Linear Algebra – Change of Basis Matrix Exercise

linear algebramatrices

The matrix of basis change from basis $B = \{b_1,b_2\}$ to the basis $C = \{(1,1),(0,2)\}$ is

$$ M = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}$$

Find basis $B$.

Well, here's what I did:

Since $M$ changes from $B$ to $C$, then it Works like this:

$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}{\begin{bmatrix}a_1\\a_2\end{bmatrix}}_B$$

In other words, a vector with coordinates in $C$ is written as the matrix $M$ multiplied by the vector written in $B$.

$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = a_1\begin{bmatrix} 1 \\2\end{bmatrix}+a_2\begin{bmatrix} 0 \\3\end{bmatrix}$$

where $\begin{bmatrix} 1 \\2\end{bmatrix}$and $\begin{bmatrix} 0 \\3\end{bmatrix}$ are the vectors $b_1,b_2$ written in $C$, so:

$$\vec{b_1} = 1(1,1) + 2(0,2) = (1,5)\\\vec{b_2} = 0(1,1) + 3(0,2) = (0,6)$$

but this is not the answer.

Could somebody tell me what I'm doing wrong? (instead os just solving in another way)

Best Answer

Here is how I would do it:

  • First remember the change of basis matrix $P_\mathcal B^{\mathcal B'}$ from an old base $\mathcal B$ of a vector space $ E $ to a new base $\mathcal B\,' $ has as column-vectors the coordinates of the newbase in the oldbase. It is the matrix of the identity map from $ (E, \mathcal B\,') $ to $ (E, \mathcal B) $.
  • The change of basis matrix the other way is just the inverse matrix of the previous one: $$P_{\mathcal B'}^{\mathcal B}=\bigl(P_\mathcal B^{\mathcal B'}\bigr)^{-1}$$
  • It allows to express the old coordinates $X$ of a vector from the new coordinates $X'$: $$X=P_\mathcal B^{\mathcal B'}X'$$
  • You can compose the change of basis matrix: $$ P_{\mathcal B}^{\mathcal B''} = P_{\mathcal B}^{\mathcal B’}P_{\mathcal B’}^{\mathcal B''}$$

Denote the canonical base as $\mathrm{Can}$. Determining $\mathcal B$ is the same as determining $P_{\mathrm{Can}}^{\mathcal B}$. By the composition formula, we have: $$P_{\mathrm{Can}}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}P_{\mathcal C}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}.$$

Now $ P_{\mathrm{Can}}^{\mathcal C}= \begin{bmatrix} 1 & 0 \\1&2\end{bmatrix} $, and we compute with the pivot method that $$\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}= \begin{bmatrix} 1 & 0 \\2&3\vphantom{\tfrac 13}\end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 \\-\tfrac23& \tfrac 13\end{bmatrix}$$ so that $$P_{\mathrm{Can}}^{\mathcal B}=\begin{bmatrix} 1 & 0 \\-\tfrac 13& \tfrac 23\end{bmatrix}$$

Hence $b_1=\begin{bmatrix} 1 \\-\tfrac 13\end{bmatrix}$, $\quad b_2=\begin{bmatrix} 0 \\\tfrac 23\end{bmatrix}$.

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