Change of basis | Chapter 13, Essence of linear algebra

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• Basis and dimension ExplanationAt 0:32
  Why this connection?

  The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$. A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$. These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$. Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$. Then $P(-1,2)^T=(-4,1)^T$: this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$. Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$, so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$. To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$, apply $A$, then convert back with $P^{-1}$. The resulting matrix is $P^{-1}AP$. For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$.

• Basis and dimension ExplanationAt 0:32
  Why this connection?

  The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$. A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$. These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$. Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$. Then $P(-1,2)^T=(-4,1)^T$: this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$. Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$, so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$. To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$, apply $A$, then convert back with $P^{-1}$. The resulting matrix is $P^{-1}AP$. For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$.