As we did in the previous Chapter, let us simultaneously refer the Euclidean space to two coordinate systems: the unprimed coordinates $Z^{i}$ and the primed coordinates $Z^{i^{\prime}}$. Let and be the equations of the coordinate transformation. Recall that the Jacobians $J_{i}^{i^{\prime}}$ and $J_{i^{\prime}}^{i}$ are defined as the collections of partial derivatives of those equations, i.e. and

The precise algebraic definition of a tensor depends on the order of the variant in question. In words, a variant is a tensor if its transformation rule is captured by contracting each of its indices with the appropriate Jacobian. In order to express this definition by an algebraic equation, we will consider a variant $T_{j}^{i}$ with a representative collection of indices and use the equation below to prescribe the rule for transforming each kind of index. It is then understood that in order to apply the definition to a variant with an arbitrary indicial signature, the appropriate rule must be applied to each index. In the future, we will introduce as many as six different kinds of indices and therefore our representative collection of indices will become larger. However, for now, a single superscript and a single subscript with suffice.

A variant $T_{j}^{i}$ is a tensor if its primed and unprimed coordinate manifestations $T_{j}^{i}$ and $T_{j^{\prime}}^{i^{\prime}}$ are related by the equation This definition is to be understood in the sense that for every index in the indicial signature, there is a contraction with the appropriate Jacobian -- specifically, the one that makes for a valid summation convention. The transformation rule that corresponds to a superscript, and therefore a contraction with $J_{i}^{i^{\prime}}$, is called contravariant. The rule that corresponds to a subscript, and therefore a contraction with $J_{i^{\prime}}^{i}$, is called covariant.

When applied to a first-order variant $T^{i}$ with a superscript, the definition reads Such a variant is called a first-order contravariant tensor. Similarly, for a first-order variant $T_{i}$ with a subscript, the definition reads Such a variant is called a first-order covariant tensor. For a higher-order example, a fourth-order variant $T_{jkl}^{i}$ is a tensor if It is best described as a fourth-order tensor of the type indicated by its indicial signature.

In the past, we determined placements of indices according to relatively informal guidelines. For example, we assigned a subscript to the covariant basis $\mathbf{Z}_{i}$ since in its definition, the index $i$ on the right appears on the bottom of a "fraction". The covariant metric tensor $Z_{ij}$, defined by received two subscripts because that was the only choice that conforms to the rules of the tensor notation. The contravariant components $U^{i}$, defined by received a superscript in order to make the summation convention work. We recently discovered that our intuitive guidelines lead to placements that proved to be accurate predictors of the manner in which variants transform under a change of coordinates. However, having now given a precise definition of a tensor, we must retroactively reverse the logic. From this point forward, it is the manner in which the tensor transforms that determines the placement of the indices.

Recall that the covariant and contravariant bases, and the covariant and contravariant metric tensors transform according to the rules Thus, the use of the terms covariant and contravariant in describing these objects is finally justified.

For tensors of order greater than one, it is cumbersome to properly assign the terms covariant and contravariant, especially if the order of the indices is not explicitly specified. It is more efficient to describe such tensors, as we did above, to be of the kind indicated by the indicial signature. On the other hand, the indicial signature speaks for itself and therefore, in most circumstances, no additional description is needed. We will find that the terms covariant and contravariant are often omitted altogether, except for objects such as the metric tensors, where these terms help to distinguish between two closely related objects.

It is clear from the definition that a variant $T_{j}^{i}$ whose elements are defined to be zero in all coordinate systems is a tensor. Furthermore, if a variant that is known to be a tensor, vanishes in one coordinate system, then it vanishes in all coordinate systems. This seemingly simple fact finds frequent applications in practice since it makes it easy to show that a tensor vanishes by finding one special coordinate system in which that is easy to see. In a spectacular application of this idea, we will demonstrate the Riemann-Christoffel identity which holds in any coordinate system in a Euclidean space. The challenging part in proving this identity is demonstrating that the expression on the left is a tensor. Once that is established, however, the identity follows immediately by observing that the Christoffel symbol vanishes in affine coordinates.

Crucially, the definition of a tensor applies to a variant of order zero, i.e. a single number or a single vector. Since variants of order zero have no indices, the definition contains no Jacobians and thus implies that the values of the variant in different coordinate systems are the same. In other words, a tensor of order zero is an invariant. This statement deserves an exclamation point: a tensor of order zero is an invariant! As it will emerge from the upcoming discussion, this observation is not an edge-case technicality but is a fact that is at the heart of the matter and the key to producing geometrically meaningful results.

Stating this fact algebraically requires a slight notational modification. Since a variant $T$ of order zero has no indices that can accept a prime, we are forced to place it on the symbol $T$ itself, as in $T^{\prime}$, to denote the value of $T$ in the primed coordinates. In this notation, the fact that $T$ is a tensor reads

Importantly, not all variants of order zero are invariants. Recall from Chapter 9, that the volume element is defined to be $\sqrt{Z}$, where $Z$ is the determinant of the matrix representing the covariant metric tensor $Z_{ij}$. Since and it is evident that $\sqrt{Z}$ is not an invariant and therefore not a tensor.

Finally, note that the term invariant is also used to describe scalar and vector fields that exist in the Euclidean space completely independently of coordinates. These are fields that, so to say, "predate" coordinates. For example, the position vector field $\mathbf{R}$, a temperature field $T$, and its gradient $\mathbb{\nabla}T$ can be rightfully described as invariants since their values are independent of coordinates. For these fields, however, the term invariant should not be understood in the sense of a coordinate-independent result of a coordinate-dependent analytical procedure. The invariance of such fields is deeper: these fields are constructed without a reference to any coordinate system.

We will soon show -- in short order! -- that all basic algebraic operations on variants, i.e. addition, multiplication, and contraction, preserve the tensor property. In other words, a sum of two tensors, a product of two tensors, and a contraction of a tensor are all tensors in their own right. Among these operations, contraction is the only one that reduces the order of a variant. It is therefore always the final step in constructing tensors of order zero, i.e. invariants.

The mechanism by which contraction preserves the tensor property is extraordinarily insightful. Later in this Chapter, we will give a general demonstration but, for now, we will reveal that mechanism by considering the narrow example of contracting a second-order tensor $T_{j}^{i}$. The resulting variant $T_{i}^{i}$ of order zero is commonly referred to as the trace of $T_{j}^{i}$ by analogy with matrices.

Suppose that $T_{j}^{i}$ is a tensor. In other words, its values $T_{j^{\prime}}^{i^{\prime}}$ in the primed coordinate system are related to $T_{j}^{i}$ by the equation Let $T$ be the result of contracting $T_{j}^{i}$, i.e. and $T^{\prime}$ be the result of contracting $T_{j^{\prime}}^{i^{\prime}}$, i.e. Since, the contraction $T_{i^{\prime}}^{i^{\prime}}$ is obtained by replacing $j^{\prime}$ with $i^{\prime}$ on the right: Note that in the previous identity the two Jacobians do not interact. In on the other hand, the two Jacobians are connected by a contraction on $i^{\prime}$. Since, crucially, the two Jacobians are the matrix inverses of each other, i.e. we find that i.e. In other words, $T$ is an invariant, as we set out to show.

This simple example is the culmination of much of our efforts so far. It shows how the tensor framework achieves invariance through the interplay between covariance and contravariance. Tensors are not themselves invariants and are therefore not in and of themselves geometrically meaningful. However, they can be combined to produce geometrically meaningful invariants.

This example also shows how the indicial notation, along with Einstein's summation convention, seamlessly assure that one adheres to operations that preserve the tensor property. The principal guideline of Tensor Calculus is to limit one's analysis to tensors and tensor-preserving operations and thus to be assured of geometrically meaningful results. This seemingly simplistic approach will prove to be extremely effective and will dictate the future course of our investigations.

A further word of appreciation for tensors is in order. What a beautiful object a tensor is! It simultaneously exhibits features that reflect its geometric origins along with some artifacts of the coordinate system. However, the artifacts of the coordinate system are not out of control. They are present in systematic fashion captured algebraically by the Jacobians. By the end of analysis, those artifacts are removed by contraction, leaving us with a geometrically meaningful invariant. Thus, tensors can be thought of as nearly-invariant. So much so that the term invariant is sometimes used in place of tensor. For example, the covariant derivative introduced in Chapter 15 is sometimes referred to as the invariant derivative. The same is the case for the $\dot{\nabla}$-derivative in the Calculus of Moving Surfaces which is also often referred to as the invariant time derivative.

The sum property states that the sum of two tensors is a tensor in its own right. This, of course, is a nearly trivial statement. After all, the tensor transformation rule amounts to a contraction with a Jacobian which is essentially an elementary sum of elementary products. Thus, the fact that the sum of two tensors is itself a tensor is equivalent to the distributive property of multiplication. Nevertheless, we will give a formal proof of the sum property as it will be help establish a template for a number of future proofs.

Since the precise form of the transformation of a tensor depends on its order, proofs of the tensor property, much like its definition, are usually given for tensors with a reasonably representative collection of indices. This approach is effective as long as it makes it clear that the proof remains valid for tensors with arbitrary indicial signatures. In that spirit, suppose that the variants $A_{jk}^{i}$ and $B_{jk}^{i}$ are tensors, i.e. Let $C_{jk}^{i}$ be the sum of $A_{jk}^{i}$ and $B_{jk}^{i}$: Our goal is to show $C_{jk}^{i}$ is a tensor, i.e. By definition, $C_{j^{\prime}k^{\prime}}^{i^{\prime}}$ is the sum of $A_{j^{\prime}k^{\prime}}^{i^{\prime}}$ and $B_{j^{\prime}k^{\prime} }^{i^{\prime}}$: Substituting the transformation rules for $A_{jk}^{i}$ and $B_{jk}^{i}$ yields By the distributive law, Since $A_{jk}^{i}+B_{jk}^{i}=C_{jk}^{i}$ we find that, indeed, as we set out to prove.

Let us now demonstrate the product property of tensors, i.e. that the product of two tensors is a tensor in its own right. As before, we will consider two tensors with specific indicial signatures and, as before, it will be clear that the argument remains valid in general. Suppose that $A_{j}^{i}$ and $B_{lm}^{k}$ are tensors, i.e. and let $C_{jlm}^{ik}$ be their product: We must show that $C_{jlm}^{ik}$ is a tensor, i.e. In the primed coordinate system, $C_{j^{\prime}l^{\prime}m^{\prime} }^{i^{\prime}k^{\prime}}$ is given by the product of $A_{j^{\prime} }^{i^{\prime}}$ and $B_{l^{\prime}m^{\prime}}^{k^{\prime}}$: Substituting the transformation rules for $A_{j}^{i}$ and $B_{lm}^{k}$, we find Rearranging the multiplicative terms yields the desired identity

Note that the product property applies to the dot product as well as to the ordinary tensor product. Demonstrating this fact is left as an exercise.

According to the contraction property, a contraction of a tensor is a tensor in its own right. Earlier, we demonstrated a special case of this property when we showed that the contraction $T_{i}^{i}$ of a second-order tensor $T_{j}^{i}$ is an invariant, i.e. a tensor of order zero. At the heart of the proof was the reciprocal relationship between the Jacobians $J_{i}^{i^{\prime}}$ and $J_{i^{\prime}}^{i}$. This result needs to be generalized only slightly to show that a contraction of any tensor is a tensor in its own right.

Once again, the proof proceeds by considering a tensor with a representative collection of indices. Suppose that $S_{jk}^{i}$ is a tensor, i.e. and let $T_{k}$ be the result of contracting $S_{jk}^{i}$ on $i$ and $j$: Or goal is to show that $T_{k}$ is a tensor, i.e. To this end, contract both sides of the identity on $i^{\prime}$ and $j^{\prime}$, i.e. replace $j^{\prime}$ with $i^{\prime}$ on both sides of the equation: Note that the first two Jacobians on the right are contracted on $i^{\prime}$. Since $J_{i}^{i^{\prime}}J_{i^{\prime}}^{j}=\delta_{i}^{j}$ and $S_{jk} ^{i}\delta_{i}^{j}=S_{ik}^{i}$, we find Since $S_{i^{\prime}k^{\prime}}^{i^{\prime}}=T_{k^{\prime}}$ and $S_{ik} ^{i}=T_{k}$, we have as we set out to prove.

As we have already discussed, the implications of the contraction property are far reaching. When we are able to contract away all the indices of a tensor, the result is a tensor of order zero, i.e. an invariant. For example, if $S_{kl}^{ij}$ is a tensor, then $S_{ij}^{ij}$ is an invariant. The quantity $S_{ji}^{ij}$ is also an invariant and, of course, it is very much different from $S_{ij}^{ij}$.

In Chapter 11, we introduced index juggling, a notational device that allows us to use the same letter for variants related by contraction with a metric tensor. Since the metric tensors are, in fact, tensors, index juggling preserves the tensor property by a combination of the product and contraction properties. Consistent with its name, index juggling invariably changes the placement of the index, e.g. and therefore the type of the tensor. More precisely, the result of performing index juggling on a tensor of the type indicated by its indicial signature is a tensor in its own right of the type indicated by the resulting signature.

It is tempting to think of the Kronecker delta $\delta_{j}^{i}$ as an invariant since it has the same values (namely, $1$ if $i=j$ and $0$ otherwise) in all coordinates. However, labeling it an invariant is not a valid application of that concept since the Kronecker delta is not a variant of order zero. It is a variant of order two and therefore a more relevant question is whether it is a tensor. As a matter of fact, this question is essential since the Kronecker delta will continue to appear frequently in our analysis. And, since we are making a commitment to operate exclusively with tensors, we must make sure that the Kronecker delta is one.

By definition, the Kronecker delta is a tensor if the identity is valid. Since $\delta_{j}^{i}J_{i}^{i^{\prime}}=J_{j}^{i^{\prime}}$ and $J_{j}^{i^{\prime}}J_{j^{\prime}}^{j}=\delta_{j^{\prime}}^{i^{\prime}}$, the two sides of the equation match and thus the above identity is confirmed. We therefore conclude that the Kronecker delta is indeed a tensor.

An alternative derivation is based on the identity Since the contravariant and the covariant bases are tensors, the tensor property of the Kronecker delta follows from the product property.

There are number of interesting invariants involving the Kronecker delta. The simplest one is which, of course, represents the dimension $n$ of the relevant Euclidean space. Other invariants involve a second-order tensor $A_{j}^{i}$. For example, which, as we have already mentioned, is referred to as the trace of $A_{j}^{i}$ by analogy with matrices. Another interesting invariant is and yet another is Note, also, that the quantities in parentheses are tensors. These expressions exhibit patterns associated with the determinant, an important concept whose tensor treatment is the subject of Chapter 16.

Suppose that $F$ is an invariant. It may be a vector field, such as the position vector $\mathbf{R}$, or a scalar field, such as a temperature distribution $T$. Then the collection of partial derivatives is a first-order covariant tensor. The symbol introduced earlier clearly treats the index as a subscript which is now warranted given the covariant tensor nature of the resulting object. Note that the symbol $\nabla_{i}$, as defined above, should not be applied to tensors of order greater than zero since, as we shall see, the result is not necessarily a tensor. This conundrum is fixed by introducing the concept of the covariant derivative which the symbol $\nabla_{i}$ will eventually denote.

The demonstration of the tensor property of $\nabla_{i}F$ is essentially a repeat of the calculation that proved the tensor property of the covariant basis $\mathbf{Z}_{i}$. This is not surprising since $\mathbf{Z}_{i}$ is a special case of $\nabla_{i}F$ for $F=\mathbf{R}$:

The variant $\nabla_{i^{\prime}}F$ in the primed coordinate system $Z^{i^{\prime}}$ is given by Represent $F\left( Z^{\prime}\right)$ as a composition of $F\left( Z\right)$ with the equations of coordinate transformation i.e. Differentiating this identity with respect to $Z^{i^{\prime}}$ yields Since we find as we set out to prove.

The fact that $\nabla_{i}F$ is a tensor gives new insight into the coordinate expression for the gradient $\mathbf{\nabla}F$ of a scalar field $F$. Since $\nabla_{i}F$ and $\mathbf{Z}^{i}$ are tensors, the contraction $\nabla_{i}F\ \mathbf{Z} ^{i}$ is an invariant.

Consider a tensor $\mathbf{U}_{k}^{ij}$ with vector elements. Then the contravariant components $U_{k}^{ijl}$ of $\mathbf{U}_{k}^{ij}$, i.e. and the covariant components $U_{kl}^{ij}$, i.e. are tensors in their own right.

This can be demonstrated in a number of ways. The shortest way is to give explicit expressions for the components with the help of the dot product: The tensor property of $U_{k}^{ijl}$ and $U_{kl}^{ij}$ then follows from the product property of tensors.

Let us also give another proof that avoids the dot product. When we substitute the transformation rules and into the identity we find Inverting the three Jacobians on the left yields Since this identity represents the decomposition of $\mathbf{U}_{k^{\prime} }^{i^{\prime}j^{\prime}}$ with respect to $\mathbf{Z}_{l^{\prime}}$, the quantity multiplying $\mathbf{Z}_{l^{\prime}}$ on the right must be the component $U_{k^{\prime}}^{i^{\prime}j^{\prime}l^{\prime}}$, i.e. as we set out to prove. It is left as an exercise to show that the covariant component $U_{kl}^{ij}$ is also a tensor.

The fact that the components of tensors with vector elements are tensors offers an important insight into a topic considered earlier in Chapters 10 and 12. Recall our discussions of the motion of a material particle given by the equations We demonstrated that the components $V^{i}$ of the velocity vector $\mathbf{V}$ are given by Due to the tensor property of the components of a vector, we can now conclude that $V^{i}$ is a tensor.

Furthermore, we demonstrated that the components $A^{i}$ of the acceleration vector $\mathbf{A}$ are given by the following expression Thus, we can conclude that the combination on the right is a tensor. This is a very intriguing realization since the Christoffel symbol, which transforms according to the rule is not a tensor. It is also easy to show, and is left as an exercise, that $dV^{i}/dt$ is not a tensor. Yet, the combination is a tensor. This observation foreshadows two important developments. The first is the recognition that differentiation is not a tensor-preserving operation, i.e. the result of differentiating a tensor is not a tensor. This fact will be discussed below in this Chapter. The second is the counterbalancing insight that a proper corrective term involving the Christoffel symbol can restore the tensor property. This is the main idea behind the operator of the covariant derivative and, in fact, of numerous other novel differential operators in Tensor Calculus and the Calculus of Moving Surfaces.

Any definition of a new concept should be scrutinized against the possibility of internal contradictions. Suppose, for a moment, that the definition of the tensor property of a variant $T_{j^{\prime}}^{i^{\prime}}$ reads Although it is similar to the actual definition $T_{j^{\prime}}^{i^{\prime} }=T_{j}^{i}J_{i}^{i^{\prime}}J_{j^{\prime}}^{j}$, the hypothetical definition above would immediately give rise to several internal contradictions.

For instance, what would be the implications of this definition if $Z^{i}$ and $Z^{i^{\prime}}$ happen to be the same coordinate system? Or, to put it another way, if the equations of coordinate transformation were Under this scenario, the Jacobians correspond to the identity matrix and this implies that the elements of $T_{j^{\prime}}^{i^{\prime}}$ equal twice their value, i.e. which is a glaring contradiction for any nonzero variant $T_{j}^{i}$.

Furthermore, the hypothetical definition would not hold up when the roles of the coordinate systems $Z^{i}$ and $Z^{i^{\prime}}$ are reversed. According to the hypothetical definition, we must have On the other hand, inverting the Jacobians in equation yields This, once again, is an internal contradiction.

Finally, a third contradiction would be reached if we considered three coordinate systems $Z^{i}$, $Z^{i^{\prime}}$, and $Z^{i^{\prime\prime}}$ and compared the transformation of a variant $T_{j}^{i}$ from $Z^{i}$ to $Z^{i^{\prime}}$ and then from $Z^{i^{\prime}}$ to $Z^{i^{\prime\prime}}$ to the transformation directly from $Z^{i}$ to $Z^{i^{\prime\prime}}$.

Of course, the actual definition of a tensor does not suffer from these contradictions. It is reflexive, meaning that it is valid when $Z^{i}$ and $Z^{i^{\prime}}$ are the same coordinate systems. It is symmetric, meaning that it consistently applies when roles of the coordinate systems $Z^{i}$ and $Z^{i^{\prime}}$ are reversed. The symmetric property of tensors was effectively demonstrated in Section 13.3.5. Finally, it is transitive, meaning that transformations among three or more coordinate systems are consistent. Specifically, the transitive property states that the equations imply The proof of the transitive property is left as an exercise.

All coordinate space variants that we have encountered so far have arisen naturally as systems associated, in one way or another, with geometric objects in the Euclidean space. However, there also exists an important and rather straightforward way of artificially constructing tensors, known as synthetic tensors, that are not associated with any geometric object.

The method works the same way for a tensor of any order. Thus, for the sake of specificity, we will illustrate it with the help of a fourth-order tensor $A_{jkl}^{i}$. In some unprimed coordinates $Z^{i}$ let $A_{jkl}^{i}$ have arbitrary values. Then, in all other coordinate systems $Z^{i^{\prime}}$, define the values of $A_{j^{\prime}k^{\prime }l^{\prime}}^{i^{\prime}}$ to be In order to confirm that the resulting object is a tensor, we must show that for two coordinate systems $Z^{i^{\prime}}$ and $Z^{i^{\prime\prime}}$, the systems $A_{j^{\prime}k^{\prime}l^{\prime}}^{i^{\prime}}$ and $A_{j^{\prime \prime}k^{\prime\prime}l^{\prime\prime}}^{i^{\prime\prime}}$ are related by This, of course, is precisely the transitive property of the tensor discussed in the previous Section. Even though the present circumstances are slightly different in that the transformation rule is stipulated as a matter of definition, the demonstration of this property would be exactly the same.

This construction of synthetic tensors is somewhat un-tensorlike in that it singles out one coordinate system, $Z^{i}$, relative to others. Nevertheless, a tensor is a tensor whether it arose from a geometric object or not. The significance of synthetic tensors is that any indexed collection of values in any coordinate system is a manifestation of some tensor. In fact, there is a perfect one-to-one correspondence between tensors and the set of all possible indexed collection of values in a given coordinate system. Synthetic tensors have a number of applications, including one found in the next Section in the proof of the quotient theorem.

We have already established a number of ways in which tensors are formed: by differentiation of invariants, by decomposing vectors with the respect to either the covariant or the contravariant basis, by a synthetic procedure described in the previous Section, and, subsequently, by addition, multiplication, and contraction of other tensors. The quotient theorem provides an additional useful way of recognizing that a particular variant is a tensor. As the name suggests, the quotient theorem is, in a way, the converse of the product property.

Suppose that $U_{jk}^{i}$ and $V^{kl}$ are tensors. Then, by the combination of the product and contraction properties, the variant is a tensor in its own right. However, what if the circumstances are such that it is known that $W_{j}^{il}$ and $V^{kl}$ are tensors. Can we conclude that $U_{jk}^{i}$ is a tensor?

If no further stipulations are added, the answer is certainly no. For example, if both $V^{kl}$ and $W_{j}^{il}$ are identically zero in all coordinate systems, then the above identity holds for any variant $U_{jk}^{i}$ and thus we cannot conclude that $U_{jk}^{i}$ is a tensor. In order to be able to conclude that $U_{jk}^{i}$ is a tensor, the combination $U_{jk}^{i}V^{kl}$ must be a tensor for all tensors $V^{kl}$. That is the content of the quotient theorem. In general, the quotient theorem states that if three variants $U_{\cdot\cdot}^{\cdot\cdot}$, $V_{\cdot\cdot} ^{\cdot\cdot}$, and $W_{\cdot\cdot}^{\cdot\cdot}$ are related by the equation that satisfies all rules of the tensor notation, where the dots indicate otherwise arbitrary indicial signatures and any number of valid contractions on the right, and if for any tensor $V_{\cdot\cdot}^{\cdot\cdot}$ the result $W_{\cdot\cdot}^{\cdot\cdot}$ is also a tensor, then $U_{\cdot\cdot} ^{\cdot\cdot}$ is a tensor in its own right.

Before we turn to the proof, let us mention an analogous problem in Linear Algebra that will serve as a structural blueprint for an argument at the end of the proof. If $A$ is an $n\times m$ matrix and $x$ is a vector in the sense of Matrix Algebra (i.e. an $m\times1$ matrix), under what conditions does the equality $Ax=0$ imply that $A=0$? It is certainly not enough for $Ax=0$ to hold for some $x$: if $x=0$ then $Ax=0$ holds for all matrices $A$. Requiring that $x$ is nonzero is also not sufficient since for any nonzero vector $x$, the equality $Ax=0$ holds for any matrix $A$ that contains the vector $x$ in its null space. In fact, $Ax=0$ can hold for all $x$ in an $\left( m-1\right)$-dimensional subspace of $\mathbb{R} ^{m}$, and $A$ could still be a nonzero, albeit rank-$1$, matrix. For instance, for all $x$ of the form Clearly, such vectors $x$ represents a $4$-dimensional subspace of $\mathbb{R} ^{5}$.

And so we come to the realization that, in order for $Ax=0$ to imply $A=0$, it must hold for all $x$. To prove that that is sufficient, take advantage of the freedom to choose any $x$ and consider Then $Ax_{1}$ is the first column of $A$ and, since $Ax_{1}=0$, we conclude that the first column of $A$ is zero. Similarly, the vector proves that the second column of $A$ is zero, and so on. This completes the proof.

We can now return to the proof of the quotient theorem based on our representative example. In the primed coordinate system, the identity reads Substituting the relationships into the primed identity, we find Invert each of the Jacobians on the left in order to isolate $W_{j}^{il}$: Subtracting this from the initial identity we find Since all the preceding identities are valid for any $V^{kl}$, so is this one. By the argument that we developed for the Linear Algebra problem above, in combination with the fact that any combination of values corresponds to some -- possibly synthetic -- tensor, we can conclude that the quantity in parentheses vanishes. In other words, as we set out to prove.

In defining the tensor property, we insisted that the variant transforms according to a specific rule under all coordinate transformations. In many situations, however, it is appropriate to consider a limited class of transformations. In those situations, we may classify a variant as a tensor if it transforms according to the classical tensor rule under that particular class of transformations. As a result, a variant that is not considered a tensor in the broad sense may be, in fact, considered a tensor in the narrow sense associated with the limited class of transformations. Such a variant is referred to as a qualified tensor.

For example, we may limit our attention to linear changes of variables, i.e. where $A_{i}^{i^{\prime}}$ is a constant second-order system. Then, and therefore Since the second-order Jacobians are related by we also have As a result, the Christoffel symbol $\Gamma_{ij}^{k}$, whose general transformation rule reads transforms according to under linear changes of coordinates. Therefore, it is a tensor with respect to linear changes of coordinates.