Kristina Mcclain

2022-02-02

Given Matrix transformation matrix
$A=\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)$
Are there vectors in ${\mathbb{R}}^{3}$ that do not change under this transformation, i.e. such that $T\left(u\right)=u$? Explain why yes or why not.

Ronald Alvarez

Step 1
Let
$u=\left(\begin{array}{c}\lambda \\ 0\\ 0\end{array}\right)$
for any $\lambda$
A little work shows that these are the only solutions to
$Au=u$
A note: no matter what linear transformation you have, $T\left(0\right)=0$ is always true, but sometimes students forget to state the whole problem, so I posted a complete set of solutions just in case

Gwendolyn Meyer

Step 1
$\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}\left(\begin{array}{c}x+y+z\\ 2y+2z\\ 3z\end{array}\right)$
$=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\phantom{\rule{0.278em}{0ex}}⟺\phantom{\rule{0.278em}{0ex}}\left(\begin{array}{c}y+z\\ y+2z\\ 2z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$
Third equation gives $z=0$ and first equation gives $y=0$. So,
$\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)\left(\begin{array}{c}x\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c}x\\ 0\\ 0\end{array}\right)$
for any $x\in \mathbb{R}.$

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