Question bank — Change of basis matrix

False. The arrow $v$ never moves; only its coordinate description changes from $B$-language to $C$-language. This is the passive view — opposite of a linear transformation which actually relocates arrows.

True. A basis is by definition a linearly independent spanning set, so both $B$ and $C$ (matrices of basis vectors) are invertible, and $P_{C\leftarrow B}=C^{-1}B$ is a product of invertibles — hence invertible.

False. The columns are the old basis vectors $b_j$ written in new coordinates, i.e. $[b_j{]}_C$. The output lives in $C$-land, so each column must already be a $C$-coordinate vector.

True. Translating a language into itself changes nothing: each $[b_j{]}_C=[b_j{]}_B=e_j$, so the columns are $e_1,\dots ,e_n$ — the identity.

False. They are inverses, not the same matrix flipped: $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}$. Reversing a translation means undoing it, which is the inverse operation.

False — that direction is the easy one: $[v{]}_E=B[v{]}_B$, just multiply by $B$ whose columns are already in standard coords. The inverse ($B^{-1}$) is needed for the reverse trip $E\to B$.

True. A diagonal $P$ means each $b_j$ expressed in $C$ uses only $c_j$: $b_j=p_{jj}{c}_j$. So each old vector is a scalar multiple of the corresponding new one — same directions, rescaled.

False. Coordinates relative to a fixed basis are unique, and different bases generally read a fixed arrow differently. Same coordinates in different bases would describe different arrows (unless the bases happen to agree on how they build $v$).

True. Since $P$ is always invertible, its determinant is nonzero. A zero determinant would mean the "basis" vectors were dependent — not a basis at all.

Error: the formula is $P_{C\leftarrow B}=C^{-1}B$. Reasoning: go $B$-coords $\to$ standard (multiply by $B$), then standard $\to C$-coords (multiply by $C^{-1}$); matrices apply right-to-left, so it's $C^{-1}B$.

Error: $P_{C\leftarrow B}$ eats $B$-coords and gives $C$-coords. To take $C$-coords in and $B$-coords out you need $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}$. Read the subscript arrow as "input on the right."

Error: stack $[b_1{]}_C\mid \cdots \mid [b_n{]}_C$ — the old $b_j$ written in the new language. Test: feeding $e_j$ must return $[b_j{]}_C$, which only works if column $j$ is $[b_j{]}_C$.

Error: the arrow — and its true length — is fixed. Coordinates are just a description; unless the new basis is orthonormal the coordinate numbers need not even reflect length. Nothing physical moved.

Error in the setup: $P_{C\leftarrow E}=C^{-1}E=C^{-1}$, not $C$. It's $P_{E\leftarrow C}=C$ that equals the matrix of $C$'s vectors. The subscript direction was flipped.

Error: they are similar, generally not equal. Same underlying linear map seen in a different basis — same eigenvalues/determinant/trace, but different entries.

Because expansion in a basis is unique: adding vectors adds their coordinate columns and scaling a vector scales its column. This linearity is exactly what lets us collapse the translation into a single matrix.

Feed the column $e_j=[b_j{]}_B$: the product selects column $j$, returning $[b_j{]}_C$. So $P$ maps each building block correctly, and by linearity it maps every combination correctly too.

Because neither basis is the reference frame ${\mathbb{R}}^n$ is built in — the standard basis $E$ is. We route through $E$: decode $B$ into standard ($B$), then encode standard into $C$ ($C^{-1}$).

Diagonalizing picks the basis of eigenvectors. In that basis the map only stretches each axis, so $P^{-1}AP=D$ is diagonal — the same map, described in its most natural language.

Both bases of the same space $V$ have the same number of elements — the dimension $n$. So $P$ maps $n$ coordinates to $n$ coordinates, forcing an $n\times n$ shape.

Because undoing "translate then translate back" must give the identity: $P_{B\leftarrow C}{P}_{C\leftarrow B}=I$, which is the definition of inverse. Transpose only coincides with inverse for orthogonal bases — a special case, not the rule.

It is the identity $I$: $E^{-1}E=I$. Standard coordinates translated into standard coordinates are unchanged.

No. A zero column would mean some $[b_j{]}_C=0$, i.e. $b_j=0$ — but a basis cannot contain the zero vector. So every column is nonzero.

Then it isn't a basis — the set is dependent, the matrix $B$ is singular ($\det =0$), and $C^{-1}B$ or $B^{-1}$ fails to exist. The change of basis is undefined; the invertibility guarantee breaks.

A single nonzero number $1\times 1$. If $B=\{b\}$ and $C=\{c\}$ with $b=kc$, then $P_{C\leftarrow B}=[k]$ — a pure rescaling of the one coordinate, and its inverse is $[1/k]$.

Not necessarily. $P_{C\leftarrow B}=C^{-1}B$; even with $C$ orthonormal ($C^{-1}=C^T$), the product is orthogonal only if $B$ is also orthonormal. One orthonormal side is not enough.

No — $P\cdot 0=0$ for any matrix, so $[v{]}_C=0$ too. The zero arrow has zero coordinates in every basis; there is nothing to translate.