Chapter 16 Quiz — Change of Basis
Twelve conceptual checks. Try each before opening the answer; every answer includes a one-line reason so you can repair a wrong mental model on the spot.
Q1. The columns of the change-of-basis matrix $P$ are:
Answer
The **new basis vectors, written in the old coordinates**. (When the old basis is standard, that is just the new basis vectors themselves.) This is what makes $P$ convert new coordinates to old: $P$ interprets a "new" instruction back in old-basis language.Q2. To convert a coordinate vector from the old basis to the new basis, you multiply by $P$ or by $P^{-1}$?
Answer
By $P^{-1}$: $[\mathbf{v}]_{\text{new}} = P^{-1}[\mathbf{v}]_{\text{old}}$. The matrix $P$ goes new $\to$ old; its inverse reverses the translation, old $\to$ new. Mixing these up is the single most common error in the chapter.Q3. A vector has standard coordinates $(4,2)$ and, in a new basis, coordinates $(3,-1)$. Which of the following changed: the vector, the coordinates, or both?
Answer
Only the **coordinates** changed. The vector — the geometric arrow — is the same object; $(4,2)$ and $(3,-1)$ are two addresses for it in two coordinate systems. This is recurring theme #1: numbers describe vectors, they are not the vectors.Q4. Define what it means for two matrices $A$ and $B$ to be similar.
Answer
$B = P^{-1}AP$ for some invertible matrix $P$. Equivalently, $A$ and $B$ represent the **same linear transformation** in two different bases. Similar matrices are generally *different* matrices that agree on every coordinate-free invariant.Q5. Why is $P$ on the right and $P^{-1}$ on the left in $B = P^{-1}AP$ (rather than the reverse)?
Answer
Because matrix products compose right-to-left (Chapter 8), and the steps happen in the order: $P$ (new $\to$ old) first, then $A$ (act), then $P^{-1}$ (old $\to$ new) last. Reading $P^{-1}AP$ right-to-left gives exactly that sequence of verbs.Q6. Name two quantities guaranteed to be equal for similar matrices, and explain in one phrase why they must be.
Answer
The **trace** and the **determinant** (also rank, characteristic polynomial, and eigenvalues). They are equal because each is a property of the underlying *transformation*, not of the coordinate system — e.g. the determinant is the area-scaling factor, which cannot depend on the grid you draw.Q7. A symmetric matrix $A$ is conjugated by an invertible $P$ that is not orthogonal. Is $P^{-1}AP$ necessarily symmetric?
Answer
No. Symmetry is **basis-dependent**: the chapter's symmetric $A = \begin{bmatrix}2&1\\1&2\end{bmatrix}$ became the non-symmetric $\begin{bmatrix}4&1\\-3&0\end{bmatrix}$ under a non-orthogonal change of basis. (Symmetry *is* preserved when $P$ is orthogonal, $P^{-1}=P^{\mathsf{T}}$ — the spectral theorem's setting, Chapter 27.)Q8. Two matrices both have trace $2$ and determinant $1$. Must they be similar?
Answer
No — equal trace and determinant are **necessary but not sufficient**. Counterexample: $I = \begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $J = \begin{bmatrix}1&1\\0&1\end{bmatrix}$ both have trace $2$, det $1$, but $J$ is not similar to $I$ (the only matrix similar to $I$ is $I$). A complete similarity invariant is the Jordan form (Chapter 36).Q9. When you re-grid the transformation $\begin{bmatrix}2&1\\1&2\end{bmatrix}$ into the basis $\{(1,1),(-1,1)\}$, the matrix becomes $\begin{bmatrix}3&0\\0&1\end{bmatrix}$. Did re-gridding change the transformation? What did it reveal?
Answer
It did **not** change the transformation — it is the same stretch of the plane. It *revealed* the transformation's intrinsic action: a scaling by $3$ along $(1,1)$ and by $1$ along $(-1,1)$. Those directions are eigenvectors and the diagonal entries are eigenvalues (Chapter 23); the right basis makes the action obvious.Q10. Does every change of basis simplify a matrix?
Answer
No. A *bad* basis makes the matrix messier: re-gridding the clean projection $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ into a skewed basis gave the cluttered $\begin{bmatrix}0.5&-0.5\\-0.5&0.5\end{bmatrix}$. Only the *right* basis (the eigenbasis) simplifies a given transformation; each transformation has its own preferred coordinate system.Q11. Why is the change-of-basis matrix $P$ always invertible?
Answer
Because its columns are a **basis**, hence linearly independent, and a square matrix with independent columns is invertible (Chapter 9). There is no change of basis without an inverse — the invertibility of $P$ *is* the independence of the basis.Q12. A change of basis between two non-standard bases $\mathcal{U}$ and $\mathcal{W}$ uses the matrix $M = P_{\mathcal{W}}^{-1}P_{\mathcal{U}}$. What role does the standard basis play here?