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.

Are there vectors in

Ronald Alvarez

Beginner2022-02-03Added 11 answers

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

Let

for any

A little work shows that these are the only solutions to

A note: no matter what linear transformation you have,

Gwendolyn Meyer

Beginner2022-02-04Added 11 answers

Step 1

$\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x\\ y\\ z\end{array}\right){\textstyle \phantom{\rule{0.278em}{0ex}}}\u27fa{\textstyle \phantom{\rule{0.278em}{0ex}}}\left(\begin{array}{c}x+y+z\\ 2y+2z\\ 3z\end{array}\right)$

$=\left(\begin{array}{c}x\\ y\\ z\end{array}\right){\textstyle \phantom{\rule{0.278em}{0ex}}}\u27fa{\textstyle \phantom{\rule{0.278em}{0ex}}}\left(\begin{array}{c}y+z\\ y+2z\\ 2z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right).$

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$x\in \mathbb{R}.$

Third equation gives

for any