Definitions as Proof Instructions
A definition does more than assign a name to an object: it tells you exactly what must be shown to use that name. A direct proof begins by identifying the definition’s assumptions and conclusion. For a statement that says “every choice with property \(P\) has property \(Q\),” the proof takes an arbitrary choice satisfying \(P\) and derives \(Q\). For a statement that says “there exists a choice,” the proof must construct a particular choice and verify it. Mixing up these tasks is a common source of incomplete arguments.
We use vector spaces over a field \(\mathbb{F}\), which may be \(\mathbb{R}\) or \(\mathbb{C}\). A list is ordered, so its entries and their coefficients have specified positions. In the definitions below, the vectors belong to a vector space \(V\) over \(\mathbb{F}\). When an example uses \(\mathbb{R}^n\), its vectors are columns.
The two definitions have different logical shapes. Independence says that every coefficient choice satisfying the zero equation must be trivial. To prove it, assume an arbitrary coefficient choice satisfies that equation and derive that each coefficient is zero. To prove dependence instead, one nontrivial coefficient choice is enough. Spanning says that for each target vector, some coefficients represent it. A proof must start with an arbitrary target and find coefficients that work for that target.
In symbols, the key distinction is between “for every choice” and “there exists a choice.” In a spanning claim, the order matters: first choose an arbitrary \(\mathbf{v}\in V\), then find coefficients that may depend on \(\mathbf{v}\). The definition does not demand one fixed coefficient tuple that represents all vectors.
Linearity and Subspace Conditions
Here the assumptions are arbitrary vectors \(\mathbf{u},\mathbf{v}\) and arbitrary scalars \(a,b\). The conclusion is the stated equality. A direct proof must use general symbols and establish the equality without choosing special vectors or scalars. Checking a few inputs can disprove linearity if one fails, but it cannot prove the equality for every input.
Linearity is also often defined by two conditions: \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v}\), and \(T(a\mathbf{u})=aT(\mathbf{u})\) for all \(a,\mathbf{u}\). These conditions and the two-vector formula above are equivalent.
Proof. Suppose first that the two-vector formula holds. Set \(a=b=1\) to obtain \(T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})\). Set \(b=0\) to obtain \(T(a\mathbf{u})=aT(\mathbf{u})\). Conversely, suppose additivity and scalar compatibility hold. Applying them in succession gives \(T(a\mathbf{u}+b\mathbf{v})=T(a\mathbf{u})+T(b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\) for all \(\mathbf{u},\mathbf{v},a,b\). Thus the two formulations imply each other. \(\square\)
For a direct proof of a subspace claim, these conditions indicate the separate goals: establish that zero belongs, take arbitrary elements of \(U\) to check addition, and take an arbitrary element and scalar to check scalar multiplication. The conditions are universal over the indicated choices; checking only a few elements does not establish them. The upcoming subject of subspace tests develops how these closure conditions can be organized into a proof.
Worked Proofs from the Quantified Forms
Worked Example: Proving Independence Directly
Show that the list \(((1,1),(1,-1))\) is linearly independent in \(\mathbb{R}^2\).
By the definition, take arbitrary \(c_1,c_2\in\mathbb{R}\) and assume their linear combination is \(\mathbf{0}\). The column-vector equation is
Equating coordinates gives \(c_1+c_2=0\) and \(c_1-c_2=0\). Adding these equations yields \(2c_1=0\), so \(c_1=0\). Substituting into \(c_1+c_2=0\) gives \(c_2=0\). This proves the required conclusion for arbitrary coefficients.
Check: substituting \(c_1=c_2=0\) into the original combination gives \(\begin{bmatrix}0\\0\end{bmatrix}\). The argument also showed that no other coefficient choice can give zero.
Worked Example: Finding Coefficients for a Spanning Claim
Find coefficients that express \(\begin{bmatrix}3\\1\end{bmatrix}\) as a linear combination of \(\begin{bmatrix}1\\1\end{bmatrix}\) and \(\begin{bmatrix}1\\-1\end{bmatrix}\).
For this target, the existence clause in the definition asks us to find \(c_1,c_2\in\mathbb{R}\) satisfying
Comparing coordinates gives \(c_1+c_2=3\) and \(c_1-c_2=1\). Adding the equations gives \(2c_1=4\), hence \(c_1=2\). The first equation then gives \(c_2=1\).
Check: \(2\begin{bmatrix}1\\1\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}2\\2\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}\). This establishes membership in the span for this particular target. To prove that the list spans a whole space, the proof must handle every target in that space, not just this one.
Worked Example: Verifying Linearity from a Formula
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T(x,y)=(x+2y,3x-y)\). Verify linearity directly.
Take arbitrary \(\mathbf{u}=(x_1,y_1)\), \(\mathbf{v}=(x_2,y_2)\) in \(\mathbb{R}^2\), and \(a,b\in\mathbb{R}\). First form the input combination: \(a\mathbf{u}+b\mathbf{v}=(ax_1+bx_2,ay_1+by_2)\). Applying the formula for \(T\) gives
The equality holds for arbitrary inputs and scalars, so the definition proves \(T\) is linear.
Check: For \(\mathbf{u}=(1,2)\), \(\mathbf{v}=(0,1)\), \(a=2\), and \(b=-1\), the input is \(2(1,2)-(0,1)=(2,3)\), and \(T(2,3)=(8,3)\). Separately, \(2T(1,2)-T(0,1)=2(5,1)-(2,-1)=(8,3)\), matching the general identity.
Quantified Claims and Direct-Proof Structure
The definitions can be converted into a practical sequence of questions before writing a proof. Identify the domain of arbitrary choices, write down the assumptions imposed on them, and state the exact conclusion. Then decide whether the claim is universal, existential, or an “if and only if.” For a universal implication, begin with an arbitrary object satisfying the assumptions. For an existence claim, construct a candidate and verify it. For an “if and only if,” prove each direction separately, because each direction has its own assumption and conclusion.
In an independence claim, the coefficients are arbitrary. In a spanning claim, the target vector is arbitrary and the coefficients are to be found.
For independence, the assumption is that a combination equals zero; the goal is that every coefficient equals zero. For linearity, the assumptions are arbitrary inputs and scalars; the goal is the defining equality.
Derive a conclusion from arbitrary assumptions for a universal claim. Produce and check a witness for an existential claim.
A calculation for one vector proves a statement about that vector. It does not, without further argument, prove a claim about every vector in a space.
These steps are not a replacement for the definitions; they are a way to use them without dropping a quantifier. In particular, “let the coefficients be arbitrary” is not a decorative phrase in an independence proof. It ensures that the argument covers every coefficient choice in the definition. Likewise, “let \(\mathbf{v}\in V\) be arbitrary” is what makes a representation argument relevant to a spanning claim about all of \(V\).
A Subspace Claim as a Proof Goal
Worked Example: Refuting a Subspace Claim with a Witness
Consider \(U=\{(x,y)\in\mathbb{R}^2:xy=0\}\). The set consists of points on the coordinate axes. Determine whether it is a subspace.
The subspace definition requires closure under addition for every pair of elements of \(U\). Choose \(\mathbf{u}=(1,0)\) and \(\mathbf{v}=(0,1)\). Since \(1\cdot0=0\), \(\mathbf{u}\in U\). Since \(0\cdot1=0\), \(\mathbf{v}\in U\). Their sum is \(\mathbf{u}+\mathbf{v}=(1,1)\), and its coordinate product is \(1\cdot1=1\ne0\). Therefore \((1,1)\notin U\), so the required universal closure condition fails.
Check: Both selected vectors satisfy the defining equation \(xy=0\), while their sum does not. A single such pair disproves a condition that requires closure for every pair, so \(U\) is not a subspace.
This example illustrates an important asymmetry. To establish a universal condition, an argument must cover arbitrary eligible choices. To refute it, one counterexample is enough. Thus, finding one pair whose sum leaves \(U\) disproves addition closure, whereas checking several pairs that happen to remain in \(U\) cannot prove closure for all pairs.
Common Quantifier Errors
Wrong: one target proves a spanning claim. The calculation expressing \((3,1)\) as a combination of two vectors proves a representation for \((3,1)\), not for every vector of \(\mathbb{R}^2\). The correction is to begin with arbitrary \((x,y)\) and find coefficients in terms of \(x,y\), or otherwise establish that every target has a representation.
Wrong: checking a few inputs proves linearity. For example, the function \(S:\mathbb{R}\to\mathbb{R}\), \(S(x)=x^2\), satisfies \(S(0)=0\) and \(S(1)=1\). Those checks do not prove linearity. In fact, using \(a=2\) and \(x=1\), the scalar condition would require \(S(2)=2S(1)\), but \(S(2)=4\) while \(2S(1)=2\). This concrete failure shows that \(S\) is not linear. To prove linearity, the defining equality must be established for arbitrary inputs and scalars.
Wrong: reverse the quantifiers in a spanning statement. “For every target, there exist coefficients” allows the coefficients to depend on the target. It does not mean “there exist coefficients that work for every target.” The latter is a much stronger and usually irrelevant claim. Keep the target fixed but arbitrary first, and then construct its coefficients.
A well-formed proof goal is therefore more than a topic label such as “independence” or “linearity.” It is a precise statement of what is assumed, which objects remain arbitrary, and what must follow. Once those parts are visible, the direct proof has a clear starting point and a checkable endpoint.
Check Your Understanding
For each question, identify the quantifiers and the direct-proof goal before calculating.
- State the quantified definition of linear independence for a list of \(p\) vectors.
- In a proof that a list spans \(V\), which object should be chosen arbitrarily first, and what must then be found?
- Write the assumptions and conclusion in the direct proof that \(T\) is linear.
- What single calculation would be enough to disprove addition closure for a proposed subspace?
- Explain why representing one particular vector does not prove that a list spans \(\mathbb{R}^2\).
