Question bank — Coordinate vectors — change of basis

False. The arrow is fixed, but $[\mathbf{v}{]}_{\mathcal{B}}$ is the description relative to a chosen basis; change the basis and the list of numbers changes even though $\mathbf{v}$ does not move — exactly as $\left[\begin{array}{c}0\\ 2\end{array}\right]$ vs $\left[\begin{array}{c}1\\ 1\end{array}\right]$ in Figure s01.

False. A single vector can happen to have identical coordinates in two different bases (e.g. any $\mathbf{v}$ fixed by the map $P$); equal coordinates for all $\mathbf{v}$ would force $P=I$ and hence $\mathcal{B}=\mathcal{C}$.

True. Its columns are $[{\mathbf{b}}_j{]}_{\mathcal{C}}$, the coordinate images of a linearly independent set, so they stay independent — an invertible matrix with inverse $\underset{\mathcal{B}\leftarrow \mathcal{C}}{P}$.

False. Orthonormality makes $P$ orthogonal ($P^{-1}=P^{\top }$ over $\mathbb{R}$), not symmetric; a rotation matrix is orthogonal but not symmetric.

False. They are inverses, so their determinants are reciprocals: $\det (P^{-1})=1/\det (P)$; equal only in the special case $\det P=\pm 1$.

True. Coefficients $c_i$ in $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ are arbitrary scalars; nothing forces positivity — Example 3 in the parent gave $[p{]}_{\mathcal{C}}$ with a $-1$.

True. Similarity is exactly the statement that $A$ and $A^{\prime }$ represent the same linear map in two bases, and $P$ is the change-of-basis matrix relating those bases.

True. Expanding $\mathbf{u}+\alpha \mathbf{w}$ in $\mathcal{B}$ adds and scales the coefficients termwise, so $[\mathbf{u}+\alpha \mathbf{w}{]}_{\mathcal{B}}=[\mathbf{u}{]}_{\mathcal{B}}+\alpha [\mathbf{w}{]}_{\mathcal{B}}$.

Wrong. Trace the machine: $P$ receives a $\mathcal{B}$-label $\left[\begin{array}{c}{c}_1\\ ⋮\end{array}\right]$ and its columns get scaled by those $c_i$; so each column must be what one $\mathcal{B}$-vector looks like in the target, i.e. $[{\mathbf{b}}_j{]}_{\mathcal{C}}$. Using the ${\mathbf{c}}_j$ would build the reverse matrix.

Wrong order. Walk the input through: start with a $\mathcal{B}$-label; $B$ multiplies it to give the arrow in standard numbers ($\mathcal{B}\to \mathcal{E}$); then $C^{-1}$ turns those standard numbers into a $\mathcal{C}$-label ($\mathcal{E}\to \mathcal{C}$). Applied right-to-left that is $C^{-1}B$; the inner "standard" step cancels. $B^{-1}C$ chains $\mathcal{C}\to \mathcal{E}\to \mathcal{B}$ — the opposite direction.

Wrong. Only in the standard basis do a vector's entries equal its coordinates; for a tilted $\mathcal{B}$ (Figure s01) the arrow needs a different count of tilted steps, so you must solve $B[\mathbf{v}{]}_{\mathcal{B}}=\mathbf{v}$, giving $[\mathbf{v}{]}_{\mathcal{B}}=B^{-1}\mathbf{v}$.

Wrong. $P$ moves the list of numbers, not the arrow. Compare Figures s01 and s02: the arrow sits in the identical spot; only its label $\left[\begin{array}{c}0\\ 2\end{array}\right]\to \left[\begin{array}{c}1\\ 1\end{array}\right]$ changes.

Wrong. Once you fix an ordered basis, any finite-dimensional space (like $P_2$) has coordinate vectors in ${\mathbb{R}}^n$, so change of basis is an ordinary matrix — Example 3 used $\{1,t,t^2\}$.

Wrong. A basis is ordered; permuting ${\mathbf{b}}_1,{\mathbf{b}}_2$ permutes the entries $c_1,c_2$ correspondingly — a different coordinate vector via a permutation change-of-basis matrix.

Wrong. $P$ is built purely from the two bases; the same $P$ works for every $\mathbf{v}$, which is exactly why it deserves to be called a matrix.

Independence guarantees the expansion $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ is one-of-a-kind; two different expansions would subtract to a nontrivial combination equal to $0$, contradicting independence.

Because $\mathbf{v}=\sum c_i{\mathbf{b}}_i$ and taking $\mathcal{C}$-coordinates (a linear map) gives $[\mathbf{v}{]}_{\mathcal{C}}=\sum c_i[{\mathbf{b}}_i{]}_{\mathcal{C}}$ — a matrix with columns $[{\mathbf{b}}_i{]}_{\mathcal{C}}$ times the $c_i$.

Translating to $\mathcal{C}$ and immediately back to $\mathcal{B}$ must leave every coordinate vector untouched, so the composite is $I$ — meaning the two matrices are inverses.

In $\mathcal{E}$ a vector's coordinates are its literal entries, so $\underset{\mathcal{E}\leftarrow \mathcal{B}}{P}=B$ (basis vectors as columns) needs no solving — the hard work is only inverting to go the other way.

You cannot chain $\mathcal{B}\to \mathcal{C}$ directly; you route through the standard basis: $B$ takes $\mathcal{B}$-numbers to standard, $C^{-1}$ takes standard to $\mathcal{C}$-numbers, giving $C^{-1}B$.

Both coordinate lists must reconstruct the same $\mathbf{v}$; equating $B[\mathbf{v}{]}_{\mathcal{B}}=\mathbf{v}=C[\mathbf{v}{]}_{\mathcal{C}}$ and solving for $[\mathbf{v}{]}_{\mathcal{C}}$ yields exactly $P=C^{-1}B$.

It is the zero column $\left[\begin{array}{c}0\\ ⋮\\ 0\end{array}\right]$ in every basis, since the only combination giving $0$ from an independent set is all-zero coefficients.

The identity $I$: re-describing a vector in the basis it already uses changes nothing, so every column $[{\mathbf{b}}_j{]}_{\mathcal{B}}$ is a standard unit vector.

Yes — if a "basis" is actually dependent, then $B$ or $C$ is singular, $C^{-1}$ doesn't exist, and no change-of-basis matrix is defined; independence is a hard prerequisite.

$P=\frac{1}{k}I$: longer basis rulers mean fewer of them are needed, so each coordinate shrinks by $1/k$ — a degenerate but instructive diagonal case.

You get the standard unit column $e_2$ (a $1$ in slot 2, zeros elsewhere), because ${\mathbf{b}}_2=0\cdot {\mathbf{b}}_1+1\cdot {\mathbf{b}}_2+\cdots$.

A single nonzero scalar: with $\mathcal{B}=\{b\}$, $\mathcal{C}=\{c\}$ and $b=\lambda c$, the "matrix" is the $1\times 1$ number $\lambda$, just a rescaling of the one coordinate.

No. Over a complex vector space the inner product is conjugate-symmetric, so orthonormal-to-orthonormal change of basis is unitary: $P^{-1}=P^{\ast }={\overline{P}}^{\top }$, the conjugate-transpose, not the plain transpose. The real orthogonal case is just the special case where entries are real and conjugation does nothing.

Numerically yes — the same array $M$ can be read as "re-describe the fixed vector" (change of basis) or "actually move the vector" (a transformation); the interpretation, not the numbers, differs.