Comparing Lengths Without Computing Them Exactly
Suppose two vectors are close in norm: their difference has a small norm. It is then natural to ask how much their lengths can differ. The triangle inequality answers one direction immediately: the length of one vector cannot exceed the length of the other by more than the norm of their difference. Reversing the roles of the vectors gives a bound on the difference in the other direction as well. Together these statements form the reverse triangle inequality.
We work over \(\mathbb{R}\) or \(\mathbb{C}\). A norm on a vector space \(V\) is a function assigning a real number \(\|\mathbf{x}\|\) to each \(\mathbf{x}\in V\), with \(\|\mathbf{x}\|\ge 0\), \(\|\mathbf{x}\|=0\) exactly when \(\mathbf{x}=\mathbf{0}\), \(\|c\mathbf{x}\|=|c|\|\mathbf{x}\|\), and \(\|\mathbf{x}+\mathbf{y}\|\le \|\mathbf{x}\|+\|\mathbf{y}\|\). A vector space equipped with a norm is called a normed vector space. In particular, the norms induced by inner products satisfy these properties, including the triangle inequality established in the earlier material.
The reverse triangle inequality compares three quantities: \(\|\mathbf{x}\|\), \(\|\mathbf{y}\|\), and \(\|\mathbf{x}-\mathbf{y}\|\). It does not say that the first two quantities are equal, or that their difference equals the third. Its conclusion is a bound, valid for every pair of vectors, including when either vector is zero.
The Reverse Triangle Inequality
Proof. Write \(\mathbf{x}=(\mathbf{x}-\mathbf{y})+\mathbf{y}\). The triangle inequality gives \(\|\mathbf{x}\|\le\|\mathbf{x}-\mathbf{y}\|+\|\mathbf{y}\|\), so \(\|\mathbf{x}\|-\|\mathbf{y}\|\le\|\mathbf{x}-\mathbf{y}\|\). Now exchange \(\mathbf{x}\) and \(\mathbf{y}\). Then \(\|\mathbf{y}\|-\|\mathbf{x}\|\le\|\mathbf{y}-\mathbf{x}\|\). By homogeneity, \(\|\mathbf{y}-\mathbf{x}\|=\|-(\mathbf{x}-\mathbf{y})\|=|-1|\|\mathbf{x}-\mathbf{y}\|=\|\mathbf{x}-\mathbf{y}\|\). Thus each of the two real numbers \(\|\mathbf{x}\|-\|\mathbf{y}\|\) and \(\|\mathbf{y}\|-\|\mathbf{x}\|\) is at most \(\|\mathbf{x}-\mathbf{y}\|\). Their maximum is \(\bigl|\|\mathbf{x}\|-\|\mathbf{y}\|\bigr|\), which proves the result. No division or assumption that either vector is nonzero was used. \(\square\)
The two one-sided inequalities are useful in different situations. If \(\|\mathbf{x}\|\) is larger, the first says how much larger it can be; if \(\|\mathbf{y}\|\) is larger, the second gives the corresponding bound. The absolute value packages both possibilities into one statement. The theorem applies to inner-product norms, but its proof only uses the norm axioms, so it applies to any normed vector space.
Worked Example: Comparing Two Lengths in \(\mathbb{R}^2\)
Let \(\mathbf{x}=(3,4)^{\top}\) and \(\mathbf{y}=(1,2)^{\top}\), with the standard Euclidean norm. Compute the difference of their lengths and compare it with the norm of their difference.
First calculate the two lengths:
Since \(5>\sqrt{5}\), the absolute difference of the lengths is \(5-\sqrt{5}\). The vector difference and its norm are
To check the claimed bound exactly, both sides are nonnegative, so they may be squared. We have \((5-\sqrt{5})^2=30-10\sqrt{5}\), and \(30-10\sqrt{5}\le 8\) is equivalent to \(11/5\le\sqrt{5}\). Both sides of this last inequality are positive, and squaring gives \(121/25\le 125/25\). Therefore \(\bigl|\|\mathbf{x}\|-\|\mathbf{y}\|\bigr|\le\|\mathbf{x}-\mathbf{y}\|\), as required.
Check: the directly computed values are \(5-\sqrt{5}\) and \(2\sqrt{2}\), and the exact comparison above confirms that the length difference is no larger than the norm of the vector difference.
Using a Bound on a Perturbation
The theorem is especially useful when a vector is changed by a known, small amount. If \(\mathbf{x}\) is obtained from \(\mathbf{y}\) by adding a perturbation \(\mathbf{e}\), then \(\mathbf{x}-\mathbf{y}=\mathbf{e}\). The reverse triangle inequality bounds the change in length directly by \(\|\mathbf{e}\|\).
Proof. The reverse triangle inequality gives \(\bigl|\|\mathbf{x}\|-\|\mathbf{y}\|\bigr|\le\|\mathbf{x}-\mathbf{y}\|\le\delta\). For any real number \(r\), \(|r|\le\delta\) implies \(-\delta\le r\le\delta\). Apply this with \(r=\|\mathbf{x}\|-\|\mathbf{y}\|\). The resulting two inequalities are \(-\delta\le\|\mathbf{x}\|-\|\mathbf{y}\|\le\delta\). Adding \(\|\mathbf{y}\|\) throughout gives the stated bounds. The lower bound remains valid even if \(\|\mathbf{y}\|-\delta\) is negative; nonnegativity of the norm then makes that lower bound automatic. \(\square\)
Worked Example: Bounding a Length After a Small Change
Let \(\mathbf{y}=(3,4)^{\top}\), and suppose \(\mathbf{x}=\mathbf{y}+\mathbf{e}\), where \(\mathbf{e}=(1/3,0)^{\top}\). Use the perturbation bound with \(\delta=1/2\), then check the actual length of \(\mathbf{x}\).
The reference length is \(\|\mathbf{y}\|=\sqrt{3^2+4^2}=5\), and \(\|\mathbf{e}\|=\sqrt{(1/3)^2+0^2}=1/3\le1/2\). Because \(\mathbf{x}-\mathbf{y}=\mathbf{e}\), the corollary gives
Here \(\mathbf{x}=(3+1/3,4)^{\top}=(10/3,4)^{\top}\), so its actual length is \[ \|\mathbf{x}\| =\sqrt{\left(\frac{10}{3}\right)^2+4^2} =\sqrt{\frac{100}{9}+\frac{144}{9}} =\frac{\sqrt{244}}{3} =\frac{2\sqrt{61}}{3}. \] To check it lies in the stated interval, the lower comparison \(2\sqrt{61}/3\ge9/2\) is equivalent to \(4\sqrt{61}\ge27\); squaring positive sides gives \(976\ge729\). The upper comparison \(2\sqrt{61}/3\le11/2\) is equivalent to \(4\sqrt{61}\le33\); squaring gives \(976\le1089\).
Check: the calculated length \(2\sqrt{61}/3\) lies between \(9/2\) and \(11/2\), in agreement with the bound obtained from \(\|\mathbf{e}\|\le1/2\).
The Same Bound in Complex and Matrix Spaces
The theorem does not depend on coordinates being real. In a complex inner-product space, norms are still nonnegative real numbers, and the absolute value on the left is the ordinary absolute value of a real number. It also applies to matrix spaces when they are equipped with an induced norm such as the Frobenius norm. In each case, the essential input remains the triangle inequality for that norm.
Worked Example: Complex Vectors
In \(\mathbb{C}^2\) with the standard inner-product norm, let \(\mathbf{x}=(1+i,1)^{\top}\) and \(\mathbf{y}=(1,i)^{\top}\). Compare their lengths with the norm of their difference.
Using \(|1+i|^2=1^2+1^2=2\) and \(|i|^2=1\), we obtain
Since \(\sqrt{3}>\sqrt{2}\), the length difference is \(\sqrt{3}-\sqrt{2}\). The vector difference is \(\mathbf{x}-\mathbf{y}=(i,1-i)^{\top}\). Since \(|i|^2=1\) and \(|1-i|^2=1^2+(-1)^2=2\), its norm is \(\|\mathbf{x}-\mathbf{y}\|=\sqrt{1+2}=\sqrt{3}\).
Check: \(0<\sqrt{3}-\sqrt{2}<\sqrt{3}\), because \(0<\sqrt{2}\). Thus the computed values satisfy the reverse triangle inequality.
Worked Example: Frobenius Norms of Matrices
For a matrix \(\mathbf{M}=(m_{ij})\), its Frobenius norm is \(\|\mathbf{M}\|_F=\sqrt{\sum_{i,j}|m_{ij}|^2}\). Take \(\mathbf{A}=\begin{bmatrix}1&2\\2&0\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\). Compare the difference of their Frobenius norms with the Frobenius norm of their difference.
Summing the squares of the entries gives \(\|\mathbf{A}\|_F=\sqrt{1+4+4+0}=3\) and \(\|\mathbf{B}\|_F=\sqrt{0+1+0+0}=1\). Hence the absolute difference of the norms is \(|3-1|=2\). Subtracting the matrices entry by entry gives
Check: \(2\le\sqrt{6}\), since both sides are nonnegative and \(2^2=4\le6=(\sqrt{6})^2\). The matrix calculation therefore satisfies the reverse triangle inequality.
Common Pitfall: Replacing a Bound with an Equality
A frequent mistake is to treat the reverse triangle inequality as an equality. For the vectors \(\mathbf{x}=(3,4)^{\top}\) and \(\mathbf{y}=(1,2)^{\top}\) from the first example, an incorrect calculation would be \(\|\mathbf{x}-\mathbf{y}\|=\|\mathbf{x}\|-\|\mathbf{y}\|=5-\sqrt{5}\). But direct subtraction gives \(\mathbf{x}-\mathbf{y}=(2,2)^{\top}\), and therefore \(\|\mathbf{x}-\mathbf{y}\|=2\sqrt{2}\), not \(5-\sqrt{5}\). The correct statement is the inequality \(5-\sqrt{5}\le2\sqrt{2}\), verified by squaring in that example. Equality may occur for some vector pairs, but the theorem guarantees only a bound, not equality for every pair.
Another error is to omit the absolute value and conclude that \(\|\mathbf{x}\|-\|\mathbf{y}\|\) is always nonnegative. The lengths have no fixed order: swapping \(\mathbf{x}\) and \(\mathbf{y}\) changes the sign of their difference, while \(\|\mathbf{x}-\mathbf{y}\|\) stays the same. Keeping the absolute value, or explicitly checking both one-sided inequalities, handles either order.
Why the Inequality Is Useful
The reverse triangle inequality gives a stability statement: changing a vector by a small norm amount can change its length by no more than that amount. Sometimes the exact length after a change is difficult to calculate, or the change is known only through an upper bound. The perturbation corollary still gives a valid interval for the new length. This makes the result useful in estimates involving induced norms of vectors, functions, and matrices, without requiring a special coordinate formula for each space.
Check Your Understanding
For each question, state the relevant inequality and show enough calculation to justify the conclusion.
- Let \(\mathbf{x}=(0,5)^{\top}\) and \(\mathbf{y}=(3,1)^{\top}\) in \(\mathbb{R}^2\). Compute both sides of the reverse triangle inequality.
- If \(\|\mathbf{x}-\mathbf{y}\|\le 2\) and \(\|\mathbf{y}\|=7\), what interval must contain \(\|\mathbf{x}\|\)?
- In \(\mathbb{C}^2\), let \(\mathbf{x}=(i,0)^{\top}\) and \(\mathbf{y}=(0,1+i)^{\top}\). Compute \(\bigl|\|\mathbf{x}\|-\|\mathbf{y}\|\bigr|\) and \(\|\mathbf{x}-\mathbf{y}\|\).
- Explain why the triangle inequality applied to \(\mathbf{x}=(\mathbf{x}-\mathbf{y})+\mathbf{y}\) gives one one-sided bound, and identify the swap that gives the other.
- Give a reason the reverse triangle inequality does not generally imply \(\|\mathbf{x}-\mathbf{y}\|=\bigl|\|\mathbf{x}\|-\|\mathbf{y}\|\bigr|\).
