Convergence One Coordinate at a Time
A vector in \(\mathbb{R}^n\) is specified by finitely many real coordinates. This gives a natural way to test whether a sequence of vectors approaches a proposed limit: examine whether each coordinate approaches the corresponding coordinate of that limit. The important point is that there are only finitely many coordinates, so the separate estimates can be made to hold from one common index onward. We will show that, in \(\mathbb{R}^n\), this coordinatewise test is equivalent to convergence in the Euclidean norm used in “Convergence in \(\mathbb{R}^n\).”
The starting index \(K_i\) in this definition may depend on the coordinate. Coordinatewise convergence does not initially give a single index that works for every coordinate and every tolerance. In finite dimensions, however, we can take the largest of the finitely many coordinate-specific starting indices. That fact is central to the equivalence below.
The Coordinate Criterion for Convergence
Proof. First suppose that \(x^{(k)}\to a\) in the Euclidean norm. For each coordinate \(i\), \[ |x_i^{(k)}-a_i|\leq\|x^{(k)}-a\|_2, \] because the square of the Euclidean norm is the sum of the squares of all coordinate differences, including the \(i\)-th one. Given \(\varepsilon>0\), convergence in the Euclidean norm provides an index \(K\) such that \(\|x^{(k)}-a\|_2<\varepsilon\) for every \(k\geq K\). The displayed inequality then gives \(|x_i^{(k)}-a_i|<\varepsilon\) for every such \(k\). This holds for each coordinate \(i\), so every coordinate sequence converges to the corresponding coordinate of \(a\).
Conversely, suppose that \(x_i^{(k)}\to a_i\) for every \(i=1,\ldots,n\). Fix \(\varepsilon>0\). For each coordinate \(i\), convergence of that real sequence gives a positive integer \(K_i\) such that \[ |x_i^{(k)}-a_i|<\frac{\varepsilon}{\sqrt{n}} \qquad\text{whenever }k\geq K_i. \] Since there are only \(n\) indices \(K_i\), their maximum \(K=\max\{K_1,\ldots,K_n\}\) is a positive integer. If \(k\geq K\), then all \(n\) coordinate estimates hold at once. Consequently, \[ \|x^{(k)}-a\|_2 =\left(\sum_{i=1}^{n}|x_i^{(k)}-a_i|^2\right)^{1/2} <\left(\sum_{i=1}^{n}\frac{\varepsilon^2}{n}\right)^{1/2} =\varepsilon. \] Thus \(x^{(k)}\to a\) in the Euclidean norm. Both implications are proved. \(\square\)
This theorem turns vector convergence into a finite collection of familiar real convergence problems. It also explains why the same vector limit must appear in the coordinate test: the \(i\)-th coordinate has to approach \(a_i\). The uniqueness of limits in \(\mathbb{R}^n\), noted in the previous tutorial, ensures there cannot be a second, different vector limit.
Worked Example: Verifying a Limit by Coordinates
Consider \(x^{(k)}=(4+2/k,\,-1-1/k,\,6+3/k)\) in \(\mathbb{R}^3\), and let \(a=(4,-1,6)\). The coordinate differences are \[ x_1^{(k)}-a_1=\frac{2}{k},\qquad x_2^{(k)}-a_2=-\frac{1}{k},\qquad x_3^{(k)}-a_3=\frac{3}{k}. \] Each tends to zero, so each coordinate sequence converges to the corresponding coordinate of \(a\). More explicitly, given \(\varepsilon>0\), choose a positive integer \(K>3/\varepsilon\). For every \(k\geq K\), \[ \left|\frac{2}{k}\right|\leq\frac{2}{K}<\varepsilon,\qquad \left|-\frac{1}{k}\right|\leq\frac{1}{K}<\varepsilon,\qquad \left|\frac{3}{k}\right|\leq\frac{3}{K}<\varepsilon. \] All three coordinates therefore converge to those of \(a\). By the Coordinate Criterion for Convergence, \(x^{(k)}\to(4,-1,6)\) in the Euclidean norm. The common index \(K\) here works for all three coordinates; in general, taking the maximum of the coordinate-specific indices provides such a common choice.
Using Coordinates to Rule Out Convergence
The criterion is useful in both directions. To establish convergence, it is enough to prove convergence in every coordinate. To rule out convergence, it is enough to find one coordinate that does not converge. There is no need to estimate the full Euclidean distance if a single real coordinate already prevents a limit.
Worked Example: One Nonconvergent Coordinate Prevents a Vector Limit
Let \(y^{(k)}=(1/k,\,(-1)^k)\) in \(\mathbb{R}^2\). The first coordinate converges to \(0\), but the second coordinate does not converge. Indeed, its even-indexed terms equal \(1\), while its odd-indexed terms equal \(-1\). If the second coordinate had a real limit \(L\), both of these subsequences would have to converge to \(L\). The constant even subsequence has limit \(1\), and the constant odd subsequence has limit \(-1\). These limits are different, contradicting uniqueness of limits in \(\mathbb{R}\). Thus the second coordinate does not converge, and the Coordinate Criterion for Convergence shows that \(y^{(k)}\) does not converge in \(\mathbb{R}^2\).
A frequent mistake is to check only some of the coordinates. In \(\mathbb{R}^n\), every one of the \(n\) coordinate sequences must converge to the corresponding coordinate of the proposed vector limit. A single coordinate that oscillates or becomes unbounded is enough to disprove convergence of the vector sequence.
Worked Example: Different Coordinate Rates Still Give Convergence
Define \(z^{(k)}=(2+1/k,\,-5+1/\sqrt{k},\,3+(-1)^k/k^2)\) in \(\mathbb{R}^3\), and propose \(b=(2,-5,3)\). The coordinate differences are \(1/k\), \(1/\sqrt{k}\), and \((-1)^k/k^2\). Their absolute values are respectively \(1/k\), \(1/\sqrt{k}\), and \(1/k^2\), and each tends to zero. Thus all three coordinates converge to the corresponding coordinates of \(b\), even though they do so at different rates and one difference changes sign. The Coordinate Criterion for Convergence gives \(z^{(k)}\to b\).
One can also make the common-index step explicit. Given \(\varepsilon>0\), choose a positive integer \(K\) larger than each of \(1/\varepsilon\), \(1/\varepsilon^2\), and \(1/\sqrt{\varepsilon}\). For \(k\geq K\), we have \(1/k<\varepsilon\), \(1/\sqrt{k}<\varepsilon\), and \(1/k^2<\varepsilon\). Hence all three coordinate differences are less than \(\varepsilon\) in absolute value from that index onward. The theorem then supplies Euclidean convergence.
Fixed Linear Transformations Preserve Coordinatewise Limits
Coordinates are also useful when a sequence is transformed by a fixed matrix. Each output coordinate is a finite linear combination of input coordinates. If the input coordinates converge, these finite combinations converge to the corresponding combinations of their limits. The next result states this precisely and includes the case of rows whose entries are all zero.
Proof. Write \(x^{(k)}=(x_1^{(k)},\ldots,x_n^{(k)})\) and \(a=(a_1,\ldots,a_n)\). For an output coordinate \(i\), matrix multiplication gives \[ (Ax^{(k)})_i-(Aa)_i =\sum_{j=1}^{n}a_{ij}(x_j^{(k)}-a_j). \] Set \(M=\max_{1\leq i\leq m}\sum_{j=1}^{n}|a_{ij}|\). If \(M=0\), every row has sum of absolute values zero, so every matrix entry is zero. Then \(Ax^{(k)}=0=Aa\) for every \(k\), and the conclusion holds.
Now suppose \(M>0\), and fix \(\varepsilon>0\). Since each input coordinate converges, for every \(j\) there is an index \(K_j\) such that \[ |x_j^{(k)}-a_j|<\frac{\varepsilon}{2M} \qquad\text{whenever }k\geq K_j. \] Let \(K=\max\{K_1,\ldots,K_n\}\). For \(k\geq K\), the triangle inequality gives, for every output coordinate \(i\), \[ |(Ax^{(k)})_i-(Aa)_i| \leq\sum_{j=1}^{n}|a_{ij}|\,|x_j^{(k)}-a_j| \leq\frac{\varepsilon}{2M}\sum_{j=1}^{n}|a_{ij}| \leq\frac{\varepsilon}{2}<\varepsilon. \] These inequalities also apply when row \(i\) is zero: in that case the output-coordinate error is directly \(0\), and the displayed upper bounds remain valid. Thus every output coordinate converges to the corresponding coordinate of \(Aa\), proving the theorem. \(\square\)
Worked Example: Computing the Limit After a Matrix Transformation
Let \(x^{(k)}=(1+1/k,\,-2+2/k,\,3-1/k)\), so that \(x^{(k)}\to a=(1,-2,3)\) coordinatewise. Take \[ A=\begin{pmatrix}1&2&-1\\0&-1&3\end{pmatrix}. \] The proposed transformed limit is \[ Aa= \begin{pmatrix}1&2&-1\\0&-1&3\end{pmatrix} \begin{pmatrix}1\\-2\\3\end{pmatrix} = \begin{pmatrix}1-4-3\\0+2+9\end{pmatrix} = \begin{pmatrix}-6\\11\end{pmatrix}. \] For the sequence itself, direct calculation gives \[ Ax^{(k)} = \begin{pmatrix} (1+1/k)+2(-2+2/k)-(3-1/k)\\ -(-2+2/k)+3(3-1/k) \end{pmatrix} = \begin{pmatrix}-6+6/k\\11-5/k\end{pmatrix}. \] The first output coordinate tends to \(-6\), and the second tends to \(11\). Hence \(Ax^{(k)}\to(-6,11)=Aa\), as the theorem predicts.
Why the Finite-Dimensional Hypothesis Matters
The proof of the Coordinate Criterion for Convergence uses the finiteness of the coordinate set when it takes the maximum of \(K_1,\ldots,K_n\). For any fixed finite \(n\), this maximum exists and is a valid common starting index. With infinitely many coordinates, coordinatewise convergence need not provide a single index that controls all coordinates in a norm measuring the largest coordinate error.
For example, consider sequences with infinitely many real coordinates and let \(u^{(k)}\) have value \(1\) in coordinate \(k\) and value \(0\) in every other coordinate. For each fixed coordinate \(i\), the \(i\)-th coordinate of \(u^{(k)}\) is eventually zero: it is zero whenever \(k\neq i\). Thus every fixed coordinate converges to zero. Yet the largest absolute coordinate of \(u^{(k)}\) is \(1\) for every \(k\), so the sequence does not approach the zero sequence in the supremum norm. This does not contradict our theorem; it illustrates why the finite-dimensional setting must not be dropped from its statement.
In \(\mathbb{R}^n\), no such gap occurs. The coordinate criterion is an exact test for Euclidean convergence: every coordinate must approach its proposed limit, and the finitely many resulting estimates can be synchronized by choosing one sufficiently large index. When applying the test, identify the candidate limit, verify every coordinate, and use finiteness to obtain a common tail.
Check Your Understanding
Use the definition and the results proved here to answer each question.
- What does it mean for a sequence in \(\mathbb{R}^n\) to converge coordinatewise to \(a\)?
- In the proof from coordinatewise convergence to Euclidean convergence, why can one choose a single index \(K\) that works for all coordinates?
- If a vector sequence has one coordinate that fails to converge, what can you conclude about convergence of the vector sequence?
- Why does Euclidean convergence imply convergence of each coordinate sequence?
- For a fixed matrix \(A\), if \(x^{(k)}\to a\) coordinatewise, what is the coordinatewise limit of \(Ax^{(k)}\)?
- Which step in the coordinate criterion proof depends on there being only finitely many coordinates?