## 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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