Reading a Vector Equation One Coordinate at a Time
In \(\mathbb{R}^n\), every vector has \(n\) coordinates, so an equation between vectors requires agreement in every coordinate. This simple fact gives a direct proof that the standard coordinate vectors are linearly independent. It also provides a useful way to check other lists: rather than row-reducing a matrix, write the zero-combination equation and inspect its coordinates.
Throughout, \(n\in\mathbb{N}\), scalars are real, and vectors in \(\mathbb{R}^n\) are columns. For \(1\leq i\leq n\), the standard coordinate vector \(\mathbf{e}_i\) has a \(1\) in coordinate \(i\) and \(0\) in every other coordinate. Thus, for example, \(\mathbf{e}_2\in\mathbb{R}^4\) is the column with entries \(0,1,0,0\). Recall that a list is linearly independent when the only linear combination of its vectors equal to \(\mathbf{0}\) has all coefficients zero.
The defining feature of \(\mathbf{e}_i\) is that its \(i\)th coordinate records its coefficient in a linear combination of the standard coordinate vectors. In the sum \(c_1\mathbf{e}_1+\cdots+c_n\mathbf{e}_n\), coordinate \(k\) receives a contribution from \(\mathbf{e}_k\), but none from the other standard coordinate vectors. The sum therefore has coordinates \(c_1,\dots,c_n\).
This coordinate identity is the entire mechanism behind the proof. In particular, if the sum is \(\mathbf{0}\), each coordinate \(c_k\) must be zero. The argument applies for every positive integer \(n\); it does not depend on a separate case for a particular dimension.
The Standard Coordinate Vectors Are Independent
Proof. Assume that real scalars \(c_1,\dots,c_n\) satisfy \(c_1\mathbf{e}_1+\cdots+c_n\mathbf{e}_n=\mathbf{0}\). By the coordinate identity above, the left side is the column with entries \(c_1,\dots,c_n\). The zero vector has \(0\) in every coordinate, so equality of the two columns gives \(c_k=0\) for each \(k\) with \(1\leq k\leq n\). Thus the only zero combination is the one with all coefficients zero, which is the definition of linear independence. \(\square\)
The proof depends on matching coordinates, not on counting how many entries in the sum are nonzero. For instance, in \(\mathbb{R}^4\), setting \(2\mathbf{e}_1-\mathbf{e}_2+5\mathbf{e}_3+0\mathbf{e}_4=\mathbf{0}\) would require the first coordinate to be \(2=0\). That requirement fails, so this particular combination is not the zero vector.
Worked Example: Checking the Coordinate Argument in \(\mathbb{R}^4\)
Suppose \(a\mathbf{e}_1+b\mathbf{e}_2+c\mathbf{e}_3+d\mathbf{e}_4=\mathbf{0}\). Find the coefficients.
Write the vectors as columns and add coordinate by coordinate:
Equality of columns gives \(a=0\), \(b=0\), \(c=0\), and \(d=0\). These are all the coefficients, so the equation has only the trivial choice.
Check: Substituting those values gives \(0\mathbf{e}_1+0\mathbf{e}_2+0\mathbf{e}_3+0\mathbf{e}_4=\mathbf{0}\), as required.
Selecting Standard Coordinate Vectors
A list need not contain every standard coordinate vector to benefit from this argument. If it contains distinct standard coordinate vectors, each vector still has a coordinate where it contributes \(1\) and all the other vectors in the list contribute \(0\). The coefficients can therefore be read off from those coordinates.
Proof. Let the chosen list be \((\mathbf{e}_{i_1},\dots,\mathbf{e}_{i_q})\), where the indices \(i_1,\dots,i_q\) are distinct. Assume \(c_1\mathbf{e}_{i_1}+\cdots+c_q\mathbf{e}_{i_q}=\mathbf{0}\). In coordinate \(i_j\), the vector \(\mathbf{e}_{i_j}\) contributes \(c_j\), and every other vector in the list contributes \(0\), because its index differs from \(i_j\). Thus coordinate \(i_j\) of the sum is \(c_j\). Since the sum is \(\mathbf{0}\), \(c_j=0\). This holds for each \(j\) from \(1\) to \(q\), so all coefficients vanish and the list is independent. \(\square\)
Worked Example: A Selected List in \(\mathbb{R}^5\)
Determine whether \((\mathbf{e}_2,\mathbf{e}_5,\mathbf{e}_1)\) is independent in \(\mathbb{R}^5\).
Set \(a\mathbf{e}_2+b\mathbf{e}_5+c\mathbf{e}_1=\mathbf{0}\). In coordinates, the equation is
Coordinate \(1\) gives \(c=0\); coordinate \(2\) gives \(a=0\); and coordinate \(5\) gives \(b=0\). Therefore the only zero combination has all coefficients zero, and the list is independent.
Check: Substitution of \(a=b=c=0\) produces the zero vector. Any nonzero choice would make at least one of coordinates \(1,2,5\) nonzero.
A Coordinate That Is Unique to Each Vector
The same reasoning works for vectors that are not standard coordinate vectors. What matters is whether a coordinate can isolate a coefficient. If, for each vector in a list, there is a coordinate where that vector is nonzero and all the other vectors are zero, then the corresponding coordinate of a zero combination forces that vector’s coefficient to vanish.
Proof. Assume that \(c_1\mathbf{v}_1+\cdots+c_p\mathbf{v}_p=\mathbf{0}\). Fix any \(j\). The \(k_j\)th coordinate of the left side is \(c_j\) times the \(k_j\)th coordinate of \(\mathbf{v}_j\), because every other vector has zero in that coordinate. The whole sum is \(\mathbf{0}\), so this coordinate is zero. The \(k_j\)th coordinate of \(\mathbf{v}_j\) is nonzero, and a nonzero real number times \(c_j\) equals zero only when \(c_j=0\). This argument applies to every \(j\), so all coefficients are zero. Therefore the list is independent. \(\square\)
Worked Example: Isolating Coefficients in \(\mathbb{R}^4\)
Use coordinates to determine whether the vectors \(\mathbf{v}_1=(2,0,1,0)\), \(\mathbf{v}_2=(0,-3,0,0)\), and \(\mathbf{v}_3=(0,0,4,1)\) are independent.
Suppose \(a\mathbf{v}_1+b\mathbf{v}_2+c\mathbf{v}_3=\mathbf{0}\). Expanding the coordinates gives
Coordinate \(1\) gives \(2a=0\), so \(a=0\). Coordinate \(2\) gives \(-3b=0\), so \(b=0\). Coordinate \(4\) gives \(c=0\). Coordinate \(3\) gives \(a+4c=0\), which is also satisfied by these values. The only zero combination is trivial, so the vectors are independent.
Check: With \(a=b=c=0\), the coordinate sum is \((0,0,0,0)\). Each coefficient was forced to zero by a coordinate where its vector was the only one with a nonzero entry.
The coordinate isolation condition is sufficient, not necessary: a list can be independent even when no coordinate belongs to only one vector. In that situation, several coefficients may appear together in each coordinate equation, and those equations must be solved together.
Worked Example: Overlapping Coordinates Can Still Determine Every Coefficient
Test whether \(\mathbf{u}_1=(1,1,0)\), \(\mathbf{u}_2=(1,0,1)\), and \(\mathbf{u}_3=(0,1,1)\) are independent.
Set \(a\mathbf{u}_1+b\mathbf{u}_2+c\mathbf{u}_3=\mathbf{0}\). Adding the coordinates gives
Thus \(a+b=0\), \(a+c=0\), and \(b+c=0\). The first equation gives \(b=-a\), and the second gives \(c=-a\). Substituting both into the third gives \(-a-a=0\), or \(-2a=0\). Since the scalars are real, \(a=0\), and hence \(b=0\) and \(c=0\). The list is independent, even though each coordinate equation initially involves two coefficients.
Check: The only coefficients found are \(a=b=c=0\), and they give the zero vector. The equations also show why checking only one coordinate would not have been enough: for example, \(a+b=0\) alone allows nonzero choices such as \(a=1\), \(b=-1\).
Common Mistake: Adding Coefficients as if They Were Coordinates
A zero combination is a vector equation, not a scalar equation formed by adding all the coefficients. For example, a mistaken calculation might claim that \(2\mathbf{e}_1-\mathbf{e}_2+4\mathbf{e}_3=\mathbf{0}\) because \(2-1+4\) is a scalar sum that can be manipulated separately from the vector entries. That does not test whether the vector is zero. The actual coordinate calculation is
The correction is to compare each coordinate with zero. Here the first coordinate is \(2\), the second is \(-1\), and the third is \(4\), so the combination is not zero. For the standard coordinate vectors, that same comparison proves independence: every coefficient appears by itself in one coordinate.
Coordinate-by-coordinate reasoning is particularly efficient when a vector list has visibly distinctive entries. If the coefficients do not separate immediately, as in the overlapping-coordinate example, the method still applies: write every coordinate equation, then solve the resulting scalar equations without discarding any of them.
Check Your Understanding
For each question, write the zero-combination equation and compare coordinates.
- Prove directly that \((\mathbf{e}_1,\mathbf{e}_3)\) is independent in \(\mathbb{R}^4\).
- For \(\mathbf{v}_1=(3,0,1)\) and \(\mathbf{v}_2=(0,2,0)\), identify a coordinate that isolates each coefficient and use it to test independence.
- Do the vectors \((1,0,1)\), \((0,1,1)\), and \((1,1,2)\) form an independent list? Show the coordinate equations and decide.
- Explain why the coordinate isolation theorem gives a sufficient condition for independence, but does not claim that every independent list must have such coordinates.
- Find the coordinates of \(3\mathbf{e}_1-2\mathbf{e}_3+\mathbf{e}_5\) in \(\mathbb{R}^5\), then state whether this combination is the zero vector.
