## Linear Algebra and Its Applications, Exercise 2.6.14

Exercise 2.6.14. Suppose that $A$, $B$, and $C$ are linear transformations, with $A$ taking vectors from $V$ to $W$, $B$ taking vectors from $U$ to $V$, and $C$ taking vectors from $X$ to $U$. Consider the product $(AB)C$ of these transformations. It starts with a vector $x$ in $X$ and produces a new vector $u = Cx$ in $U$. It then follows 2V on page 123 to apply the product linear transformation $AB$ to $u$ producing a final vector $w$ in $W$ representing the result $(AB)Cx$.

i) Is the result of this process the same as separately applying $C$ followed by $B$ followed by $A$?

ii) Is the final result of this process the same as applying $BC$ followed by $A$? In other words, does the associative law apply to linear transformations, so that $(AB)C = A(BC)$?

Answer: i) In computing $(AB)Cx$ we start with a vector $x$ in $X$ and apply the linear transformation $C$ to produce a new vector $u = Cx$ in $U$. We then apply the product transformation $AB$ to the vector $u$ to produce a vector $w = (AB)u$ in $W$.

But by rule 2V the linear transformation $AB$ is simply the composition of the two linear transformations $A$ and $B$. So applying the product transformation $AB$ to the vector $u$ to produce a vector $w = (AB)u$ in $W$ is equivalent to first applying the transformation $B$ to the vector $u$ to produce a vector $v = Bu$ in $V$ and then applying the transformation $A$ to the vector $v$ to produce a vector $w = Av$ in $W$. In other words, $(AB)u = A(Bu)$.

So the final result $w$ is the same whether we apply $AB$ to $u$ or apply $B$ to $u$ and then apply $A$ to $Bu$. Since the first step, namely applying $C$ to $x$ to produce $u$, is the same in both cases, the result of $(AB)Cx$ (applying $C$ and then $AB$) is the same as applying first $C$, then $B$, and then $A$. In other words, $(AB)Cx = A(B(Cx))$.

ii) Just as $A$ and $B$ can be composed together to form a linear transformation $AB$ from $V$ into $W$, so can $B$ and $C$ can be composed together to create a linear transformation $BC$ from $X$ into $V$. By rule 2V we know that this is the same as applying $C$ to a vector $x$ in $X$ to produce a vector $u$ in $U$ and then applying $B$ to $u$ to produce a vector $v$ in $V$. In other words, $(BC)x = B(Cx)$.

In either case we can then take the resulting vector $v$ and transform it using $A$ to produce a vector $w$ in $W$. In other words, $w = A(BC)x = A(B(Cx))$. But from the previous answer we also know that $(AB)Cx = A(B(Cx))$. From the two equations together we see that $(AB)Cx = A(BC)x$ and thus that the linear transformations $A$, $B$ and $C$ obey the law of associativity.

BONUS: As noted in the text, associativity of linear transformations implies associativity of matrix multiplication.

Suppose that we choose sets of basis vectors in the vector spaces $X$, $U$, $V$, and $W$. (Our choice is arbitrary; any basis sets will do.) We can then represent the transformation $C$ by a matrix $[C]$ constructed using the chosen basis vectors in $X$ and $U$. We can represent the transformation $B$ by a matrix $[B]$ constructed using the chosen basis vectors in $U$ and $V$. Finally, we can represent the transformation $A$ by a matrix $[A]$ constructed using the chosen basis vectors in $V$ and $W$.

By rule 2V composition of two linear transformations creates a new linear transformation that can be represented by a matrix that is the product of the matrices representing the original linear transformations. Since we have $(AB)Cx = A(BC)x$ we then also have $([A][B])[C]x = [A]([B][C])x$. Any matrix can represent a linear transformation, so we then have $([A][B])[C] = [A]([B][C])$ for any three matrices $[A]$, $[B]$, and $[C]$ that can be multiplied together.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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