ kA=(kI)A. $$ Of course one can write $$ kA=k(IA)=(kI)A, $$ if one knows

  • how $kI$ is defined
  • why one has "associativity" in the second identity.