The Central Idea
In ordinary Euclidean space, the dot product of two vectors
\[ u= \begin{bmatrix} u_1\\ \vdots\\ u_n \end{bmatrix}, \qquad v= \begin{bmatrix} v_1\\ \vdots\\ v_n \end{bmatrix} \]is
\[ \boxed{ u\cdot v = u_1v_1+\cdots+u_nv_n}. \]This single operation lets us define length, angle, orthogonality, projection, and distance.
What Is an Inner Product?
An inner product is a rule
\[ \langle u,v\rangle \]that assigns a scalar to a pair of vectors while satisfying certain structural properties.
In \(\mathbb{R}^n\), the standard inner product is the dot product:
\[ \boxed{ \langle u,v\rangle = u^Tv = u_1v_1+\cdots+u_nv_n}. \]Properties of the Standard Inner Product
For vectors \(u,v,w\) and scalar \(c\),
\[ \langle u,v\rangle=\langle v,u\rangle, \] \[ \langle cu,v\rangle=c\langle u,v\rangle, \] \[ \langle u+v,w\rangle = \langle u,w\rangle+\langle v,w\rangle, \]and
\[ \boxed{ \langle v,v\rangle\ge0} \]with equality only when \(v=0\).
Example: Compute a Dot Product
Let
\[ u= \begin{bmatrix} 2\\ -1\\ 3 \end{bmatrix}, \qquad v= \begin{bmatrix} 4\\ 5\\ -2 \end{bmatrix}. \]Then
\[ u\cdot v = 2(4)+(-1)(5)+3(-2). \]Therefore,
\[ \boxed{u\cdot v=-3}. \]Norm and Length
The norm induced by the standard inner product is
\[ \boxed{ \|v\| = \sqrt{\langle v,v\rangle}}. \]In \(\mathbb{R}^n\),
\[ \|v\| = \sqrt{v_1^2+\cdots+v_n^2}. \]This generalizes the familiar distance formula.
Example: Compute a Norm
Let
\[ v= \begin{bmatrix} 3\\ 4 \end{bmatrix}. \]Then
\[ \|v\| = \sqrt{3^2+4^2} = \boxed{5}. \]Unit Vectors
A vector \(u\) is a unit vector when
\[ \boxed{\|u\|=1}. \]If \(v\neq0\), then
\[ \boxed{ \frac{v}{\|v\|}} \]is a unit vector pointing in the same direction as \(v\).
This process is called normalization.
Example: Normalize a Vector
For
\[ v= \begin{bmatrix} 3\\ 4 \end{bmatrix}, \]we found
\[ \|v\|=5. \]Therefore the corresponding unit vector is
\[ \boxed{ \frac15 \begin{bmatrix} 3\\ 4 \end{bmatrix} = \begin{bmatrix} 3/5\\ 4/5 \end{bmatrix}}. \]Distance Between Vectors
The distance between vectors \(u\) and \(v\) is
\[ \boxed{d(u,v)=\|u-v\|}. \]This is the length of the displacement from one vector endpoint to the other.
Angles Between Vectors
For nonzero vectors \(u\) and \(v\), the angle \(\theta\) between them satisfies
\[ \boxed{ \cos\theta = \frac{u\cdot v}{\|u\|\|v\|}}. \]Therefore,
\[ \theta = \arccos \left( \frac{u\cdot v}{\|u\|\|v\|} \right). \]Sign of the Dot Product
The sign of
\[ u\cdot v \]gives a quick angular interpretation:
- positive: the angle is acute;
- zero: the vectors are perpendicular;
- negative: the angle is obtuse.
Orthogonal Vectors
Two vectors are orthogonal when their inner product is zero:
\[ \boxed{\langle u,v\rangle=0}. \]In Euclidean space, this means the vectors meet at a right angle.
Example: Test Orthogonality
Let
\[ u= \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad v= \begin{bmatrix} 2\\ -1 \end{bmatrix}. \]Then
\[ u\cdot v = 1(2)+2(-1) = 0. \]Therefore,
\[ \boxed{u\perp v}. \]Orthogonal Sets
A set of nonzero vectors
\[ \{v_1,\ldots,v_k\} \]is orthogonal if
\[ \boxed{ \langle v_i,v_j\rangle=0 \quad\text{whenever }i\neq j}. \]Every distinct pair must be orthogonal.
Orthogonal Nonzero Vectors Are Independent
Any orthogonal set of nonzero vectors is linearly independent.
Suppose
\[ c_1v_1+\cdots+c_kv_k=0. \]Take the inner product with \(v_j\):
\[ c_j\langle v_j,v_j\rangle=0. \]Since \(v_j\neq0\),
\[ \langle v_j,v_j\rangle>0, \]so \(c_j=0\).
This holds for every \(j\).
Orthonormal Sets
A set is orthonormal if its vectors are mutually orthogonal and each has norm \(1\).
Thus
\[ \boxed{ \langle q_i,q_j\rangle = \begin{cases} 1,&i=j,\\ 0,&i\neq j. \end{cases}} \]Orthonormal Bases
An orthonormal basis is a basis whose vectors form an orthonormal set.
The standard basis
\[ e_1,\ldots,e_n \]of \(\mathbb{R}^n\) is orthonormal.
Orthonormal bases make coordinate calculations especially simple.
Coordinates in an Orthonormal Basis
Suppose
\[ B=(q_1,\ldots,q_n) \]is an orthonormal basis.
Every vector \(v\) can be written
\[ v=c_1q_1+\cdots+c_nq_n. \]The coefficients are simply
\[ \boxed{c_i=\langle v,q_i\rangle}. \]Orthonormal Expansion Formula
Therefore,
\[ \boxed{ v = \langle v,q_1\rangle q_1 +\cdots+ \langle v,q_n\rangle q_n}. \]No system of equations is required to recover the coordinates.
Projection onto a Vector
The projection of \(v\) onto a nonzero vector \(u\) is
\[ \boxed{ \operatorname{proj}_u(v) = \frac{\langle v,u\rangle}{\langle u,u\rangle}u}. \]This gives the component of \(v\) pointing along \(u\).
Projection onto a Unit Vector
If \(\|u\|=1\), then
\[ \langle u,u\rangle=1. \]Therefore,
\[ \boxed{ \operatorname{proj}_u(v) = \langle v,u\rangle u}. \]Example: Projection onto a Vector
Let
\[ v= \begin{bmatrix} 3\\ 4 \end{bmatrix}, \qquad u= \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]Then
\[ v\cdot u=3, \qquad u\cdot u=1. \]Hence
\[ \boxed{ \operatorname{proj}_u(v) = \begin{bmatrix} 3\\ 0 \end{bmatrix}}. \]Orthogonal Decomposition
A vector \(v\) can be decomposed into a part parallel to \(u\) and a part orthogonal to \(u\):
\[ \boxed{ v = \operatorname{proj}_u(v) + \left( v-\operatorname{proj}_u(v) \right)}. \]The second term is orthogonal to \(u\).
Projection onto a Subspace
Suppose
\[ W=\operatorname{span}\{q_1,\ldots,q_k\} \]where the \(q_i\) form an orthonormal basis for \(W\).
Then
\[ \boxed{ \operatorname{proj}_W(v) = \sum_{i=1}^k \langle v,q_i\rangle q_i}. \]The Orthogonal Error
If
\[ p=\operatorname{proj}_W(v), \]then the error vector
\[ e=v-p \]is orthogonal to every vector in \(W\):
\[ \boxed{e\in W^\perp}. \]Orthogonal Complement
The orthogonal complement of a subspace \(W\subseteq \mathbb{R}^n\) is
\[ \boxed{ W^\perp = \{v\in\mathbb{R}^n: \langle v,w\rangle=0 \text{ for every }w\in W\}}. \]It contains all vectors perpendicular to the entire subspace.
Orthogonal Complements Are Subspaces
If \(u,v\in W^\perp\), then for every \(w\in W\),
\[ \langle au+bv,w\rangle = a\langle u,w\rangle+b\langle v,w\rangle = 0. \]Therefore,
\[ \boxed{W^\perp\text{ is a subspace}}. \]Dimension of an Orthogonal Complement
If \(W\subseteq\mathbb{R}^n\), then
\[ \boxed{ \dim(W)+\dim(W^\perp)=n}. \]The subspace and its orthogonal complement together account for all independent directions in the ambient space.
Orthogonal Direct-Sum Decomposition
Every vector \(v\in\mathbb{R}^n\) can be written uniquely as
\[ \boxed{v=w+z} \]where
\[ w\in W, \qquad z\in W^\perp. \]This is written
\[ \boxed{\mathbb{R}^n=W\oplus W^\perp}. \]Row Space and Null Space
For a matrix \(A\), the null space is orthogonal to the row space:
\[ \boxed{ \operatorname{Null}(A) = \operatorname{Row}(A)^\perp}. \]This follows because
\[ Ax=0 \]means that \(x\) has zero dot product with every row of \(A\).
Column Space and Left Null Space
The orthogonal complement of the column space is the null space of \(A^T\):
\[ \boxed{ \operatorname{Col}(A)^\perp = \operatorname{Null}(A^T)}. \]This relationship becomes central in least-squares theory.
The Four Fundamental Subspaces
A matrix naturally produces four subspaces:
- column space;
- null space;
- row space;
- left null space.
Orthogonality pairs them:
\[ \operatorname{Row}(A)^\perp = \operatorname{Null}(A), \] \[ \operatorname{Col}(A)^\perp = \operatorname{Null}(A^T). \]The Pythagorean Theorem for Vectors
If
\[ u\perp v, \]then
\[ \boxed{ \|u+v\|^2 = \|u\|^2+\|v\|^2}. \]This is the vector-space form of the Pythagorean theorem.
Why the Pythagorean Identity Works
Expand:
\[ \|u+v\|^2 = \langle u+v,u+v\rangle. \]This becomes
\[ \langle u,u\rangle + 2\langle u,v\rangle + \langle v,v\rangle. \]Since \(u\perp v\),
\[ \langle u,v\rangle=0. \]Therefore,
\[ \|u+v\|^2 = \|u\|^2+\|v\|^2. \]Cauchy-Schwarz Inequality
For any vectors \(u,v\),
\[ \boxed{ |\langle u,v\rangle| \le \|u\|\|v\|}. \]Equality occurs exactly when the vectors are linearly dependent.
This inequality guarantees that the cosine formula for angles produces a value between \(-1\) and \(1\).
Triangle Inequality
Norms satisfy
\[ \boxed{ \|u+v\| \le \|u\|+\|v\|}. \]Geometrically, traveling directly from one point to another is no longer than traveling through an intermediate point.
Orthogonal Matrices
A square matrix \(Q\) is orthogonal if its columns form an orthonormal basis.
Equivalently,
\[ \boxed{Q^TQ=I}. \]Therefore,
\[ \boxed{Q^{-1}=Q^T}. \]Orthogonal Matrices Preserve Length
If \(Q\) is orthogonal, then
\[ \|Qx\|=\|x\|. \]Indeed,
\[ \|Qx\|^2 = (Qx)^T(Qx) = x^TQ^TQx = x^Tx. \]Orthogonal Matrices Preserve Inner Products
For orthogonal \(Q\),
\[ \boxed{ \langle Qu,Qv\rangle = \langle u,v\rangle}. \]Therefore orthogonal transformations preserve lengths and angles.
Rotations and reflections are standard examples.
Determinant of an Orthogonal Matrix
Since
\[ Q^TQ=I, \]taking determinants gives
\[ \det(Q)^2=1. \]Therefore,
\[ \boxed{\det(Q)=\pm1}. \]Gram-Schmidt Orthogonalization
The Gram-Schmidt process converts a linearly independent set into an orthogonal or orthonormal set spanning the same subspace.
Starting from
\[ v_1,\ldots,v_k, \]we remove from each new vector the components lying in directions already constructed.
First Gram-Schmidt Step
Set
\[ \boxed{u_1=v_1}. \]If an orthonormal basis is desired, normalize:
\[ q_1=\frac{u_1}{\|u_1\|}. \]Second Gram-Schmidt Step
Remove from \(v_2\) its projection onto \(u_1\):
\[ \boxed{ u_2 = v_2 - \operatorname{proj}_{u_1}(v_2)}. \]Then \(u_2\perp u_1\).
Normalize if needed:
\[ q_2=\frac{u_2}{\|u_2\|}. \]General Gram-Schmidt Step
For \(j\ge2\),
\[ \boxed{ u_j = v_j - \sum_{i=1}^{j-1} \operatorname{proj}_{u_i}(v_j)}. \]After normalization,
\[ q_j=\frac{u_j}{\|u_j\|}. \]The vectors \(q_1,\ldots,q_k\) form an orthonormal basis for the same span.
Example: Gram-Schmidt in \(\mathbb{R}^2\)
Let
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]Set
\[ u_1=v_1. \]Then
\[ \operatorname{proj}_{u_1}(v_2) = \frac{v_2\cdot u_1}{u_1\cdot u_1}u_1. \]Since
\[ v_2\cdot u_1=1 \]and
\[ u_1\cdot u_1=2, \]we get
\[ \operatorname{proj}_{u_1}(v_2) = \frac12 \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Finish the Gram-Schmidt Example
Therefore,
\[ u_2 = \begin{bmatrix} 1\\ 0 \end{bmatrix} - \frac12 \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 1/2\\ -1/2 \end{bmatrix}. \]One normalized pair is
\[ \boxed{ q_1= \frac1{\sqrt2} \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad q_2= \frac1{\sqrt2} \begin{bmatrix} 1\\ -1 \end{bmatrix}}. \]Gram-Schmidt Preserves Span
Gram-Schmidt changes the generating vectors but not the subspace they generate:
\[ \boxed{ \operatorname{span}\{v_1,\ldots,v_k\} = \operatorname{span}\{u_1,\ldots,u_k\}}. \]The result is a geometrically cleaner basis for the same space.
QR Factorization
Applying Gram-Schmidt to the independent columns of a matrix \(A\) produces an orthonormal matrix \(Q\) and an upper triangular matrix \(R\) such that
\[ \boxed{A=QR}. \]QR factorization is a major computational tool in least squares and eigenvalue algorithms.
Inner Products Beyond \(\mathbb{R}^n\)
Inner products can be defined on many vector spaces.
For continuous functions on an interval, one common example is
\[ \boxed{ \langle f,g\rangle = \int_a^b f(x)g(x)\,dx}. \]This lets us discuss orthogonal functions just as we discuss orthogonal vectors.
Orthogonal Functions
Functions \(f\) and \(g\) are orthogonal under the integral inner product when
\[ \boxed{ \int_a^b f(x)g(x)\,dx=0}. \]Orthogonal function systems appear in Fourier analysis, differential equations, approximation theory, and signal processing.
Inner Products on Matrix Spaces
For real matrices \(A\) and \(B\) of the same size, the Frobenius inner product is
\[ \boxed{ \langle A,B\rangle = \operatorname{tr}(A^TB)}. \]This equals the sum of the products of corresponding entries.
It turns matrix spaces into geometric inner-product spaces.
Weighted Inner Products
Geometry can be modified by assigning different importance to different directions.
If \(M\) is symmetric positive definite, then
\[ \boxed{ \langle u,v\rangle_M = u^TMv} \]defines an inner product.
Such weighted inner products appear in optimization, statistics, mechanics, and numerical analysis.
Orthogonality Depends on the Inner Product
Two vectors may be orthogonal under one inner product and not under another.
Therefore orthogonality is not only a property of the vectors; it also depends on the geometric structure chosen for the space.
Projection as Best Approximation
Let \(W\) be a subspace and let
\[ p=\operatorname{proj}_W(v). \]Then \(p\) is the vector in \(W\) closest to \(v\):
\[ \boxed{ \|v-p\| \le \|v-w\| \quad \text{for every }w\in W}. \]This is the central geometric idea behind least squares.
Why Orthogonality Produces the Closest Point
For any \(w\in W\),
\[ v-w = (v-p)+(p-w). \]Here
\[ v-p\in W^\perp \]while
\[ p-w\in W. \]Therefore the two pieces are orthogonal.
By the Pythagorean theorem,
\[ \|v-w\|^2 = \|v-p\|^2+\|p-w\|^2 \ge \|v-p\|^2. \]Projection Matrices
If the columns of \(Q\) form an orthonormal basis for a subspace \(W\), then the projection onto \(W\) is
\[ \boxed{P=QQ^T}. \]For any vector \(v\),
\[ \operatorname{proj}_W(v)=Pv. \]Properties of Projection Matrices
An orthogonal projection matrix satisfies
\[ \boxed{P^2=P} \]and
\[ \boxed{P^T=P}. \]Applying the projection twice has the same effect as applying it once.
Symmetric Matrices and Orthogonality
Real symmetric matrices have orthogonal eigenspaces corresponding to distinct eigenvalues.
This fact allows a real symmetric matrix to be diagonalized by an orthogonal matrix:
\[ \boxed{A=QDQ^T}. \]This is one form of the spectral theorem.
Why Orthonormal Coordinates Are Numerically Useful
Orthonormal bases avoid unnecessary stretching in coordinate conversion.
Because
\[ Q^{-1}=Q^T, \]inversion becomes inexpensive and stable.
Lengths and angles are also preserved exactly in ideal arithmetic.
Common Error: Confusing Orthogonal with Independent
Orthogonal nonzero vectors are always independent.
But independent vectors do not need to be orthogonal.
Orthogonality is a stronger geometric relationship.
Common Error: Forgetting to Normalize
An orthogonal set is not automatically orthonormal.
Each vector must also have norm \(1\).
Normalize using
\[ q=\frac{v}{\|v\|}. \]Common Error: Dividing a Projection by \(\|u\|\)
The general projection formula is
\[ \boxed{ \operatorname{proj}_u(v) = \frac{v\cdot u}{u\cdot u}u}. \]The denominator is
\[ u\cdot u=\|u\|^2, \]not merely \(\|u\|\).
Common Error: Using the Unit-Vector Formula on a Nonunit Vector
The simplified formula
\[ \operatorname{proj}_u(v) = (v\cdot u)u \]is valid only when
\[ \|u\|=1. \]Common Error: Assuming Orthogonal Complement Means Set Difference
The notation
\[ W^\perp \]does not mean the vectors outside \(W\).
It means the vectors orthogonal to every vector in \(W\).
Common Error: Including the Zero Vector in an Orthogonal Basis
The zero vector is orthogonal to every vector, but it cannot belong to a basis because it makes the set linearly dependent.
Orthogonal bases consist of nonzero vectors.
Common Error: Skipping Earlier Directions in Gram-Schmidt
At step \(j\), the vector \(v_j\) must have all components along
\[ u_1,\ldots,u_{j-1} \]removed.
Subtracting only the most recent projection generally fails to produce a fully orthogonal set.
A Reliable Procedure for Dot Products and Angles
- Multiply corresponding vector components.
- Add the products to obtain \(u\cdot v\).
- Compute \(\|u\|\) and \(\|v\|\) if an angle is needed.
- Use \(\cos\theta=(u\cdot v)/(\|u\|\|v\|)\).
- Interpret the sign of the dot product when a qualitative angle test is enough.
A Reliable Procedure for Projection onto a Vector
- Compute \(v\cdot u\).
- Compute \(u\cdot u\).
- Form the scalar \((v\cdot u)/(u\cdot u)\).
- Multiply that scalar by \(u\).
- Subtract from \(v\) if the orthogonal residual is needed.
A Reliable Procedure for Projection onto a Subspace
- Find an orthonormal basis \(q_1,\ldots,q_k\) for the subspace.
- Compute each coefficient \(\langle v,q_i\rangle\).
- Multiply each coefficient by its basis vector.
- Add the components.
- Check that the residual \(v-\operatorname{proj}_W(v)\) is orthogonal to the subspace.
A Reliable Procedure for Gram-Schmidt
- Start with a linearly independent list \(v_1,\ldots,v_k\).
- Set \(u_1=v_1\).
- For each new \(v_j\), subtract its projections onto all earlier \(u_i\).
- Verify that the resulting \(u_j\) is nonzero.
- Normalize each \(u_j\) if an orthonormal basis is desired.
- Check pairwise inner products when needed.
Example: Full Projection Workflow
Let
\[ v= \begin{bmatrix} 4\\ 2\\ 3 \end{bmatrix} \]and let
\[ W= \operatorname{span} \left\{ e_1,e_2 \right\}. \]The vectors \(e_1,e_2\) form an orthonormal basis for the \(xy\)-plane.
Therefore,
\[ \operatorname{proj}_W(v) = (v\cdot e_1)e_1 + (v\cdot e_2)e_2. \]Since
\[ v\cdot e_1=4, \qquad v\cdot e_2=2, \]we obtain
\[ \boxed{ \operatorname{proj}_W(v) = \begin{bmatrix} 4\\ 2\\ 0 \end{bmatrix}}. \]The residual is
\[ \boxed{ v-\operatorname{proj}_W(v) = \begin{bmatrix} 0\\ 0\\ 3 \end{bmatrix}}, \]which is orthogonal to the \(xy\)-plane.
Example: Full Orthogonal-Basis Workflow
Let
\[ v_1= \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}. \]Start with
\[ u_1=v_1. \]Compute
\[ \operatorname{proj}_{u_1}(v_2) = \frac{v_2\cdot u_1}{u_1\cdot u_1}u_1. \]Here
\[ v_2\cdot u_1=1, \qquad u_1\cdot u_1=2. \]Thus
\[ u_2 = v_2-\frac12u_1 = \begin{bmatrix} 1/2\\ -1/2\\ 1 \end{bmatrix}. \]The vectors \(u_1,u_2\) are orthogonal and span the same subspace as \(v_1,v_2\).
Why Inner Products Matter in Multidimensional Systems
Inner products let us compare directions, measure error, define distance, and identify independent geometric components in high-dimensional spaces.
Orthogonality makes those components easier to separate and interpret. Projection then turns geometric approximation into an algebraic process.
These ideas support least squares, QR factorization, Fourier methods, principal directions, spectral analysis, numerical optimization, and dimensional reduction.
Practice 1: Dot Product
Compute
\[ \begin{bmatrix} 2\\ 3 \end{bmatrix} \cdot \begin{bmatrix} 4\\ -1 \end{bmatrix}. \]Practice 2: Norm
Find the norm of
\[ \begin{bmatrix} 5\\ 12 \end{bmatrix}. \]Practice 3: Orthogonality
Are
\[ \begin{bmatrix} 1\\ 2 \end{bmatrix} \quad\text{and}\quad \begin{bmatrix} 4\\ -2 \end{bmatrix} \]orthogonal?
Practice 4: Normalize a Vector
Normalize
\[ \begin{bmatrix} 6\\ 8 \end{bmatrix}. \]Practice 5: Projection
Project
\[ v= \begin{bmatrix} 3\\ 4 \end{bmatrix} \]onto
\[ u= \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]Practice 6: Orthogonal Complement Dimension
If \(W\subseteq\mathbb{R}^7\) has dimension \(3\), what is \(\dim(W^\perp)\)?
Practice 7: Orthonormal Coordinates
If \(q_1,q_2\) form an orthonormal basis and
\[ v=5q_1-2q_2, \]what are
\[ \langle v,q_1\rangle \quad\text{and}\quad \langle v,q_2\rangle? \]Practice 8: Pythagorean Identity
If \(u\perp v\), \(\|u\|=3\), and \(\|v\|=4\), find \(\|u+v\|\).
Practice 9: Orthogonal Matrix
If \(Q\) is orthogonal, what is \(Q^{-1}\)?
Practice 10: Projection Matrix
If the columns of \(Q\) form an orthonormal basis for \(W\), what matrix projects vectors onto \(W\)?
Practice 11: Gram-Schmidt
Let
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Use one Gram-Schmidt step to find a vector \(u_2\) orthogonal to \(v_1\).
Practice 12: Best Approximation
If \(p=\operatorname{proj}_W(v)\), what geometric property does the error
\[ e=v-p \]satisfy?
Practice Answers
Practice 1
\[ 2(4)+3(-1)=8-3=\boxed{5}. \]Practice 2
\[ \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = \boxed{13}. \]Practice 3
\[ 1(4)+2(-2)=4-4=0. \]Therefore,
\[ \boxed{\text{the vectors are orthogonal}}. \]Practice 4
The norm is
\[ \sqrt{6^2+8^2}=10. \]Therefore,
\[ \boxed{ \frac1{10} \begin{bmatrix} 6\\ 8 \end{bmatrix} = \begin{bmatrix} 3/5\\ 4/5 \end{bmatrix}}. \]Practice 5
\[ \operatorname{proj}_u(v) = \frac{v\cdot u}{u\cdot u}u = \frac{3}{1} \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]Thus
\[ \boxed{ \begin{bmatrix} 3\\ 0 \end{bmatrix}}. \]Practice 6
\[ \dim(W^\perp) = 7-3 = \boxed{4}. \]Practice 7
In an orthonormal basis, coefficients are inner products:
\[ \boxed{\langle v,q_1\rangle=5}, \qquad \boxed{\langle v,q_2\rangle=-2}. \]Practice 8
\[ \|u+v\|^2 = 3^2+4^2 = 25. \]Therefore,
\[ \boxed{\|u+v\|=5}. \]Practice 9
\[ \boxed{Q^{-1}=Q^T}. \]Practice 10
\[ \boxed{P=QQ^T}. \]Practice 11
Since
\[ \operatorname{proj}_{v_1}(v_2) = \begin{bmatrix} 1\\ 0 \end{bmatrix}, \]we obtain
\[ u_2 = v_2-\operatorname{proj}_{v_1}(v_2) = \boxed{ \begin{bmatrix} 0\\ 1 \end{bmatrix}}. \]Practice 12
The error is orthogonal to the entire subspace:
\[ \boxed{e\in W^\perp}. \]This is why \(p\) is the closest vector in \(W\) to \(v\).
What to Remember
The standard inner product in \(\mathbb{R}^n\) is
\[ \boxed{\langle u,v\rangle=u^Tv}. \]- Norm is defined by \(\|v\|=\sqrt{\langle v,v\rangle}\).
- Orthogonal vectors have zero inner product.
- Orthogonal nonzero vectors are automatically independent.
- Orthonormal vectors are orthogonal unit vectors.
- Projection extracts the component of one vector along another vector or subspace.
- The projection error lies in the orthogonal complement.
- \(\dim(W)+\dim(W^\perp)=n\) in \(\mathbb{R}^n\).
- Gram-Schmidt converts an independent set into an orthogonal or orthonormal basis.
- Orthogonal matrices satisfy \(Q^{-1}=Q^T\).
- Projection onto an orthonormal-column space has matrix \(QQ^T\).
The central geometric principle is:
\[ \boxed{ \text{orthogonality separates a vector into independent geometric components}}. \]Where We Go Next
Projection onto a subspace gives the closest possible vector in that subspace to a target. This becomes especially useful when a linear system has no exact solution.
The next lesson, Least Squares, uses orthogonal projection to find best-fit solutions to inconsistent systems and develops the normal equations, residual vectors, regression geometry, and QR-based solution methods.