Norms Induced by Inner Products

The induced norm is the working notion of "size" throughout mathematics and its applications. In numerical linear algebra, the Euclidean norm $\|x\| = \sqrt{x^Tx}$ (or $\sqrt{x^*x}$ for complex vectors) measures residual errors in solving linear systems and underlies the QR and least-squares algorithms. In statistics and machine learning, the induced norm on a space of random variables with inner product $\langle X,Y\rangle = \mathbb{E}[XY]$ gives the root-mean-square error, and orthogonality of random variables (uncorrelatedness after centering) is exactly the Pythagorean identity used in variance decompositions such as ANOVA. In quantum mechanics, state vectors live in a complex Hilbert space, and $\|\psi\| = \sqrt{\langle \psi,\psi\rangle}$ normalizing to $1$ encodes the requirement that total probability sums to one; orthogonal states correspond to perfectly distinguishable outcomes, and the Pythagorean identity underlies the additivity of probabilities for orthogonal measurement outcomes. In signal processing, the norm induced by the inner product $\langle f,g\rangle = \int f\bar g$ measures the energy of a signal, and Parseval's theorem — a Pythagorean identity for infinitely many mutually orthogonal Fourier modes — expresses total signal energy as the sum of energies in each frequency component. In computer graphics and robotics, induced norms on $\mathbb{R}^3$ measure distances and are combined with orthogonal projections for collision detection and control.

Ask the tutor: How does the Cauchy–Schwarz inequality follow from these same inner product axioms, and how does it lead to the triangle inequality?

Think of the inner product $\langle u, v \rangle$ as encoding both the sizes of $u$ and $v$ and the angle between them, all at once — in $\mathbb{R}^n$ with the standard dot product, $\langle u,v\rangle = \|u\|\|v\|\cos\theta$. Setting $v = u$ removes the angle from the picture (since a vector makes angle $0$ with itself, so $\cos\theta = 1$) and leaves exactly the squared length, $\langle u, u\rangle = \|u\|^2$. Taking a square root then recovers the length itself. This is why $\|v\| = \sqrt{\langle v,v\rangle}$ is the natural and essentially forced definition: it is the one operation that converts the abstract pairing $\langle \cdot,\cdot\rangle$ into an honest measure of magnitude, agreeing with ordinary Euclidean length whenever $V = \mathbb{R}^n$ with the dot product. The Pythagorean identity is the algebraic shadow of a picture you already know from plane geometry: a right triangle with legs $u$ and $v$ has hypotenuse $u+v$, and the sum of squares of the legs equals the square of the hypotenuse exactly when the legs are perpendicular. In an abstract inner product space, "perpendicular" is defined by the vanishing of the inner product, and the identity $\|u+v\|^2 = \|u\|^2+\|v\|^2$ becomes a theorem provable directly from the axioms, valid for polynomials, matrices, or any other object living in an inner product space — not just arrows in the plane.

Example 1: The norm on $\mathbb{C}^3$ with the standard inner product. Let $V = \mathbb{C}^3$ with the standard inner product $\langle x, y \rangle = x_1\overline{y_1} + x_2 \overline{y_2} + x_3\overline{y_3}$. Let

First compute $\|u\|$. We have

so $\|u\| = \sqrt{15}$. Now check orthogonality of $u$ and $v$:

Compute each term: $\overline{1-i} = 1+i$, so $(2+i)(1+i) = 2 + 2i + i + i^2 = 2 + 3i - 1 = 1 + 3i$. Next, $\overline{2i} = -2i$, so $(-1)(-2i) = 2i$. The last term is $0$. Summing,

so $u$ and $v$ are not orthogonal — the Pythagorean identity does not apply directly to this pair. Let's instead illustrate homogeneity and positivity concretely. Take $\lambda = 3 - 4i$, so $|\lambda| = \sqrt{9+16} = 5$. Then $\lambda u = (3-4i)(2+i, -1, 3i)$. Compute each coordinate:

So $\lambda u = (10-5i,\ -3+4i,\ 12+9i)$. Compute $\|\lambda u\|^2$ directly:

Sum: $125+25+225 = 375$, so $\|\lambda u\| = \sqrt{375}$. Now check homogeneity: $|\lambda|\|u\| = 5\sqrt{15} = \sqrt{25 \cdot 15} = \sqrt{375}$, matching exactly. This confirms $\|\lambda u\| = |\lambda|\|u\|$ as guaranteed by part (b) of the theorem, and since $u \neq 0$ we also confirm $\|u\| = \sqrt{15} > 0$, consistent with part (a).

Example 2: An orthogonal pair in the polynomial inner product space $\mathcal{P}_2(\mathbb{R})$ with $\langle f, g\rangle = \int_{-1}^1 f(x)g(x)\,dx$. Let $V = \mathcal{P}_2(\mathbb{R})$, the real polynomials of degree at most $2$, with $\langle f,g\rangle = \int_{-1}^1 f(x)g(x)\,dx$. This is an inner product: it is symmetric and bilinear because integration is linear, and $\langle f,f\rangle = \int_{-1}^1 f(x)^2\,dx \geq 0$, vanishing only when $f \equiv 0$ on $[-1,1]$ (a continuous function whose square integrates to zero must be identically zero, since $f^2 \geq 0$ everywhere). Take $f(x) = x$ and $g(x) = 3x^2 - 1$. Check orthogonality:

Both $3x^3$ and $x$ are odd functions, and the integral of an odd function over a symmetric interval $[-1,1]$ is zero. Hence $\langle f,g\rangle = 0$, so $f \perp g$. Compute the norms. First,

so $\|f\| = \sqrt{2/3}$. Next,

Compute term by term: $\int_{-1}^1 9x^4\,dx = 9\cdot\frac{2}{5} = \frac{18}{5}$; $\int_{-1}^1 6x^2\,dx = 6\cdot\frac23 = 4$; $\int_{-1}^1 1\,dx = 2$. So

Now form $h = f+g = 3x^2 + x - 1$ and verify the Pythagorean identity directly, as a check on part (c):

Expand: $(3x^2+x-1)^2 = 9x^4 + x^2 + 1 + 6x^3 - 6x^2 - 2x = 9x^4 + 6x^3 -5x^2 -2x + 1$. Integrate term by term over $[-1,1]$, discarding odd-power terms ($6x^3$ and $-2x$ integrate to $0$):

Common denominator $15$: $\frac{18}{5} = \frac{54}{15}$, $\frac{10}{3} = \frac{50}{15}$, $2 = \frac{30}{15}$. So

Compare with $\|f\|^2 + \|g\|^2 = \frac23 + \frac85 = \frac{10}{15} + \frac{24}{15} = \frac{34}{15}$. The two computations agree exactly, confirming the Pythagorean identity $\|f+g\|^2 = \|f\|^2 + \|g\|^2$ for this orthogonal pair.