# Change of basis matrix confusion

There seems to be a contradiction in two different sets of Linear Algebra lecture notes I have.

One says that the columns of the change of basis matrix $P$ is the coordinates of the 'old' basis vectors $e_i$ with respect to the 'new' basis vectors $e_i^'$.

The other one says that the columns of $P$ are the new basis vectors $e_i^'$ written as column vectors in the old basis vectors $e_i$.

Am I missing something here or is this a contradiction?

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There actually is an issue of nomenclature, with the same matrix being called different things depending on the textbook. See for example this old question, where the quoted nomenclature from Schaum conflicts with the nomenclature from Friedberg/Insel/Spence. – Arturo Magidin Mar 24 '12 at 2:33

The difference is which base you're switching from and to. Suppose $\beta = \{v_1,\dots,v_n \}$ and $\gamma = \{w_1,\dots,w_n \}$. Then say $v_i = c_{1i}w_1+\cdots+c_{ni}w_n$. We have $[v_i]_\gamma = [c_{1i} \;\; c_{2i} \;\; \cdots \;\; c_{ni}]^T$ (the $\gamma$-coordinates of $v_i$. Then

$$P = [I]_\beta^\gamma = \begin{bmatrix} [v_1]_\gamma & [v_2]_\gamma & \cdots & [v_n]_\gamma \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & \vdots & & \vdots \\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{bmatrix}$$

Here $P$ changes from $\beta$ to $\gamma$ coordinates.

Changing from $\gamma$ to $\beta$ coordinates would yield $Q=[I]_\gamma^\beta = \begin{bmatrix} [w_1]_\beta & [w_2]_\beta & \cdots & [w_n]_\beta \end{bmatrix}$.

Of course, $P=Q^{-1}$ and $Q=P^{-1}$.

In the end both of the matrices your notes refer to are change of basis matrices. The first changes from the old to the new and the second changes from the new to the old. If I had to venture a guess, then reason for the discrepancy has to do with which way they write conjugations (i.e. $PAP^{-1}$ versus $P^{-1}AP$). In more detail, suppose $A$ is the coordinate matrix of some linear operator written in $\beta$ coordinates. Then $B=PAP^{-1}=Q^{-1}AQ$ is the coordinate matrix in $\gamma$ coordinates. If you want inverses to appear on the left, then you refer to a change of basis as going from new to old coordinates (i.e. $Q$). If you want the inverse to appear on the right, you go with the other.

So in the end, no contradiction just different conventions. I hope this helps!

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