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We make a change of base with the matrix $$S=\left[\begin{matrix}p & q \\ 1 & 1\end{matrix}\right]$$ so the vector $x= (8, 3)$ of $\mathbb{R}^{2\times 1}$ becomes $x= (1, 2)$ and the vector $y=(5, 2)$ of $\mathbb{R}^{2\times 1}$ becomes $y=(1, 1)$.

How can I get to know what $z=(-1, 0)$ of $\mathbb{R}^{2\times 1}$ becomes?

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1 Answer 1

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You may find two numbers $a$ and $b$ such that $z=ax+by$, where the $x, y$ and $z$ here are the ones under the old basis. Now, the linear relationship between $x,y$ and $z$ is preserved under a change of basis. Thus the new $z$ can be computed as $z=ax+by$, where the $x,y$ are the new ones. In this way you don't need to touch $S$.

Alternatively, you may solve $S^{-1}\begin{bmatrix}8&5\\3&2\end{bmatrix}=\begin{bmatrix}1&1\\2&1\end{bmatrix}$ -- or equivalently $\begin{bmatrix}8&5\\3&2\end{bmatrix}=S\begin{bmatrix}1&1\\2&1\end{bmatrix}$ -- for $p$ and $q$ first, and then compute the new $z$ as $S^{-1}z$.

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