Change of Bases Problem

Algebra Level pending

Consider the matrix A: $\mathbb{R}^{3} \rightarrow \mathbb{R}^{2}$ Given by,

$\begin{bmatrix}1&3&5 \\ 4&1&-1 \end{bmatrix}$

Let $e_{1} \, e_{2} \, e_{3}$ denote the standard basis and consider a new basis $\{v_{1} , v_{2} , v_{3} \}$ for the domain and $\{w_{1} , w_{2} \}$ for the co-domain which are defined in terms of the standard basis as,

$\{ v_{1} , v_{2} , v_{3} \}= \{ e_{1}+e_{2}+e_{3},e_{2}+e_{3}, e_{3} \}$ $\{ w_{1} , w_{2} \}= \{ e_{1}-e_{2}, e_{2} \}$

What is equivalent representation of L in these new bases?

\begin{bmatrix}-2&3&5 \\ 5&8&-4 \end{bmatrix}

\begin{bmatrix}9&8&5 \\ 13&8&6 \end{bmatrix}

\begin{bmatrix}1&3\\ 4&1\\ 4 & -9 \end{bmatrix}

\begin{bmatrix}1&3 \\ 4&1& \\ 9&8 \end{bmatrix}

1 solution

Samuel Hansen
Apr 11, 2017

First construct the representation of the transformation from the u basis to the e basis, that is find the matrix P such that the ith column of $\hat{P}$ is $v_{i}$ expressed in the $v_{i}$ basis. Call this matrix P, then

$P = \begin{bmatrix}1&0&0 \\ 1&1&0 \\ 1&1&1 \end{bmatrix}$

Note that indeed, the vector $[ 1\, 0 \, 0]'$ expressed in this basis is $e_{1}+e_{2}+e_{3}$ . Similarly we construct another matrix Q to go from the basis $\{w \}$ to basis $\{ e\}$ .

$W =\begin{bmatrix}1&0 \\ -1&1 \end{bmatrix}$

Now to get from {v} to {w} via L we go from $\{v\} \in \mathbb{R}^{3} \rightarrow \{e\} \in \mathbb{R}^{3}$ then use to the transform L to go from $\{e\} \in \mathbb{R}^{3} \rightarrow \{e\} \in \mathbb{R}^{2}$ and then finally $\{e\} \in \mathbb{R}^{2} \rightarrow \{w\} \in \mathbb{R}^{2}$ Thus the equivalent expression for L, call it $\hat{L}$ is,

$\hat{L}=Q^{-1} L P$

Performing the matrix multiplication yields the desired result.

Change of Bases Problem

1 solution

0 pending reports