Chapter 15 Quiz — Dimension, Basis, and Coordinates
Twelve conceptual checks. Try each before opening the answer. One-line explanations follow each.
Q1. A set is a basis for $V$ if and only if it is _ and _.
Answer
**Linearly independent** and **spanning**. Both conditions are required: independence rules out redundancy (giving uniqueness of coordinates), spanning guarantees every vector is reachable (giving existence).Q2. The set $\{(1, 0), (0, 1), (2, 3)\}$ spans $\mathbb{R}^2$. Is it a basis?
Answer
**No.** It spans but is linearly *dependent* (three vectors in $\mathbb{R}^2$ must be), so it fails the independence condition. A spanning set that isn't independent gives non-unique coordinates.Q3. What is the dimension of a space, in one sentence?
Answer
The **number of vectors in any basis** of the space — a count that is the same for every basis (by the replacement theorem), equal to the number of independent directions, i.e. the degrees of freedom.Q4. True or false: a vector space can have one basis with 3 vectors and another basis with 4 vectors.
Answer
**False.** By the invariance-of-dimension theorem, every basis of a given space has exactly the same number of vectors. That common number is the dimension.Q5. The coordinate vector $[\mathbf{v}]_{\mathcal{B}}$ depends on what two things?
Answer
The **vector $\mathbf{v}$** and the **ordered basis $\mathcal{B}$**. Change the basis and the same $\mathbf{v}$ gets a different coordinate list; the underlying vector is unchanged.Q6. To find $[\mathbf{v}]_{\mathcal{B}}$ for a basis $\mathcal{B}$ of $\mathbb{R}^n$, you solve which equation, and what plays the role of the matrix?
Answer
Solve $B\mathbf{c} = \mathbf{v}$, where $B$ has the **basis vectors as its columns** and $\mathbf{c} = [\mathbf{v}]_{\mathcal{B}}$. Because the columns form a basis, $B$ is invertible and the solution is unique.Q7. Why are a vector's standard coordinates equal to its raw entries?
Answer
Because the standard basis vectors $\mathbf{e}_i$ have a $1$ in slot $i$ and $0$ elsewhere, so $\mathbf{v} = v_1\mathbf{e}_1 + \dots + v_n\mathbf{e}_n$ — the entries *are* the coefficients. Equivalently, $B = I$ and solving $I\mathbf{c} = \mathbf{v}$ gives $\mathbf{c} = \mathbf{v}$.Q8. A plane through the origin in $\mathbb{R}^3$ has each of its points written with three numbers. What is its dimension, and why isn't it three?
Answer
Its dimension is **two**. The three numbers are coordinates inherited from the ambient $\mathbb{R}^3$, but the plane itself needs only two basis vectors (two rulers) to reach every point — two degrees of freedom. Dimension is intrinsic to the space, not to how it's embedded.Q9. In the existence-and-uniqueness proof for coordinates, which condition gives existence and which gives uniqueness?
Answer
**Spanning** gives existence (every vector is *some* combination of the basis). **Independence** gives uniqueness (two combinations equal to the same vector must be identical, since their difference is zero and only the all-zero coefficients do that for independent vectors).Q10. You compute the coordinates of $\mathbf{v}$ in a basis $\mathcal{B}$ by solving a $3\times 2$ system, and it comes out inconsistent. What does that tell you?
Answer
That $\mathbf{v}$ is **not in the subspace** spanned by $\mathcal{B}$ — it has no coordinates relative to that basis. Coordinates exist only for vectors inside the span; an inconsistent system is the algebra reporting "$\mathbf{v}$ is off the plane."Q11. Why is computing coordinates in an orthonormal basis easier than in a general basis?
Answer
In an orthonormal basis each coordinate is just a **dot product** $\mathbf{v}\cdot\mathbf{q}_i$ (a projection), so no system needs solving. For a *non-orthonormal* basis the dot-product shortcut fails and you must solve $B\mathbf{c} = \mathbf{v}$. (This is the payoff of Part IV.)Q12. The space $P_3$ of polynomials of degree $\leq 3$, the $2\times 2$ matrices, and $\mathbb{R}^4$ all have dimension 4. What does that imply about them?