Here's the question I got stuck on:
Suppose $S$ is a matrix that performs a similarity transformation. Applying $S$ to the point $(0, 0)$ takes it to the point $(3, 2)$. Applying $S$ to the point $(1, 2)$ takes it to the point $(2, 5)$. What is $S$?
My intuition was using the formula: $A = P^{-1}BP$ to approach this question.
The definition of the formula is that let $A$ and $B$ be $n \times n$ matrices. If we can find a non-singular $n \times n$ matrix $P$ as the formula shows, then $A$ and $B$ are similar to each other.
I just started learning linear algebra, so any help would be appreciated!
This is a bit of trick question. Assuming that there are no typos in the given vectors, there is no such linear transformation $S$. Here's why:
Notice how $$ \begin{pmatrix} 2 \\ 2 \end{pmatrix} = 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ so if $$ S \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}, $$ then by linearity, $$ S \begin{pmatrix} 2 \\ 2 \end{pmatrix} = 2 S \begin{pmatrix} 1 \\ 1 \end{pmatrix} = 2 \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix} $$ But, we are also given that $$ S \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}, $$ a contradiction! Therefore, no such linear transformation $S$ exists.