## Linear Algebra and Its Applications, Exercise 2.6.15

Exercise 2.6.15. Suppose that $T$ is a linear transformation from $\mathbb{R}^3$ to itself, or more generally from any vector space $V$ to itself. Show that $T^2$ is also a linear transformation.

Answer: If $T$ is a linear transformation from some vector space $V$ to itself then by rule 2V on page 123 the composition $TT$ of two linear transformations is also a linear transformation, again from $V$ to itself. So $T^2 = TT$ is a linear transformation if $T$ is.

Note that the assumption that $T$ maps from $V$ to itself is critical. If $T$ were a linear transformation from $V$ to a different vector space $W$ then it would not make sense to apply $T$ again to the result $w = Tv$ since $w$ is in $W$ not $V$.

Another slightly longer proof that does not rely on rule 2V:

For clarity we use function notation, so that $T(v)$ is the result of applying $T$ to a vector $v$ in $V$. We also take $T^2(x)$ to mean $T(T(x))$. In other words, $T^2$ is a composition of $T$ with itself. Since $T$ maps from $V$ to itself $T^2$ does also.

Since $T$ is a linear transformation we have $T(ax+by) = aT(x)+ bT(y)$ for any $x$ and $y$ in $V$ and any two scalars $a$ and $b$. Combining this with the definition of $T^2$ we have

$T^2(ax+by) = T(T(ax+by)) = T(aT(x)+bT(y))$

$= aT(T(x))+bT(T(y)) = aT^2(x) + bT^2(y)$

Since $T^2(ax+by) = aT^2(x) + bT^2(y)$ we see that $T^2$ is also a linear transformation.

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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