The Zero-Combination Proof Strategy
To prove that a list of vectors is linearly independent, begin with an arbitrary linear combination equal to the zero vector. Then use the coordinates of that equation to show that each coefficient is zero. This approach mirrors the definition, but the important work is in choosing and organizing equations so that the conclusion follows without skipping a step.
Throughout this tutorial the scalars are real numbers. A list \(\mathbf{v}_1,\dots,\mathbf{v}_p\) in \(\mathbb{R}^n\) is independent when its only zero combination has all coefficients zero. The statement concerns a list, so positions matter: the coefficients \(c_1,\dots,c_p\) correspond, in order, to \(\mathbf{v}_1,\dots,\mathbf{v}_p\).
This definition gives a proof plan, not an automatic solution. In a vector equation, equality to \(\mathbf{0}\) means equality in every coordinate. Those coordinate equations may immediately isolate one coefficient, or they may need to be combined. In either case, state the equations and show how they force the coefficients to be zero.
A Coordinate Pattern That Guarantees Independence
A useful situation occurs when each vector has a coordinate that is nonzero for it but zero for all vectors appearing later in the list. This creates a sequence of equations that can be handled in order. The condition below allows other coordinates to be nonzero; it only specifies enough zeros to make the argument work.
Proof. Assume \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). At coordinate \(r_1\), every vector \(\mathbf{v}_j\) with \(j>1\) contributes zero, because \(r_1<r_j\). Thus the equation in that coordinate is \(c_1(\mathbf{v}_1)_{r_1}=0\). The coordinate \((\mathbf{v}_1)_{r_1}\) is nonzero, so \(c_1=0\). Now suppose \(c_1,\dots,c_{k-1}=0\), where \(2\le k\le p\). At coordinate \(r_k\), the terms with index less than \(k\) vanish because their coefficients are zero, and the terms with index greater than \(k\) vanish because \(r_k<r_j\). The remaining equation is \(c_k(\mathbf{v}_k)_{r_k}=0\). Since \((\mathbf{v}_k)_{r_k}\ne0\), we have \(c_k=0\). Applying this argument for \(k=2,\dots,p\) proves that every coefficient is zero. \(\square\)
Worked Example: Using Successive Coordinates
Prove that the following vectors are linearly independent:
Assume \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_3=\mathbf{0}\). Writing out all four coordinates gives
The first equation gives \(a=0\). Substituting this into the second gives \(3b=0\), hence \(b=0\). Substituting \(a=b=0\) into the third gives \(-2c=0\), hence \(c=0\). Therefore the assumed zero combination has only zero coefficients, and the list is independent.
Check: Substituting \(a=b=c=0\) gives \(\mathbf{0}\), as required. The last coordinate equation also becomes \(3(0)+0+5(0)=0\), so it is consistent with the values forced by the first three equations.
Here the decisive coordinates are \(1,2,3\), and their nonzero entries are \(2,3,-2\). The fourth coordinate does not need to be used to prove independence. It is still part of the original vector equation, but a proof may use only the equations needed to force all coefficients to vanish.
When the Equations Need to Be Combined
Not every independent list displays a convenient sequence of zeros. The same strategy still applies: write each coordinate equation, then combine equations when necessary. Any operation on equations must preserve the information needed to solve for the coefficients.
Worked Example: Combining Coordinate Equations
Prove that the following list in \(\mathbb{R}^3\) is linearly independent:
Assume \(a\mathbf{u}_1+b\mathbf{u}_2+c\mathbf{u}_3=\mathbf{0}\). The three coordinate equations are
Subtract twice the first equation from the second. The left side becomes \((2a+4b+c)-2(a+2b)=c\), so \(c=0\). The third equation now gives \(a+b=0\). Subtracting this equation from \(a+2b=0\) gives \(b=0\), and then \(a+b=0\) gives \(a=0\). Thus all three coefficients vanish.
Check: With \(a=b=c=0\), the combination is \(0\mathbf{u}_1+0\mathbf{u}_2+0\mathbf{u}_3=\mathbf{0}\). Also, the equations themselves force these values: the second minus twice the first forces \(c=0\); the remaining equations force \(a=b=0\).
The order of operations is part of the reasoning. It would not be enough to say that the equations “solve to zero.” Showing how one equation combination gives \(c=0\), and how the remaining equations then give \(a=b=0\), makes the implication verifiable.
Proving Independence of Polynomials
Polynomials can be treated as vectors, but their coordinates are coefficients rather than entries in a fixed column. In \(\mathbb{R}[x]\), a polynomial is a finite expression \(a_0+a_1x+\cdots+a_dx^d\) with real coefficients. Two polynomials are equal exactly when the coefficients of each power of \(x\) agree. In particular, a polynomial is the zero polynomial exactly when all of its coefficients are zero.
Proof. Suppose \(c_1x^{d_1}+\cdots+c_px^{d_p}\) is the zero polynomial. Since the exponents \(d_1,\dots,d_p\) are distinct, the coefficient of \(x^{d_j}\) in this sum is exactly \(c_j\): none of the other terms contributes to that power. Every coefficient of the zero polynomial is zero, so \(c_j=0\) for each \(j\). Therefore the only zero combination is the one with every coefficient zero, which proves independence. \(\square\)
Worked Example: Comparing Polynomial Coefficients
Prove that \(1+x\), \(1-x\), and \(x^2\) are linearly independent in \(\mathbb{R}[x]\).
Assume a real linear combination is the zero polynomial:
Expand and collect powers of \(x\):
Comparing the constant, \(x\), and \(x^2\) coefficients gives \(a+b=0\), \(a-b=0\), and \(c=0\). Adding the first two equations gives \(2a=0\), so \(a=0\). Then \(a+b=0\) gives \(b=0\), and the third equation already gives \(c=0\). Hence the three polynomials are independent.
Check: The only coefficients found are \(a=b=c=0\), and substituting them gives the zero polynomial. Conversely, the coefficient equations show that no nonzero choice can produce the zero polynomial.
Worked Example: Independence Despite Overlapping Terms
Prove that \(1+x\), \(x+x^2\), and \(1+x^2\) are linearly independent in \(\mathbb{R}[x]\).
Assume \(a(1+x)+b(x+x^2)+c(1+x^2)=0\). Expanding gives
Coefficient comparison produces the system \(a+c=0\), \(a+b=0\), and \(b+c=0\). The first equation gives \(c=-a\), and the second gives \(b=-a\). Substituting both into the third gives \(-a-a=-2a=0\). Since the scalars are real, \(2\ne0\), so \(a=0\); consequently \(b=0\) and \(c=0\). This proves independence.
Check: Substituting \(a=b=c=0\) gives the zero polynomial. The coefficient equations also show why there is no nonzero solution: they imply \(-2a=0\), then force both remaining coefficients to vanish.
This argument relies on the scalar field. Here the coefficients are real, so dividing by \(2\) is valid. In a field where \(2=0\), that particular step would not establish \(a=0\); a proof must respect the field in which the vectors or polynomials are being considered.
A Common Mistake: Treating Summands as Separate Equations
A zero sum does not mean that each summand is zero. For example, from \(a(1+x)+b(1-x)+cx^2=0\), it is invalid to claim separately that \(a(1+x)=0\), \(b(1-x)=0\), and \(cx^2=0\). Different summands can cancel one another. The valid information comes from expanding the whole expression and equating coefficients of matching powers.
A related error is to claim that coordinate equations force all coefficients to zero without actually solving them. For the list \(\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\), the zero-combination equations are \(a+c=0\) and \(b+c=0\). A mistaken argument might assert that these imply \(a=b=c=0\). But taking \(c=1\) gives \(a=-1\) and \(b=-1\), which satisfies both equations and is nontrivial. Indeed, \(-1\begin{bmatrix}1\\0\end{bmatrix}-1\begin{bmatrix}0\\1\end{bmatrix}+1\begin{bmatrix}1\\1\end{bmatrix}=\mathbf{0}\). The list is dependent, not independent.
Before concluding independence, check that the equations really force each coefficient to zero. If they do not, a nonzero solution disproves independence. If they do, the proof should state the deductions in a sequence that can be checked line by line.
A Reliable Proof Checklist
For a list of vectors or polynomials, a concise independence proof can follow this sequence:
- Write an arbitrary linear combination with scalar coefficients and set it equal to zero.
- Expand the combination, then write the coordinate equations or polynomial coefficient equations.
- Solve those equations, showing any combinations or substitutions used.
- State explicitly that every coefficient is zero, so the list is independent by definition.
The method is useful because it proves the exact statement required by the definition. It also identifies where a proposed proof can fail: perhaps an equation was omitted, a cancellation was assumed without justification, or the coefficient equations have a nonzero solution. The final conclusion is only as strong as the deductions that establish it.
Check Your Understanding
For each question, begin with a general zero combination and show the equations that determine its coefficients.
- Prove that \(\begin{bmatrix}1\\1\\0\end{bmatrix}\), \(\begin{bmatrix}0\\2\\1\end{bmatrix}\), and \(\begin{bmatrix}0\\0\\3\end{bmatrix}\) are linearly independent.
- Are \(1+x\), \(1-x\), and \(1\) linearly independent in \(\mathbb{R}[x]\)? Justify your answer by coefficient comparison.
- For the list \(x\), \(x^2\), \(x^4\) in \(\mathbb{R}[x]\), explain why each coefficient in a zero combination must vanish.
- Find a nontrivial zero combination of \(\begin{bmatrix}1\\0\end{bmatrix}\), \(\begin{bmatrix}0\\1\end{bmatrix}\), and \(\begin{bmatrix}1\\1\end{bmatrix}\), and use it to state whether the list is independent.
- In a proof over \(\mathbb{R}\), why is it valid to infer \(a=0\) from \(2a=0\)?
