Change of base in $R^2$

Question

In the official solution of an university exam I see the following question:

Given the base vectors

$$ B = \{ \begin{pmatrix} -1 \\ 1 \end{pmatrix} , \begin{pmatrix} 1 \\ 1 \end{pmatrix} \} $$

and a second set of base vectors: $$ B^* = \{ \begin{pmatrix} 1 \\ 2 \end{pmatrix} , \begin{pmatrix} 2 \\ 3 \end{pmatrix} \} $$

give the equations of the change from B to B* basis and apply it to vector of components (3,-1) in B basis.

The official solution is:

$$ C(B,B^*) = \begin{pmatrix} -5 & 1 \\ -3 & -1 \end{pmatrix} $$

and vector in B* basis is (-16,-8)

The method has been to place side by side the two matrices and apply some row multiplication and additions till make diagonal first one.

This result surprises me. If I call T the matrix formed by the base vectors in B and $T^*$ the one for $B^*$, I think the result should be ${(T^*)}^{-1}T$, by the simple logic of first T to pass to cartesian coordinates and back to B^* with the inverse of its matrix. That gives me a result of:

$$ C(B,B^*) = \begin{pmatrix} 5 & -1 \\ -3 & 1 \end{pmatrix} $$

and vector in B* equal to (16,-10).

Please, could you help me to known which one is the correct solution?

Answer 1

Your solution is correct. You can tell because the columns of the change of basis matrix should give the coordinates of the basis elements in $B$ in terms of the basis $B^{*}$.

Indeed, we have $$\begin{pmatrix}-1\\ 1\end{pmatrix}=5\begin{pmatrix}1\\2\end{pmatrix}-3\begin{pmatrix}2\\3\end{pmatrix}$$ and $$\begin{pmatrix}1\\ 1\end{pmatrix}=-\begin{pmatrix}1\\2\end{pmatrix}+\begin{pmatrix}2\\3\end{pmatrix}$$

Answer 2

You're right and the official solution is wrong.

The vector with components $3$ and $-1$ with respect to $B$ is $$3\begin{pmatrix}-1\\1\end{pmatrix} - 1\begin{pmatrix}1\\1\end{pmatrix}= \begin{pmatrix}-4\\2\end{pmatrix}$$ in standard coordinates.

The vector with components $-16$ and $-8$ with respect to $B^*$ is $$-16\begin{pmatrix}1\\2\end{pmatrix} - 8\begin{pmatrix}2\\3\end{pmatrix}= \begin{pmatrix}-32\\-56\end{pmatrix}$$ in standard coordinates - definitely not a match.

The vector with components $16$ and $-10$ with respect to $B^*$ is $$16\begin{pmatrix}1\\2\end{pmatrix} - 10\begin{pmatrix}2\\3\end{pmatrix}= \begin{pmatrix}-4\\2\end{pmatrix}$$ in standard coordinates. Confirmed.

Answer 3

The method used in the solution is basically sound. As you say, the change-of-basis matrix should be $(T^*)^{-1}T$, which you can obtain by concatenating the two matrices and row-reducing: $$\left[\begin{array}{c|c} T^* & T \end{array}\right] \to \left[\begin{array}{c|c} I & (T^*)^{-1}T \end{array}\right].$$ Comparing the correct and incorrect results, I would guess that a simple sign error was made somewhere along the way in the given solution.