In order to use formula (3), we need to find \[ \mathbf{u} \bullet \mathbf{v} = 3 \cdot 6 - 7 \cdot 4 - 2 \cdot 5 = -20 . \]

The first approach is based on projection of the taget vector onto normal vector to the plane. Since this vector is known to be n = (2, 5, −3). Now we compute the projection of the position vector (5, −3, 2) onto n using Eq.*3): \[ \mbox{proj}_n (\mathbf{v}) = \frac{\mathbf{v} \bullet \mathbf{n}}{\| \mathbf{n} \|^2} \,\mathbf{n} = \frac{11}{38} \begin{pmatrix} -2 \\ -5 \\ 3 \end{pmatrix} . \]

Now we demonstrate another approach using    orthogonal projection:   

Recall that the orthogonal complement of W, denoted by W^⊥ is \[ W^{\perp} = \left\{ \mathbf{y} \in \mathbb{R}^3 \ : \ \mathbf{x} \bullet \mathbf{y} \quad \forall \mathbf{x} \in W \right\} , \] which leads to direct sum decomposition ℝ³ = W ⊕ W^⊥. With this approach, we need to determine a basis of subspace W. From plane equation, we find \[ 2x = -5y + 3z. \] Setting y = 0, z = 2, and then y = 2, z = 0, we obtain two linearly independent (because they have zero components) vectors \[ \mathbf{u}_1 = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} , \qquad \mathbf{u}_2 = \begin{pmatrix} -5 \\ 2 \\ 0 \end{pmatrix} . \] With these vectors at hand, we build the matrix \[ \mathbf{A} = \begin{bmatrix} 3 & -5 \\ 0 & 2 \\ 2&0 \end{bmatrix} . \] Using formula (8) we find the projection vector \[ \mathbf{p} = \mathbf{A} \left( \mathbf{A}^{\mathrm T} \mathbf{A} \right)^{-1} \mathbf{A}^{\mathrm T} \mathbf{v} = \frac{1}{38}\begin{bmatrix} 34 & -106 \\ -10&13&15 \\ 6&15&29 \end{bmatrix} \begin{pmatrix} 5 \\ -3 \\ 2 \end{pmatrix} = \frac{1}{38}\begin{pmatrix} 212 \\ -59 \\ 43 \end{pmatrix} . \]

p from v, we obtain the same vector q ∈ W^⊥.    ■

Example A: Verification of Eq.(9) in ℝ³

We consider a two-dimensional subspace W ⊂ ℝ³ spanned on two orthogonal vectors \[ \mathbf{e} _1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} , \qquad \frac{1}{\sqrt{14}} \mathbf{e} _2 = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \] We build 3 × 2 matrix of these vectors \[ \mathbf{Q} = \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 \end{bmatrix} . \] Then we form a projection matrix according to Eq.(9): \[ \mathbf{P} = \mathbf{Q}\,\mathbf{Q}^{\mathrm T} = \frac{1}{42} \begin{bmatrix} 26 & -4 & 20 \\ -4& 41& 5 \\ 20& 5& 17 \end{bmatrix} . \]

1. Using Mathematica, we generate two three-dimensional vectors
         u1 = RandomInteger[{-9, 9}, 3]
         {-6, 1, 9}
         u2 = RandomInteger[{-9, 9}, 3]
         {-7, 1, 5}
   Then we build 3×2 matrix having these column vector
         A = Transpose[{u1, u2}]
   \[ \mathbf{A} = \begin{bmatrix} -6&-7 \\ \phantom{-}1&\phantom{-}1 \\ \phantom{-}9&\phantom{-}5 \end{bmatrix} . \] Using this matrix, we form a projection matrix on space W = span{u₁, u₂} based on formular (8): \[ \mathbf{P}_W = \mathbf{A} \left( \mathbf{A}^{\mathrm T} \mathbf{A} \right)^{-1} \mathbf{A}^{\mathrm T} = \frac{1}{1106} \begin{bmatrix} 1090 & -132 & 4 \\ -132 & 17& 33 \\ 4 & 33 & 1105 \end{bmatrix} . \]
         P = A . Inverse[Transpose[A] . A] . Transpose[A]
         {{545/553, -(66/553), 2/553}, {-(66/553), 17/1106, 33/1106}, {2/553, 33/1106, 1105/1106}}
   It is well-known that any linear transformation is equivalent to matrix multiplication and vice versa. Therefore, to matrix P corresponds a linear transformation.
2. Let W be a two-dimensional subspace of ℝ³ that is spanned on two vectors u = [-9, 8, 5] and v = [0, -3, 0]:
         u = RandomInteger[{-9, 9}, 3]
         {-9, 8, 5}
         v = RandomInteger[{-9, 9}, 3]
         {0, -3, 0}
   We organize these vectors in matrix form: \[ \mathbf{A} = \begin{bmatrix} -9&0 \\ 8&-3 \\ 5&0 \end{bmatrix} . \] Projection transformation on subspace W is equivalent to matrix multiplication \[ \mathbf{P}_W = \mathbf{A} \left( \mathbf{A}^{\mathrm T} \mathbf{A} \right)^{-1} \mathbf{A}^{\mathrm T} = \frac{1}{106} \begin{bmatrix} 81 & 0 & -45 \\ 0 & 106& 0 \\ -45 & 0 & 25 \end{bmatrix} . \]
         P = A . Inverse[Transpose[A] . A] . Transpose[A]
        {{81/106, 0, -(45/106)}, {0, 1, 0}, {-(45/106), 0, 25/106}}
   We check that matrix P is singular
        Det[P]
         0
   Let us take x = u + 2 v = [-9, 2, 5] ∈ W. Then its projection on W is \[ {\bf x}_W = {\bf P}\,\mathbf{x} = \mathbf{x} . \]
        P . {-9, 2, 5}
         {-9, 2, 5}
3. In this example, we use W from the previous example, which is a span of two vectors u = [-9, 8, 5] and v = [0, -3, 0]. Then their cross product will be orthogonal to W: \[ \mathbf{x} = \mathbf{u} \times \mathbf{v} = \begin{bmatrix} 15&0&27 \end{bmatrix} . \]
         u = {-9, 8, 5}; v = {0, -3, 0};
         x = Cross[u, v]
         {15, 0, 27}
   Its projection on W becomes \[ \mathbf{x}_W = {\bf P}\,\mathbf{x} = \frac{1}{106} \begin{bmatrix} 81 & 0 & -45 \\ 0 & 106& 0 \\ -45 & 0 & 25 \end{bmatrix} \begin{pmatrix} 15 \\ 0 \\ 27 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \\ 0 \end{pmatrix} . \]
         P . x
         {0, 0, 0}
4. Composition of two linear transformations corresponds the product of their corresponding matrices. So we need to check that P P = P. So for our example we take the projection matrix from example 2. Then the product is \[ {\bf P}\,{\bf P} = \frac{1}{106} \begin{bmatrix} 81 & 0 & -45 \\ 0 & 106& 0 \\ -45 & 0 & 25 \end{bmatrix} = {\bf P}. \]
         P.P
         {{81/106, 0, -(45/106)}, {0, 1, 0}, {-(45/106), 0, 25/106}}
5. This property is equivalent to property 2.

1. The column space of projection matrix P is spanned on four vectors \[ W = \mbox{span} \left\{ \mathbf{u}_1 = \begin{pmatrix} 182 \\ 164 \\ 76 \\ 124 \end{pmatrix} , \ \mathbf{u}_2 = \begin{pmatrix} 164 \\ 236 \\ 142 \\ -50 \end{pmatrix} , \ \mathbf{u}_3 = \begin{pmatrix} 76 \\ 142 \\ 93 \\ -83 \end{pmatrix} , \ \mathbf{u}_4 = \begin{pmatrix} 124 \\ -50 \\-83 \\ 381 \end{pmatrix} \right\} . \] First, we check matrix rank of projection matrix, which is 2.
          MatrixRank[P]
          2
   Next, we verify that first two vectors u₁ and u₂ are linearly independent:
         MatrixRank[{{182, 164}, {164, 236}, {76, 142}, {124, -50}}]
         2
   Hence, \[ W = \mbox{span} \left\{ \mathbf{u}_1 = \begin{pmatrix} 182 \\ 164 \\ 76 \\ 124 \end{pmatrix} , \ \mathbf{u}_2 = \begin{pmatrix} 164 \\ 236 \\ 142 \\ -50 \end{pmatrix} \right\} . \] and we need to show that \[ W = \mbox{span} \left\{ \mathbf{u} = \begin{pmatrix} 4 \\ 4 \\ 2 \\ 2 \end{pmatrix} , \ \mathbf{v} = \begin{pmatrix} 5 \\ 2 \\ 0 \\ 8 \end{pmatrix} \right\} . \] This means that there exist four constants c₁, c₂, c₃, and c₄, not all equal to zero, so that \[ \begin{split} c_1 \mathbf{u} + c_2 \mathbf{v} &= \mathbf{u}_1 , \\ c_3 \mathbf{u} + c_4 \mathbf{v} &= \mathbf{u}_2 . \end{split} \] We ask Mathematica to solve the corresponding system of linear equations.
         LinearSolve[Transpose[A], {{182, 164, 76, 124}, {164, 236, 142, -50}}]
         {{73/3, 154/3, 104/3, -(112/3)}, {127/6, -(31/3), -(47/3), 205/3}, {0, 0, 0, 0}, {0, 0, 0, 0}}
   Since this system has a nontrivial solution, Mathematica confirms that column space of P coincides with W.
2. Let \[ W = \mbox{span}\left\{ \mathbf{w}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} , \quad \mathbf{w}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right\} \subset \mathbb{R}^{3\times 1} . \] This is a 2-dimensional subspace of ℝ^3×1. We need an orthonormal basis for W. First, we normalize w₁: \[ \mathbf{u}_1 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} . \] Then we make w₂ orthogonal to u₁: \[ w'_2 = w_2 - \left( \mathbf{u}_1 \bullet \mathbf{w}_2 \right) \mathbf{u}_1 = \begin{pmatrix} -1/2 \\ 1/2 \\ 1 \end{pmatrix} . \]
         u1 = {1, 1, 0}/Sqrt[2];        w2 = {0, 1, 1};        w2p = w2 - Dot[u1, w2]*u1
         {-(1/2), 1/2, 1}
   We normalize this vector
          u2 = w2p/Norm[w2p]
          {-(1/Sqrt[6]), 1/Sqrt[6], Sqrt[2/3]}
   \[ \mathbf{u}_2 = \frac{\mathbf{w}'_2}{\| \mathbf{w}'_2 \|} = \frac{1}{\sqrt{6}} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} . \] So an orthonormal basis for W is {u₁, u₂}.

   Now we build the projection matrix \[ \mathbf{P} = \mathbf{Q}\,\mathbf{Q}^{\mathrm T} = \frac{1}{3} \begin{bmatrix} 2 & 1 & -1 \\ 1&2&1 \\ -1 &1 & 2 \end{bmatrix} , \] where \[ \mathbf{Q} = \left\{ \mathbf{u}_1 , \mathbf{u}_2 \right\} = \frac{1}{\sqrt{6}} \begin{bmatrix} \sqrt{3}& -1 \\ \sqrt{3} & 1 \\ 0 & 2 \end{bmatrix} . \]

          Q = Transpose[{u1, u2}]
          {{1/Sqrt[2], -(1/Sqrt[6])}, {1/Sqrt[2], 1/Sqrt[6]}, {0, Sqrt[2/3]}}
          P = Q . Transpose[Q]
          {{2/3, 1/3, -(1/3)}, {1/3, 2/3, 1/3}, {-(1/3), 1/3, 2/3}}
   This is the projection matrix onto W. Of course, you can use formula (8), but it provides the same answer (Mathematica confirms)
          PP = Q . Inverse[Transpose[Q] . Q] . Transpose[Q]
          {{2/3, 1/3, -(1/3)}, {1/3, 2/3, 1/3}, {-(1/3), 1/3, 2/3}}
   To compute the null space of P, we need to solve the following system of linear equations: \[ \mathbf{P}\,\mathbf{x} = \mathbf{0} \qquad \iff \qquad \frac{1}{3} \begin{bmatrix} 2 & 1 & -1 \\ 1&2&1 \\ -1 &1 & 2 \end{bmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} . \] Fortunately, Mathematica has a dedicated comamnd to determine the null space:
          NullSpace[P]
          {{1, -1, 1}}
   Therefore, \[ \mbox{NullSpace}[\mathbf{P}] = \mbox{span} \left\{ \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \right\} . \] Now we compute W^⊥. This space consists of all vectors x = [x, y, z] that are orthogonal to w₁ and w₂: \[ \mathbf{w}_1 \bullet \mathbf{x} = 0 \quad \& \quad \mathbf{w}_2 \bullet \mathbf{x} = 0 \] So we need to solve the homogeneous system of equations \[ \begin{split} x + y &= 0 , \\ y + z & = 0 . \end{split} \] So y = −x = −z. Hence, W is spanned on one vector: [1, −1, 1], which is exactly the null space of matrix P.
3. Let \[ W = \mbox{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} , \quad \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} \right\} . \] Using formula (9), we find its standard matrix of the orthogonal projection onto W to be \[ \mathbf{P} = \frac{1}{6} \begin{bmatrix} 5 & 2 & -1 \\ 2&2&2 \\ -1&2&5 \end{bmatrix} . \]
          e1 = {1, 1, 1}/Sqrt[3]; e2 = {1, 0, -1}/Sqrt[2];
          Q = Transpose[{e1, e2}];
          P = Q . Transpose[Q]
          {{5/6, 1/3, -(1/6)}, {1/3, 1/3, 1/3}, {-(1/6), 1/3, 5/6}}
   Of course, you can use formula (8), but the answer will be the same:
          Q . Inverse[Transpose[Q] . Q] . Transpose[Q]
          {{5/6, 1/3, -(1/6)}, {1/3, 1/3, 1/3}, {-(1/6), 1/3, 5/6}}
   Verify that P² = P.
          P . P == P
         True
   Since cross product e₁ × e₂ is orthogonal to these vectors, the orthogonal complement W^⊥ is spanned on the vector \[ W^{\perp} = \mbox{span} \left\{ \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \right\} . \]
          Cross[e1, e2]
         {-(1/Sqrt[6]), Sqrt[2/3], -(1/Sqrt[6])}
4. We use matrix P from Eq.(9.1). We use Mathematica to determine eigenvalues and eigenvectors of matrix P:
         P = {{91/223, 82/223, 38/223, 62/223}, {82/223, 118/223, 71/223, -(25/223)}, {38/223, 71/223, 93/446, -(83/446)}, {62/223, -(25/223), -(83/446), 381/446}}
         Eigenvalues[P]
         {1, 1, 0, 0}
   So we see that projection matrix P has two eigenvalues λ = 1 and λ = 0, both of multiplicity 2 (not defective). The eigenvectors corresponding λ = 1 are \[ \mathbf{v}_1 = \begin{pmatrix} 5 \\ 2 \\ 0 \\ 8 \end{pmatrix} , \qquad \mathbf{v}_2 = \begin{pmatrix} 11 \\ 14 \\ 8 \\ 0 \end{pmatrix} \]
         Eigenvectors[P]
         {{5/8, 1/4, 0, 1}, {11/8, 7/4, 1, 0}, {-(7/3), 11/6, 0, 1}, {1/ 3, -(5/6), 1, 0}}
   We check that these vectors v₁ and v₂ are eigenvectors of matrix P:
         v1 = {5, 2, 0, 8};
         P . v1
         {5, 2, 0, 8}
         v2 = {11, 14, 8, 0};
         P . v2
         {11, 14, 8, 0}
   We need to show that vectors {v₁, v₂} span subspace W. So we need to show that vectors u = [4, 4, 2, 2] and v = 5, 2, 0, 8] are expressed as linear combination of vectors {v₁, v₂}. Since v = v₁, it is sufficient to show that vector u is a linear combination of {v₁, v₂}. So we need to show that the system of algebraic equations \[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{u} \] has a nontrivial solution. We deligate this job to mathematica:
         u = {4, 4, 2, 2};
         Solve[{c1*v1 + c2*v2 == u}, {c1, c2}]
         {{c1 -> 1/4, c2 -> 1/4}}
   Since c₁ = ¼ and c₂ = ¼ are not zero, eigenvectors {v₁, v₂} span subspace W.
5. From the previous example, we know that projection matrix (9.1) has double eigenvalue λ = 0 to which corresponds two eigenvectors \[ \mathbf{u}_1 = \begin{pmatrix} -14 \\ 11 \\ 0 \\ 6 \end{pmatrix} , \qquad \mathbf{u}_2 = \begin{pmatrix} 2 \\ -5 \\ 6 \\ 0 \end{pmatrix} . \] We check with Mathematica that these vectors belong to W^⊥:
         u1 = {-14, 11, 0, 6};
         P . u1
         {0, 0, 0, 0}
         u2 = {2, -5, 6, 0};
         P . u2
         {0, 0, 0, 0}
   Since eigenvectors {u₁, u₂} are linearly independent, they span W^⊥.
6. Using projection matrix (9.1), we build the matrix from its eigenvectors: \[ \mathbf{S} = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{u}_1 & \mathbf{u}_2 \end{bmatrix} = \begin{bmatrix} 5& 11& -14&2 \\ 2&14& 11 & -5 \\ 0&8&0&6 \\ 8&0&6&0 \end{bmatrix} . \] Now we calculate the product:
         S = {{5, 11, -14, 2}, {2, 14, 11, -5}, {0, 8, 0, 6}, {8, 0, 6, 0}};
         Inverse[S] . P . S
         {{1, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}}
   \[ \mathbf{S}^{-1} \mathbf{P}\,\mathbf{S} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} . \]

Example A: Reflection across a line in ℝ²

Example B: Reflection across a plane in ℝ³

Let plane U have unit normal vector n, written as a colum vector. Then the reflection matrix is \[ {\bf R}_U = {\bf I} - 2 \mathbf{n}\,\mathbf{n}^{\mathrm T} . \] For instance, reflection across the plane x₂ = 0, with normal vector n = [0, 1, 0] is \[ {\bf R}_U = \begin{bmatrix} 1& \phantom{-}0 & 0 \\ 0&-1&0 \\ 0& \phantom{-}0 & 1 \end{bmatrix} . \] So n = [0, 1, 0], and we have \[ {\bf R}_U \left( x_1 , x_2 , x_3 \right) = \left[ x_1 , - x_2 , x_3 \right] . \]

Now we consider a general plane in ℝ³.

Example C: General Reflection across a subspace in ℝⁿ

For instance, let us consider a 2-dimensional subspace U in ℝ⁴, which is spanned on orthonormal vectors \[ {\bf e}_1 = \frac{1}{\sqrt{5}} \left[ 2, \ 1, \ 0 , 0 \right] , \qquad {\bf e}_2 = \frac{1}{\sqrt{10}} \left[ 0, \ 1, \ 3, \ 0 \right] . \] The reflection matrix across the subspace generated by these unit vectors is defined by Eq.\eqref{EqDot.9}. So we use Mathematica to generate the subroutine:

1. For a given target point (4, 3, 2, 1) in ℝ⁴, find the point in \( \displaystyle \quad \mbox{span} \left\{ \begin{pmatrix} 1 \\ -1 \\ 2 \\ 3 \end{pmatrix} , \ \begin{pmatrix} 0 \\ 1 \\ 3 \\ 2 \end{pmatrix} \right\} , \quad \) which is closest to the traget point.
2. For a given target point (4, 3, 2, 1) in ℝ⁴, find the point in \( \displaystyle \quad \mbox{span} \left\{ \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \end{pmatrix} , \ \begin{pmatrix} 0 \\ 3 \\ 2 \\ 0 \end{pmatrix} \right\} , \quad \) which is closest to the traget point.
3. For a given target point (4, 3, 2, 1) in ℝ⁴, find the point on the plane described by 2x + y −3z + 4w = 0 which is closest to the traget point.

1. Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
2. Beezer, R., A First Course in Linear Algebra, 2015.
3. Beezer, R., A Second Course in Linear Algebra, 2013.