Answer: false.

Clarification:

• The scalar multiplication $\cdot: K \times V \rightarrow V$ doesn’t even have the same sets in its two arguments, i.e. $\mathbf v \cdot a$ is not even defined.
• The vector space has the commutativity of addition as an axiom: for any $\mathbf u,\mathbf v \in V$, ${\mathbf u+\mathbf v} = {\mathbf v + \mathbf u}$.
• The underlying field $K$ does have the commutativity of multiplication as a field axiom: for any $a,b \in K$, $a \textcolor{red}{\cdot} b = b \textcolor{red}{\cdot} a$.
• As a consequence, for any $\mathbf v \in V$ and $a, b \in K$,

$$a (b \mathbf v) = (a\textcolor{red}{\cdot} b)\mathbf v = (b \textcolor{red}{\cdot} a)\mathbf v = b(a \mathbf v).$$

Answer: false.

Clarification:

• By definition, a vector is an element of a vector space. If we fix a basis for the vector space, then any vector can be represented by an ordered set of numbers, which could be treated as a column vector, i.e. a matrix with one column. However, this representation depends on the choice of basis.
• The official answer brings up as a counterexample the vector space of polynomials up to some finite degree. However, here again we could represent the vectors as a column vector with any choice of a basis. E.g. using the standard basis, $p(x) = 0x^2 + 4x + 5$ could be represented as $\mathbf p = [0, 4, 5]^T$.

Answer: false.

Clarification:

• As above, a linear map $f: V \rightarrow W$ can be represented as a unique matrix only once bases are chosen for its domain $V$ and codomain $W$.
• This matrix is quadratic only if the dimensions of $V$ and $W$ are equal.

Answer: true.

Clarification:

• The dual space of a vector space $V$ is defined as the set of linear maps from $V$ to $K$: $V^* \coloneqq Hom(V,K) \coloneqq \{φ\ \vert \ φ: V \linmap K\}$.

Answer: false.

Clarification:

• This set doesn’t have a commutative identity element: by the field axioms of $\mathbb R$, it could only be the constant zero function, but that’s not an element of the set.
• This set doesn’t have a commutative inverse for any element.
• For the scalar multiplication we’d need to know the underlying field. Usually it would be $\mathbb R$, but then S-multiplication with a negative number wouldn’t result in an everywhere positive function. (Although one can construct a field from $\mathbb R^+$, I wonder how well that would combine with the above attempt at a vector space.)

Answer: true.

Clarification:

• The lecture didn’t mention tensor products, so a definition is in order. The product of an $(l,k)$-tensor $S$ and an $(n,m)$-tensor $T$ is an $(l+n,k+m)$-tensor $S \otimes T$, whose $(i_1, \ldots, i_{l+n}, j_1, \ldots, j_{k+m})$-th component is the product of the relevant components of $S$ and $T$:

$$(S \otimes T)^{i_1, \ldots, i_l, i_{l+1}, \ldots, i_{l+n}}_ {j_1, \ldots, j_k, j_{k+1}, \ldots, j_{k+m} } = S^{i_1, \ldots, i_l}_ {j_1, \ldots, j_k} T^{i_{1}, \ldots, i_{n}}_ {j_{1}, \ldots, j_{m}}.$$

Source: Wikipedia

This means that if the arguments of $S \otimes T$ are

• the $l+n$ linear maps $φ^{(p)} = \sum^{dim V}_{i=1} \varphi^{(p)}_i \epsilon^i$ for $1 \le p \le l+n$, and
• the $k+m$ vectors $\v_{(q)} = \sum^{dim V}_{j=1} v_{(q)}^j \e_j$ for $1 \le q \le k+m$

(with some particular choice of basis vectors $\{\e_i\}_i$ and basis covectors $\{\epsilon^i\}_i$ ), then

$$\begin{aligned} (S\otimes T) &(φ^{(1)}, \ldots, φ^{(l+n)}, \v_{(1)}, \ldots, \v_{(k+m)}) = \\ &= S (φ^{(1)}, \ldots, φ^{(l)}, \v_{(1)}, \ldots, \v_{(k)})\,\cdot\, T (φ^{(l+1)}, \ldots, φ^{(l+n)}, \v_{(k+1)}, \ldots, \v_{(k+m)})\\ &= \Bigg( \sum_{i_1}^{\dim V} \cdots \sum_{i_l}^{\dim V} \sum_{j_1}^{\dim V} \cdots \sum_{j_k}^{\dim V} \varphi^{(1)}_{i_1} \ldots \varphi^{(l)}_{i_l} v_{(1)}^{j_1} \ldots v_{(k)}^{j_k} S^{i_1, \ldots, i_l}_{j_1, \ldots, j_k} \Bigg) \cdot \phantom.\\ &\phantom{=} \Bigg( \sum_{i_{l+1}}^{\dim V} \cdots \sum_{i_{l+n}}^{\dim V} \sum_{j_{k+1}}^{\dim V} \cdots \sum_{j_{k+m}}^{\dim V} \varphi^{(l+1)}_{i_{l+1}} \ldots \varphi^{(l+n)}_{i_{l+n}} v_{(k+1)}^{j_{k+1}} \ldots v_{(k+m)}^{j_{k+m}} T^{i_{l+1}, \ldots, i_{l+n}}_{j_{k+1}, \ldots, j_{k+n}} \Bigg) \\ &= \sum_{i_1}^{\dim V} \cdots \sum_{i_{l+n}}^{\dim V} \sum_{j_1}^{\dim V} \cdots \sum_{j_{k+m}}^{\dim V} \varphi^{(1)}_{i_1} \ldots \varphi^{(l+n)}_{i_{l+n}} v_{(1)}^{j_1} \ldots v_{(k+m)}^{j_{k+m}} S^{i_1, \ldots, i_l}_{j_1, \ldots, j_k} T^{i_{l+1}, \ldots, i_{l+n}}_{j_{k+1}, \ldots, j_{k+n}}. \end{aligned}$$

These $(l+n+k+m)$ summations are quite a mess, but the above derivation shows that the Einstein summation convention works for tensor products as well:

$$\begin{aligned} (S\otimes T) &(φ^{(1)}, \ldots, φ^{(l+n)}, v_{(1)}, \ldots, v_{(k+m)}) =\\ &= S (φ^{(1)}, \ldots, φ^{(l)}, v_{(1)}, \ldots, v_{(k)})\,\cdot\, T (φ^{(l+1)}, \ldots, φ^{(l+n)}, v_{(k+1)}, \ldots, v_{(k+m)})\\ &= \Big( \varphi^{(1)}_{i_1} \ldots \varphi^{(l)}_{i_l} v_{(1)}^{j_1} \ldots v_{(k)}^{j_k} S^{i_1, \ldots, i_l}_{j_1, \ldots, j_k} \Big) \Big( \varphi^{(l+1)}_{i_{l+1}} \ldots \varphi^{(l+n)}_{i_{l+n}} v_{(k+1)}^{j_{k+1}} \ldots v_{(k+m)}^{j_{k+m}} T^{i_{l+1}, \ldots, i_{l+n}}_{j_{k+1}, \ldots, j_{k+n}} \Big) \\ &= \varphi^{(1)}_{i_1} \ldots \varphi^{(l+n)}_{i_{l+n}} v_{(1)}^{j_1} \ldots v_{(k+m)}^{j_{k+m}} S^{i_1, \ldots, i_l}_{j_1, \ldots, j_k} T^{i_{l+1}, \ldots, i_{l+n}}_{j_{k+1}, \ldots, j_{k+n}}. \end{aligned}$$

Answer: true.

Clarification:

• If we know the basis vectors for the vector space and the dual vector space, then the components of the vector and covector arguments are uniquely determined, and we can apply the tensor to the arguments using the components of the tensor (or some relevant finite subset in case $V$ is not finite dimensional).

Answer: false.

Clarification:

• A tensor component usually has one index for each argument, e.g. for a $(2,1)$-tensor $T$, the components are $T^{i_1,i_2}_{j_1}$.
• The range of these indices does depend on the dimension: each index ranges from $1$ to $\dim V$. Therefore an $(n,m)$-tensor $T$ has $(\dim V)^{n+m}$ many components.

Answer: false.

Clarification: see above.

Answer: false.

Clarification:

• the tensor components are defined with respect to a given basis.

Answer: false.

Clarification:

• Given a basis of $V$, $E = \{\mathbf{e}_i\}_{i=1}^d \subset V$, there is a unique dual basis of $V^*$, namely $E^* = \{\epsilon_i\}_{i=1}^d$, where $\epsilon_i(\e_i) = 1$ and $\epsilon_i(\e_j) = 0$ for $i ≠ j$.

Answer: true.

Clarification:

• The resulting set of $(d-1)$ vectors are still linearly independent, and their span is a $(d-1)$-dimensional subspace of $V$.

Answer: false.

Clarification:

• The continuity of a map is defined for topological spaces, not for vector spaces.
• $f$ is continuous iff the preimage of every open set in $W$ is open in $V$. Note that no term in this definition depends on the choice of basis for either $V$ or $W$.
• Assuming that $V$ and $W$ are real vector spaces, it is customary to equip them with the standard topology. A set $A$ is open in $V$ iff either it is the union of open $ε$-balls, or of Cartesian products of open intervals. While these definitions assume a basis for $V$, they all result in the exact same topologies. (Meaning a set can be covered with open balls iff it can be covered with open cuboids iff it can be covered with open cubes – an interesting but easy-to-prove result.)
• It’s easy to see that every linear map between real vector spaces (equipped with the standard topology) is continuous.

Answer: true.

Clarification:

• a basis for $V$ uniquely determines a dual basis for $V^*$, which uniquely determines the components of any covector.

Answer: true.

Clarification:

• Given the basis vectors $\mathbf{e}_i$ and components $v^i$ for $1 \leq i \leq d$, $\mathbf{v} = \sum_{i=1}^d v^i \mathbf{e}_i$.