Notes on Euclidean Spaces

Real euclidean spaces⌗

Real euclidean spaces have definitions of inner product and norm. Examples in \(\mathbb R^n\):

• The usual inner product
• The unit-radius circumference when considering an unusual inner product
• Cauchy-Schwarz inequality

Let \(V\) be a real vector space. A form or real function \[ \begin{aligned} \langle\cdot,\cdot\rangle\colon V\times V &\rightarrow \mathbb R \\ (x, y) &\mapsto \langle x,y\rangle \end{aligned}\ \] is said to be an inner product if, for all \(x, y, z \in V\) and all \(\alpha \in \mathbb R\),

1. \(\langle x,y \rangle = \langle y, x \rangle\)
2. \(\langle \alpha x, y\rangle = \alpha\langle x, y\rangle\)
3. \(\langle x+y, z\rangle = \langle x, z\rangle + \langle y, z\rangle\)
4. \(\langle x, x\rangle \geq 0 \wedge (\langle x, x\rangle = 0 \implies x = 0)\)

A real linear space \(V\) equipped with an inner product is called an (real) Euclidean Space.

Examples⌗

1. Usual inner product in \(\mathbb R^n\)
   • \(\mathbb R^2\) \[ \begin{aligned} \langle x, y\rangle &= \lVert x\rVert\lVert y\rVert \cos\theta \\ &= x_1 y_1 + x_2 y_2 \quad\text{in }\mathbb R^2 \end{aligned}\ \] where \(\theta\in[0, \pi]\) is the angle between the vectors \(x\) and \(y\). Note that the norm of the vector \(x\) satisfies \[ \lVert x\rVert^2 = \langle x, x\rangle \]
   • \(\mathbb R^n\) \[ \begin{aligned} \langle x, y\rangle &= x_1 y_1 + x_2 y_2 + \cdots + x_n y_n \\ \langle x, y\rangle &= y^Tx = x^Ty \end{aligned}\ \]
2. Another inner product in \(\mathbb R^2\)
   • Exercise. Determine the circumference \(C\) of radius \(1\) and centered at \((0, 0)\) \[ C = {(x_1, x_2)\in\mathbb R^2\colon \lVert(x_1, x_2)\rVert = 1} \] considering
     • The usual inner product
     • The inner product \[\langle (x_1, x_2), (y_1, y_2) \rangle = \frac{1}{9} x_1y_1 + \frac{1}{4} x_2y_2 \]

Norm and the triangle inequality⌗

For all vectors \(x\in V\), we define the norm of \(x\) as \[ \lVert x\rVert = \sqrt{\langle x, x\rangle} \] such that, for all \(x\in V\) and all \(\alpha\in\mathbb R\) we have

1. \(\lvert x\rVert\geq 0\qquad \text{and}\qquad \lVert x\rVert \iff x = 0\)
2. \(\lVert \alpha x\rVert = \lvert\alpha\rvert\lVert x\rVert\)
3. \(\lVert x + y\rVert\leq\lVert x\rVert + \lVert y\rVert \qquad\qquad \text{(triangle inequality)} \)

A function \(V \to \mathbb R\) that satisfies the above conditions is said to be a norm defined in \(V\).

The proof that the function we defined earlier and called “inner product” satisfies the triangle inequality will be done at a later point, since it relies on the Cauchy-Schwarz inequality.

Cauchy-Schwarz Inequality⌗

Theorem 1. Let \(V\) be an euclidean space. For all \(x, y \in V\) we have \[ \lvert\langle x, y\rangle\rvert \leq \lVert x\rVert \lVert y\rVert \] Note that in \(\mathbb R^2\) and \(\mathbb R^3\) we have: \[ \langle x, y\rangle = \lVert x\rVert\lVert y\rVert\cos\theta \] where \[ \lvert\langle x, y\rangle\rvert = \lVert x\rVert\lVert y\rVert\lvert \cos\theta\rvert \leq \lVert x\rVert \lVert y\rVert \]

Distance⌗

For all \(x, y \in V\), we define the distance from \(x\) to \(y\) as \[ d(x, y) = \lVert x - y\rVert \]

Parallelogram Law⌗

For all vectors \(x, y \in V\), we have \[ \lVert x + y\rVert^2 + \lVert x - y\rVert^2 = 2(\lVert x\rVert^2 + \lVert y\rVert^2) \]

Example⌗

An inner product in \(\mathbb M_{2\times 2}(\mathbb R)\).

For all matrices \(A, B \in \mathbb M_{2\times 2}(\mathbb R)\) we define \[ \begin{aligned} \langle A, B\rangle &= tr(B^T A) \\ &= \sum^2_{i, j=1}{a_{ij}b_{ij}} \end{aligned}\] with \(A = [a_{ij}]\) and \(B = [b_{ij}]\)¹. Note that, letting \(B_c\) be the canonical basis of \(\mathbb M_{2\times 2}(\mathbb R)\)

\[\langle A, B \rangle_{\mathbb M_{2\times 2}(\mathbb R)} = \langle (A)_{B_c}, (B)_{B_c}\rangle_{\mathbb R^4}\]

Meaning that the inner product defined above respects the isomorphism \( A \mapsto (A)_{B_c}\) between \( \mathbb{M}_{2 \times 2}(\mathbb R) \) and \(\mathbb R^4\)

Proof of the triangle inequality⌗

\[ \begin{aligned} \lVert x + y\rVert^2 &= \langle x+y, x+y\rangle \\ &= \langle x, x\rangle + 2\langle x, y\rangle + \langle y, y\rangle \\ &= \lVert x\rVert^2 + 2\langle x, y\rangle + \lVert y\rVert^2 \qquad (\text{Inner product in terms of the norm})\\ &\leq \lVert x\rVert^2 + 2\lvert\langle x, y\rangle\rvert + \lVert y\rVert^2 \\ &\leq \lVert x\rVert^2 + 2\lVert x\rVert\lVert y\rVert + \lVert y\rVert^2 \qquad (\text{Cauchy-Schwartz inequality}) \\ &=(\lVert x\rVert + \lVert y\rVert)^2 \end{aligned}\]

Where \[ \lVert x + y\rVert \leq \lVert x\rVert + \lVert y\rVert \qquad_\blacksquare\]

Gram Matrix⌗

Let \(V\) be a real euclidean space, and \(B = (b_1, b_2, \ldots, b_n)\) a basis of \(V\). With \(x, y \in V\) such that \(x_B = (\alpha_1, \alpha_2, \ldots, \alpha_n)\) and \(y_B = \beta_1, \beta_2, \ldots, \beta_n\), we have \[\begin{aligned} \langle x, y\rangle &= \langle\alpha_1 b_1 + \alpha_2 b_2 + \cdots + \alpha_n b_n, \beta_1 b_1 + \beta_2 b_2 + \cdots + \beta_n b_n\rangle \\ &= \begin{bmatrix}\beta_1 & \beta_2 & \ldots & \beta_n \end{bmatrix} \underbrace{\begin{bmatrix}\langle b_1, b_1\rangle & \langle b_2, b_1\rangle &\ldots & \langle b_n, b_1\rangle \\ \langle b_1, b_2\rangle & \langle b_2, b_2\rangle &\ldots & \langle b_n, b_2\rangle \\ \vdots \\ \langle b_1, b_n\rangle & \langle b_2, b_n\rangle &\ldots & \langle b_n, b_n\rangle \\ \end{bmatrix}}_G \begin{bmatrix}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix} \end{aligned}\ \]

Therefore, given an inner product in \(V\) and a basis \(B\), it is possible to determine a matrix \(G\) such that \[\langle x,y\rangle = y_B^T Gx_B\]

This matrix \(G = [g_{ij}]\), where for all \(i, j = 1, \ldots, n\) we have \(g_{ij} = \langle b_j, b_i \rangle\) is called the Gram matrix of the set of vectors \(\{b_1, b_2, \ldots, b_n\}\).

Note that:

• \(G\) is a symmetric (\(G = G^T\)) \(n\times n\) real matrix.
• For all non-null vectors \(x\in V\) \[x_B^T Gx_b > 0\]

A square real matrix \(A\) of order \(k\) is said to be positive definite if, for all non-null vectors \(x\in\mathbb R^n\), \(x^T Ax > 0\)

Proposition 1. A symmetric real matrix is positive definite iff all your proper values are positive.

Theorem 2. Let \(A\) be a symmetric real matrix of order \(n\). The following statements are equivalent.

1. The expression \[\langle x, y\rangle = y^T Ax\] defines an inner product in \(\mathbb R^n\)
2. \(A\) is positive definite.

Exercise⌗

Consider that \(\mathbb R^n\) is equipped with the canonical basis \(\mathcal{e}_n\). What is the Gram matrix \(G\) that corresponds to the usual inner product in \(\mathbb R^n\)? Also, which Gram matrix corresponds to the inner product in item (2) of the previous exercise?

Complex euclidean spaces and orthogonal vectors⌗

Example of complex euclidean space: usual inner product in \(\mathbb C^n\).

Let \(V\) be a complex vector space. A complex function or form \[ \begin{aligned}\langle\cdot,\cdot\rangle\colon V \times V &\to \mathbb C \\ (x, y) &\mapsto \langle x, y\rangle\end{aligned} \] is said to be an inner product if, for all \(x, y, z \in V\) and all \(\alpha \in \mathbb C\)

1. \(\langle x, y\rangle = \overline{\langle y, x\rangle}\)
2. \(\langle \alpha x, y\rangle = \alpha\langle x,y\rangle\)
3. \(\langle x + y, z\rangle = \langle x, z\rangle + \langle y, z\rangle\)
4. \(\langle x, x\rangle \geq 0 \wedge (\langle x, x\rangle = 0 \implies x = 0)\)

A complex vector space \(V\) equipped with an inner product is called a (complex) euclidean space.

Much like with real euclidean spaces, we define the norm of a vector as \[\lVert x\rVert = \sqrt{\langle x, x\rangle}\] and the distance from \(x\) to \(y\) as \[d(x, y) = \lVert x - y\rVert\]

Example. Usual inner product in \(\mathbb C^n\). Let \(x = (x_1, x_2, \ldots, x_n)\) and \(y_1, y_2, \ldots, y_n\) be vectors in \(\mathbb C^n\), we define \[\langle x, y\rangle = x_1\overline{y}_1 + x_2\overline{y}_2 + \cdots + x_n\overline{y}_n\] and therefore \[\langle x, y\rangle = \overline{y}^T x\] Regarding the norm we have \[\lvert x\rVert^2 = \langle x, x\rangle = x_1\overline{x}_1 + x_2\overline{x}_2 + \cdots + x_n\overline{x}_n\] or \[\lVert x\rVert = \sqrt{\lVert x, x\rVert} = \sqrt{\lvert x_1\rvert^2 + \lvert x_2\rvert^2 + \cdots + \lvert x_n\rvert^2}\]

All the remaining results that were presented regarding real euclidean spaces are also true for complex euclidean spaces (Cauchy-Schwartz, triangle inequality, parallelogram law, …).

Complex Gram Matrix⌗

Let \(V\) be a complex euclidean space, and let \(B = (b_1, b_2, \ldots, b_n)\) be a basis of \(V\). With \(x, y \in V\) such that \(x_B=(\alpha_1, \alpha_2, \dots, \alpha_n)\) and \(y_B=(\beta_1, \beta_2, \dots, \beta_n)\), we have \[ \begin{aligned} \langle x, y\rangle &= \langle\alpha_1 b_1 + \alpha_2 b_2 + \cdots + \alpha_n b_n, \beta_1 b_1 + \beta_2 b_2 + \cdots + \beta_n b_n\rangle \\ &= \begin{bmatrix}\overline{\beta}_1 & \overline{\beta}_2 & \ldots & \overline{\beta}_n\end{bmatrix} \underbrace{\begin{bmatrix}\langle b_1, b_1\rangle & \langle b_2, b_1\rangle &\ldots & \langle b_n, b_1\rangle \\ \langle b_1, b_2\rangle & \langle b_2, b_2\rangle &\ldots & \langle b_n, b_2\rangle \\ \vdots \\ \langle b_1, b_n\rangle & \langle b_2, b_n\rangle &\ldots & \langle b_n, b_n\rangle \\ \end{bmatrix}}_G \begin{bmatrix}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix} \end{aligned}\ \]

Therefore, given an inner product in \(V\) and a basis \(B\), it is possible to determine a matrix \(G\) such that \[\langle x,y\rangle = \overline{y}_B^T Gx_B\]

This matrix \(G = [g_{ij}]\), where for all \(i, j = 1, \ldots, n\) we have \(g_{ij} = \langle b_j, b_i \rangle\) is called the Gram matrix of the set of vectors \(\{b_1, b_2, \ldots, b_n\}\).

Note that:

• \(G\) is an \(n\times n\) complex matrix such that (\(G = \overline{G}^T\)).
• For all non-null vectors \(x\in V\) \[\overline{x}_B^T Gx_b > 0\]

A complex square matrix \(A\) of order \(k\) is said to be hermitian if \(A = \overline{A}^T\). Note that the spectrum \(\sigma(A)\) of a hermitian is contained in \(\mathbb R\).

A hermitian matrix \(A\) of order \(k\) is said to be positive definite if, for all non-null vectors \(x\in\mathbb C^n\), \(\overline{x}^T Ax > 0\).

Proposition 2. A hermitian matrix is positive definite iff all of it’s proper values are positive.

Theorem 3. Let \(A\) be a hermitian matrix of order \(n\). The following statements are equivalent.

1. The expression \[\langle x, y\rangle = \overline{y}^T Ax\] defines an inner product in \(\mathbb C^n\)
2. A is positive definite.

Angle between two vectors⌗

Let \(x\) and \(y\) be non-null vectors belonging to some real euclidean space \(V\). We define the angle between the vectors \(x\) and \(y\) as being the angle \(\theta\), with \(0\leq\theta\leq\pi\), such that \[\cos\theta = \frac{\langle x, y\rangle}{\lVert x\rVert \lVert y\rVert}\] With Cauchy-Schwartz we can see that \(\lvert\cos\theta\rvert\leq 1\).

Let \(x\) and \(y\) be (possibly null) vectors belonging to some real or complex euclidean space \(V\). The vectors \(x\) and \(y\) are said to be orthogonal, written \(x \perp y\), if \[\langle x, y\rangle = 0\]

Exercise. What are the orthogonal vectors to \(v = (1, 1, 0)\) considering \(\mathbb R^3\) with the usual inner product?

Theorem 4. (Pythagoras Theorem) Let \(x\) and \(y\) be orthogonal vectors of some euclidean space \(V\). Then \[\lVert x + y\rVert^2 = \lVert x\rVert^2 + \lVert y \rVert^2\]

Proof. Exercise

Orthogonal complement⌗

Let \(X\) be a subspace of an euclidean space \(V\). We say that \(u\) is orthogonal to \(X\) if \(u\) is orthogonal to all elements of \(X\). We write this \(u \perp W\).

For example, \((1, 1, 0)\) is orthogonal to the plane \(S\) of the previous exercise.

Let \(W\) be a subspace of \(V\). The orthogonal complement of \(W\), written \(W^\perp\), is defined as \[W^\perp = \{u\in V\colon u\perp W\}\]

Exercise. Determine the orthogonal complement of the line generated by the vector \((1, 1, 0)\).

Proposition 3. \(W^\perp\) is a subspace of V.

Proposition 4. Let \(W\) be a linear subspace of an euclidean space \(V\) and let \(\{u_1, u_2, \ldots, u_k\}\) be a generator set of \(W\). Then, \(e\in V\) is orthogonal to \(W\) iff it is orthogonal to \(\{u_1, u_2, \ldots, u_k\}\).

Corollary 1. In the conditions of the previous proposition, \(u\in V\) is orthogonal to \(W\) iff it is orthogonal to a basis of \(W\).

Exercise. Determine the orthogonal complement of the plane \(W\in\mathbb R^3\) with the cartesian equation \(x=y\).

Solution. \(W^\perp\) is the line described by the equations \[\begin{cases}x = -y \\ z = 0\end{cases}\qquad\qquad\textbf{cartesian equations}\] or \[(x, y, z) = t(-1, 1, 0)\qquad(t\in\mathbb R)\qquad\qquad\textbf{vector equation}\] or \[\begin{cases}x = -t \\ y = t \\ z = 0\end{cases}\qquad(t\in\mathbb R)\qquad\qquad\textbf{parametric equations}\]

Proposition 5. Let \(W\) be a subspace of an euclidean space \(V\).

1. \(W\cap W^\perp = 0\)
2. \(W^{\perp\perp} = W\)

A subset \(X\) of an euclidean space \(V\) is said to be an orthogonal set if, for all \(x, y\in X\) with \(x \neq y\) we have \(x \perp y\).

Question. Let \(X\) be an orthogonal set not containing the null vector.

• If \(X\subseteq \mathbb R^2\), how many vectors does \(X\) have at most?
• If \(X\subseteq \mathbb R^3\), how many vectors does \(X\) have at most?

Proposition 6. Let \(V\) be an euclidean space. Let \(X = \{v_1, v_2, \ldots, v_k\}\) be an orthogonal set such that \(v_j\neq 0\) for all \(j\in[1,\ldots,k]\). Then \(X\) is linearly independent.

Proof. \[\langle\alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_k v_k, v_j\rangle = \alpha_j^2 \lVert v_j \rVert^2 = 0 \implies \alpha_j = 0\]

Corollary 2. Let \(V\) be an euclidean space of dimension \(n\), and let \(X = \{v_1, v_2, \ldots, v_k\}\) be an orthogonal set such that \(v_j \neq 0\) for all \(j\in[1, k]\). Then \(k \leq n\).

Corollary 3. Let \(V\) be an euclidean space of dimension \(n\), and let \(X = \{v_1, v_2, \ldots, v_n\}\) be an orthogonal set such that \(v_j \neq 0\) for all \(j\in[1, n]\). Then \(X\) is a basis of \(V\).

Orthogonal complements of the subspaces of a real matrix⌗

Proposition 7. Let \(A\) be a \(n \times k\) matrix with real elements. Then, considering in \(\mathbb R^n\) and \(\mathbb R^k\) the usual inner products we have:²

1. \(L(A)^\perp = N(A)\)
2. \(N(A)^\perp = L(A)\)
3. \(C(A)^\perp = N(A^T)\)
4. \(N( A^T )^\perp = C(A)\)

Orthogonal Projections⌗

Orthogonal bases and orthonormal bases⌗

A basis \(\mathcal{B}\) of an euclidean space \(V\) is said to be:

• An orthogonal basis if it is an orthogonal set;
• An orthonormal basis if it is an orthogonal set, and all it’s elements have unitary norm.

Let \(x\in V\) some vector, and let \[(x)_\mathcal{B} = (\alpha_1, \alpha_2, \ldots, \alpha_n)\] be the coordinate vector of \(x\) in the basis \(\mathcal B\).

Coordinate vector in an orthogonal basis \(\mathcal B\)⌗

\[\alpha_j = \frac{\langle x, b_j\rangle}{\lVert b_j\rVert^2}\]

Coordinate vector in an orthonormal basis \(\mathcal B\)⌗

\[ \alpha_j = \langle x, b_j\rangle\]

Question. Will there always be an orthogonal and/or an orthonormal basis?

Answer. Yes -> Orthogonalization through Gram-Schmidt method.

Orthogonal projections⌗

We define the orthogonal projection of \(x\) over \(b_j\) as the vector \[\begin{aligned}\text{proj}_{b_j} x &= \frac{\langle x, b_j \rangle}{\lVert b_j \rVert^2}b_j \\ &= \alpha_j b_j \end{aligned}\]

In a more general sense, given two vectors \(u\) and \(v\) from an euclidean space \(V\), with \(v\neq 0\) the orthogonal projection of \(u\) over \(v\) is the vector \[\text{proj}_v u = \frac{\langle u, v \rangle}{\lVert v \rVert}^2 v\]

Example. Considering that \(\mathbb R^2\) is equipped with the canonical basis \(\mathcal{E}_2 = (e_1, e_2)\), any vector \(u\in\mathbb R^2\) can be expressed as a sum \[\begin{aligned}u &= \text{proj}_{ e_1} u + \text{proj}_{ e_2} u \\ &= u_W + u_{W^\perp}\end{aligned}\] Where \(W\) is the \(x\) axis.

Theorem 5. Let \(W\) be a linear subspace of some euclidean space \(V\). All vectors \(u\) of \(V\) can be decomposed uniquely as \[u = u_W + u_{W^\perp}\] where \(u\in W\) and \(u_{W^\perp}\in E^\perp\).

In these conditions, we say that \(V\) is the direct sum of \(W\) with \(W^\perp\) and write \[V = W \oplus W^\perp\] Which, by definition, is to say:

• \(V = W + W^\perp\)
• \(W \cap W^\perp = \{0\}\)

We define the orthogonal projection of \(u\) over \(W\) as being the vector \(u_W\).

If we consider that \(W\) is equipped with the ordered orthogonal basis \(\mathcal{B} = (b_1, b_2, \ldots, b_k)\), we have \[\text{proj}_W u = \text{proj}_{b_1} u + \text{proj}_{b_2} u + \cdots + \text{proj}_{b_k} u\]

Question. How can we compute the vector \(u_{W^\perp}\) or, in other words, \(\text{proj}_{W^\perp} u\)?

Answer. \[\text{proj}_{W^\perp} u = u - u_W\] or, if we consider that \(W^\perp\) is equipped with the ordered orthogonal basis \(\mathcal{B}’ = (b_1’, b_2’, \ldots, b_l’)\), we have \[\text{proj}_{W^\perp} u = \text{proj}_{b’_1} u + \text{proj}_{b’_2} u + \cdots + \text{proj}_{b’_l} u\]

Question. What is the number \(l\) of vectors in the basis of \(\mathcal{B}’\)?

Answer. Assuming that \(V\) has dimension \(n\), we have \(l = n - k\) since

1. \(\mathcal{B} \cup \mathcal{B}’\) is linearly independent.³
2. Theorem 5 guarantees that \(\mathcal{B}\cup\mathcal{B}’\) generates \(V\).

Therefore \(\mathcal{B} \cup \mathcal{B}’\) is a basis of \(V\) and the solution becomes trivial.

Distance from a point to a subspace & \(k\)-plane cartesian equations⌗

Optimal approximation⌗

Given \(u\in V\) and some subspace \(W\) of \(V\) we hope to answer the following question:

Which element \(x\) of \(W\) is closest to \(u\)?

\[\begin{aligned}d(u, x)^2 = \lVert u - x\rVert^2 &= \lVert(u - \text{proj}_W u) + (\text{proj}_W u - x)\rVert^2 \\ &= \lVert u - \text{proj}_W u\rVert^2 + \lVert \text{proj}_W u - x\rVert^2 \qquad \text{(Pythagoras)}\\ &= \lVert \text{proj}_{W^\perp} u\rVert^2 + \lVert\text{proj}_W u - x\rVert^2\end{aligned}\]

Whereby we conclude that

The optimal approximation coincides with \(\text{proj}_W u\) ⁴

With that, we define the distance from \(u\) to a subspace \(W\) as

\[d(u, W) = \lVert proj_{W^\perp} u \rVert\]

\(k\)-plane cartesian equations⌗

A \(k\)-plane of \(\mathbb R^n\) is any subset \(S\) of \(\mathbb R^n\) which can be expressed as

\[S = W + p\]

Where \(W\) is a subspace of \(\mathbb R^n\) with dimension \(k\) and \(p\) is an element of \(\mathbb R^n\). Depending on the dimension of \(W\), we have the following nomenclature:

• If \(k = 0\), \(S\) is said to be a point.
• If \(k = 1\), \(S\) is said to be a line.
• If \(k = 2\), \(S\) is said to be a plane.
• If \(k = n - 1\), \(S\) is said to be a hyperplane.⁵

Let \(x = (x_1, x_2, \ldots, x_n)\) be an elements of \(S\), there exists \(y\) in \(W\) such that

\[x = y + p\]

Or equivalently

\[y = x - p\]

The last equation show that, using vector, cartesian, or parametric equations of \(W\) we can easily obtain (substituting \(y\) for \(x-p\)) vector, cartesian, or parametric equations of \(S\), respectively.

Analogously, using the subspace \(W^\perp\) we can also obtain equations of \(S\). If \(B_{W^\perp} = (v_1, v_2, \ldots, v_{n-k})\) is a basis for the orthogonal complement of \(W\), with \(\text{dim} W = k\), we have \(x - p \in W\) or, equivalently

\[\underbrace{\begin{bmatrix} v^T_1 \\ v^T_2 \\ \vdots \\ v^T_{n-k} \end{bmatrix}}_{(n-k)\times n} \underbrace{\begin{bmatrix} x_1 - p_1 \\ x_2 - p_2 \\ \vdots \\ x_n - p_n \end{bmatrix}}_{n\times 1} = \underbrace{\begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}}_{(n-k)\times 1}\]

Defining the matrix \(A\) as

\[A = \begin{bmatrix}v^T_1 \\ v^T_2 \\ \vdots \\ v^T_{n-k} \end{bmatrix}\]

We obtain the homogeneous linear equation system \(A(x -p) = 0\). Consequently, from a vector equation of \(N(A)\), or cartesian equations of \(N(A)\), or parametric equations of \(N(A)\), we can obtain the corresponding equations of \(S\).

Exercise. Determine a vector equation, the cartesian equations, and the parametric equations of the plane passing the point \(p = (1, 2, 0)\) which is perpendicular to the line passing this same point with direction \(n=(5, 1, -2)\)

Distance from a point to a \(k\)-plane⌗

Let \(S=W+p\) and consider a point \(q\in\mathbb R^n\). Given \(x\) in \(S\),

\[\begin{aligned}d(q, x) &= \lVert q - x \rVert \\ &= \lVert (q - p) + (p - x) \\ &= \lVert (q - p) - y \rVert \\ &= d(q-p, y) \\ \end{aligned}\]

The minimal value for this distance can be obtained for \(y = \text{proj}_{W}(q - p)\), as previously described. We then define the distance from point \(q\) to the plane \(S\) as

\[\begin{aligned}d(q, S) &= d(q - p, W) \\ &= \lVert \text{proj}_{W^\perp}(q - p )\rVert\end{aligned}\]

Exercise. Compute the distance from \((3, 2, -1)\) to the plane \(S\) from the previous exercise.