Week 3: Inner-Product Spaces

Week 3: Inner-Product Spaces#

Key Terms#

  • Vector spaces with inner product and norm

  • \(\mathbb{R}^n\) and \(\mathbb{C}^n\)

  • Projections on the line

  • Orthonormal bases

  • The Gram-Schmidt procedure

  • Orthogonal and unitary matrices

Preparation and Syllabus#

Exercises – Long Days#

1: Orthonormal Basis (Coordinates). By Hand.#

Question a#

Do the vectors

\[\begin{equation*} \pmb{u}_1=\left( \frac{1}{3},\frac{2}{3},\frac{2}{3}\right), \quad \pmb{u}_2=\left( \frac{2}{3}, \frac{1}{3}, -\frac{2}{3}\right),\quad \pmb{u}_3=\left( \frac{2}{3}, -\frac{2}{3}, \frac{1}{3}\right) \end{equation*}\]

constitute an orthonormal basis in \(\mathbb{R}^3\)?

Hint

The vectors must be pairwise orthogonal, meaning each of them must be perpendicular to each of the others. How is it that we check for that?

Hint

Two vectors are orthogonal precisely when their scalar product is 0, but more is needed to ensure orthonormality.

Hint

A basis is orthonormal if the vectors are mutually orthogonal and each of them has a length (norm) of 1.

Answer

\(\langle \pmb{u}_1, \pmb{u}_2 \rangle = \langle \pmb{u}_1, \pmb{u}_3 \rangle = \langle \pmb{u}_2, \pmb{u}_3 \rangle = 0\) and all three vectors have a norm of 1, so the three vectors constitute an orthonormal basis for \(\mathbb{R}^3\).

Question b#

Consider the vector \(\pmb{x} = [1,2,3]^T\). Compute the inner products \(\langle \pmb{x}, \pmb{u}_k \rangle\) for \(k=1,2,3\).

Hint

This is just the usual dot products \(\pmb{x} \cdot \pmb{u}_k\)

Question c#

Let us denote the basis by \(\beta = \pmb{u}_1, \pmb{u}_2, \pmb{u}_3\). State the coordinate vector \({}_{\beta} \pmb{x}\) for \(\pmb{x}\) with respect to \(\beta\). Compute the norm of both \(\pmb{x}\) and the coordinate vector \({}_{\beta} \pmb{x}\).

Question d#

Create the \(3\times 3\) matrix \(U = [\pmb{u}_1 \vert \pmb{u}_2 \vert \pmb{u}_3]\) with \(\pmb{u}_1, \pmb{u}_2, \pmb{u}_3\) as the three columns. Compute \(U^T \pmb{x}\) and compare the result with the previous exercise.

2: Orthonormal Basis (Construction). By Hand.#

Create an orthonormal basis in \(\mathbb{R}^3\), in which

\[\begin{equation*} \left(\frac {\sqrt 2}2,\frac {\sqrt 2}2,0\right) \end{equation*}\]

is the first basis vector.

Hint

You can perhaps guess a unit vector that is perpendicular to the given vector.

Hint

What about \(\left(0,0,1\right)\)?

Hint

Can you guess another one? Try starting out by finding a vector that is perpendicular to both the given vector and to \(\left(0,0,1\right)\).

Hint

What about \(\left(1,-1,0 \right)\)? And now, what are missing?

Hint

\(\left(1,-1,0 \right)\) is to be normalized.

Answer

A suggestion of an orthonormal basis can be

\[\begin{equation*} \left(\frac {\sqrt 2}2,\frac {\sqrt 2}2,0\right), \left(0,0,1\right), \left(\frac {\sqrt 2}2,-\frac {\sqrt 2}2,0\right), \end{equation*}\]

but others are possible! Think about why that is the case.

3: Orthonormalizing. By Hand.#

Determine the solution set to the homogeneous equation

\[\begin{equation*} x_1+x_2+x_3=0 \end{equation*}\]

and justify for it being a subspace in \(\mathbb{R}^3\). Find an orthonormal basis for this solution space.

Hint

If we are to find an orthonormal basis for the solution space, we must first find a basis for the solution space, which means we must solve the equation.

Hint

Let \(x_2=t_1\) and \(x_3=t_2\).

Hint

With \(x_2=t_1\) and \(x_3=t_2\) we find that

\[\begin{equation*} \begin{bmatrix} x_1 \newline x_2 \\ x_3 \end{bmatrix}=t_1\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}+t_2\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix},\quad (t_1,t_2)\in\mathbb{R}^2, \end{equation*}\]

but what is then the solution space?

Hint

The solution space is \(\mathrm{span}\lbrace(-1,1,0),(-1,0,1)\rbrace\), so a basis can be constituted by the vectors

\[\begin{equation*} \pmb{v}_1=\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\,\,\,\mathrm{and}\,\,\,\pmb{v}_2=\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}. \end{equation*}\]

Now they just need orthonormalization.

Hint

When two vectors are to be orthonormalized, we must turn them relative to one another within the space they span, such that they become orthogonal (are perpendicular to one another) while still spanning the same space, and we must also normalize them, meaning extend or shorten them so that they all have a length of 1.

Hint

Luckily we can do that by simply following the Gram-Schmidt method.

Answer

An orthonormal basis for the solution space to the equation can be constituted by the vectors

\[\begin{equation*} \pmb{u}_1=\begin{bmatrix} -\frac{\sqrt 2}{2} \\ \frac{\sqrt 2}{2} \\ 0 \end{bmatrix}\quad\mathrm{and}\quad \pmb{u}_2=\begin{bmatrix} -\frac{\sqrt 6}{6} \\ -\frac{\sqrt 6}{6} \\ \frac{\sqrt 6}{3} \end{bmatrix}. \end{equation*}\]

4: Orthogonal Projections#

Let \(\boldsymbol{y}=(2,1,2) \in \mathbb{R}^3\) be given. Then the projection of \(\boldsymbol{x} \in \mathbb{R}^3\) onto the line \(Y = \mathrm{span}\{\boldsymbol{y}\}\) is given by:

\[\begin{equation*} \operatorname{Proj}_Y(\boldsymbol{x}) = \frac{\left<\boldsymbol{x},\boldsymbol{y} \right>}{\left<\boldsymbol{y},\boldsymbol{y} \right>}\boldsymbol{y} = \left<\boldsymbol{x},\boldsymbol{u}\right>\boldsymbol{u}, \end{equation*}\]

where \(\boldsymbol{u} = \frac{\boldsymbol{y}}{||\boldsymbol{y}||}\).

Question a#

Let \(\boldsymbol{x} = (1,2,3) \in \mathbb{R}^3\). Compute \(\operatorname{Proj}_Y(\boldsymbol{x})\), \(\operatorname{Proj}_Y(\boldsymbol{y})\) and \(\operatorname{Proj}_Y(\boldsymbol{u})\).

Question b#

We, as usual, consider all vectors as column vectors. Now set up the \(3 \times 3\) matrix \(P = \boldsymbol{u} \boldsymbol{u}^T\) and compute both \(P\boldsymbol{x}\) and \(P\boldsymbol{y}\).

5: An Orthonormal Basis for a Subspace of \(\mathbb{C}^4\)#

Question a#

Find an orthonormal basis \(\pmb{u}_1, \pmb{u}_2\) for the subspace \(Y = \mathrm{span}\{\pmb{v}_1, \pmb{v}_2\}\) spanned by the vectors:

v1 = Matrix([I, 1, 1, 0])
v2 = Matrix([0, I, I, sqrt(2)])

Hint

It can be found using the Gram-Schmidt algorithm. Check your calculation with the SymPy command GramSchmidt([v1,v2], orthonormal=True).

Question b#

Let

\[\begin{equation*} \pmb{x} = \left[\begin{matrix}3 i\\3 - 2 i\\3 - 2 i\\-2\sqrt{2}\end{matrix}\right]. \end{equation*}\]

Compute \(\langle \pmb{x}, \pmb{u}_1 \rangle\), \(\langle \pmb{x}, \pmb{u}_2 \rangle\) as well as

\[\begin{equation*} \langle \pmb{x}, \pmb{u}_1 \rangle \pmb{u}_1 + \langle \pmb{x}, \pmb{u}_2 \rangle \pmb{u}_2 . \end{equation*}\]

What does this linear combination give? Does \(\pmb{x}\) belong to the subspace \(Y\)?

6: A Python Algorithm#

Question a#

Consider the following code and explain what it does. Before you run the code in a Jupyter Notebook you must explain what will be the output.

from sympy import *
from dtumathtools import *
init_printing()

x1, x2, x3 = symbols('x1:4', real=True)
eqns = [Eq(1*x1 + 2*x2 + 3*x3, 1), Eq(4*x1 + 5*x2 + 6*x3, 0), Eq(5*x1 + 7*x2 + 8*x3, -1)]
eqns

A, b = linear_eq_to_matrix(eqns,x1,x2,x3)
T = A.row_join(b)  # augmented matrix

A, b, T

Question b#

We continue the Jupyter Notebook with the following code (do not run it yet). Go through the code by hand (calculate the for-loop iterations manually). Which \(T\) matrix will be the output? Copy/paste the code into a chatbot such as https://copilot.microsoft.com/ (log in with your DTU account) and ask the chatbot to explain the code line by line. Now check the result by running the code in a Python Notebook. Remember that T.shape[0] gives the number of rows in matrix \(T\).

for col in range(T.shape[0]):
    for row in range(col + 1, T.shape[0]):
        T[row, :] = T[row, :] - T[row, col] / T[col, col] * T[col, :]
    T[col, :] = T[col, :] / T[col, col]

T

Question c#

Write Python code that ensures zeroes above the diagonal in matrix \(T\) such that \(T\) ends out being in reduced-row echelon form.

Note

You should not take into account any potential divisions by zero (for general \(T\) matrices). We will assume that the calculations will work well.

Hint

You will need two for-loops, for instance:

for col in range(T.shape[0] - 1, -1, -1):
    for row in range(col - 1, -1, -1):

Remember that index \(-1\) in Python refers to the last element.

Hint

Ask a chatbot such as https://copilot.microsoft.com/ (log in with your DTU account) for help if you get stuck. You should first work with the exercise on your own, though.

Answer

# Backward Elimination: Create zeros above the diagonal
for col in range(T.shape[0] - 1, -1, -1):
    for row in range(col - 1, -1, -1):
        T[row, :] = T[row, :] - T[row, col] * T[col, :]

T

Question d#

What algorithm is it that we have implemented? Test the same algorithm on:

x1, x2, x3, x4 = symbols('x1:5', real=True)
eqns = [Eq(1*x1 + 2*x2 + 3*x3, 1), Eq(4*x1 + 5*x2 + 6*x3, 0), Eq(4*x1 + 5*x2 + 6*x3, 0), Eq(5*x1 + 7*x2 + 8*x3, -1)]
A, b = linear_eq_to_matrix(eqns,x1,x2,x3,x4)
T = A.row_join(b)  # augmented matrix

Answer

It is Gaussian elimination for solving linear equation systems, see also Mat1ha. The output is the reduced row-echelon form of the input matrix. We have not taken into account column switches, nor row switches, and division by zero is thus a risk in case the first \(n\) columns of an \(n \times m\) matrix \(T\) are not linearly independent.

7: Orthogonal Polynomials#

This is an exercise from the textbook. You can find help there.

Consider the list \(\alpha=1,x,x^{2},x^{3}\) of polynomials in \(P_{3}([-1,1])\) constructed with the \(L^{2}\)-inner product.

Question a#

Argument for why \(\alpha\) is a list of linearly independent vectors.

Hint

The assumption is that \(c_1 + c_2 x + c_3 x^2 + c_4 x^3 = 0\) for all \(x \in [-1,1]\). We want to show that \(c_1 = c_2 = c_3 = c_4 = 0\), since we then will have shown that the polynomials in \(\alpha\) are linearly independent.

Hint

Choose four different \(x\) values and evaluate the assumption for that choicethere. That will give four equations of the four unknowns \(c_1, c_2, c_3, c_4\). The \(x\) values must be chosen such that the equation system only has the zero solution. (Why?)

Hint

We choose for instance \(x = 0, 1, -1, 1/2\). This results in the following equations: \(c_{1} = 0, c_{1} + c_{2} + c_{3} + c_{4} = 0, c_{1} - c_{2} + c_{3} - c_{4} = 0, c_{1} + \frac{c_{2}}{2} + \frac{c_{3}}{4} + \frac{c_{4}}{8} = 0\)

Hint

Check that the four equations only have the zero solution. Conclude that the four polynomials are linearly independent.

Question b#

Utilize the Gram-Schmidt procedure on \(\alpha\) and show that the procedure gives a normalized version of the Legendre polynomials.

Hint

To compute the norms and the inner products you need to use integrate in SymPy (or compute the integrals by hand). For example x = symbols('x', real=True) and integrate(x**2, (x,-1,1)).

Hint

The polynomial \(u_1\) can be computed by:

x = symbols('x', real=True)
v1 = 1
v2 = x
v3 = x**2
v4 = x**3
v1_norm = sqrt(integrate(v1**2, (x,-1,1)))
u1 = v1/v1_norm
u1

Hint

The polynomial \(u_2\) can be computed by:

w2 = v2 - integrate(v2*u1, (x,-1,1)) * u1
u2 = w2/sqrt(integrate(w2**2, (x,-1,1)))
u2

Hint

The polynomial \(u_3\) can be computed by:

w3 = v3 - integrate(v3*u1, (x,-1,1)) * u1 - integrate(v3*u2, (x,-1,1)) * u2
u3 = w3/sqrt(integrate(w3**2, (x,-1,1)))

Hint

The polynomial \(u_4\) can be computed by:

w4 = v4 - integrate(v4*u1, (x,-1,1)) * u1 - integrate(v4*u2, (x,-1,1)) * u2 - integrate(v4*u3, (x,-1,1)) * u3
u4 = w4/sqrt(integrate(w4**2, (x,-1,1)))

Answer

The answer is achieved by running the code given in the above hints. Compare your answer with legendre(0,x), legendre(1,x), legendre(2,x).factor(), legendre(3,x).factor(). They should differ by a scaling factor.

Exercises – Short Day#

1: Matrix Multiplications. By Hand.#

Define

\[\begin{equation*} A = \left[\begin{matrix}1 & 2 & 3 & 4\\5 & 6 & 7 & 8\\4 & 4 & 4 & 4\end{matrix}\right], \quad \pmb{x} = \left[\begin{matrix} x_1\\ x_2\\ x_3\\ x_4\end{matrix}\right] = \left[\begin{matrix}1\\2\\-1\\1\end{matrix}\right]. \end{equation*}\]

Let \(\pmb{a}_1, \pmb{a}_2,\pmb{a}_3,\pmb{a}_4\) denote the columns of \(A\). Let \(\pmb{b}_1, \pmb{b}_2,\pmb{b}_3\) denote the rows of \(A\). We will now calculate \(A\pmb{x}\) in two different ways.

Question a#

Method 1: As a linear combination of the columns. Compute the linear combination

\[\begin{equation*} x_1 \pmb{a}_1 + x_2 \pmb{a}_2 + x_3 \pmb{a}_3 + x_4 \pmb{a}_4. \end{equation*}\]

Question b#

Method 2: As a “dot product” of the rows of \(A\) by \(\pmb{x}\). Compute

\[\begin{equation*} \left[\begin{matrix} \pmb{b}_1 \pmb{x} \\ \pmb{b}_2 \pmb{x} \\ \pmb{b}_3 \pmb{x} \end{matrix}\right]. \end{equation*}\]

Note

Since \(\pmb{b}_k\) is a row vector, \((\pmb{b}_k)^T\) is a column vector. The product \(\pmb{b}_k \pmb{x}\) hence corresponds to the dot product of \(\pmb{x}\) and \((\pmb{b}_k)^T\).

Question c#

Compute \(A\pmb{x}\) using SymPy and compare with your calculations in the previous exercises.

2: A Subspace in \(\mathbb{C}^4\) and its Orthogonal Complement#

In \(\mathbb{C}^4\,\) the following vectors are given

\[\begin{equation*} \pmb{v}_1=(1,1,1,1),\,\pmb{v}_2=(3 i ,i,i,3 i),\,\pmb{v}_3=(2,0,-2,4)\,\,\,\,\mathrm{and}\,\,\,\,\pmb{v}_4=(4-3i,2-i,-i,6-3i). \end{equation*}\]

A subspace \(Y\) in \(\mathbb{C}^4\) is determined by \(Y=\mathrm{span}\lbrace\pmb{v}_1,\pmb{v}_2,\pmb{v}_3,\pmb{v}_4\rbrace\).

Question a#

v1 = Matrix([1,1,1,1])
v2 = Matrix([3*I,I,I,3*I])
v3 = Matrix([2,0,-2,4])
v4 = Matrix([4-3*I,2-I,-I,6-3*I])

Run the command GramSchmidt([v1,v2,v3,v4], orthonormal=True) in Python. What does Python tell you?

# GramSchmidt([v1, v2, v3, v4], orthonormal = True)

Question b#

Now show that \((\pmb{v}_1,\pmb{v}_2,\pmb{v}_3)\,\) is a basis for \(Y\), and determine the coordinate vector of \(\pmb{v}_4\,\) with respect to this basis.

Answer

The vectors \(\pmb{v}_1\), \(\pmb{v}_2\), and \(\pmb{v}_3\) are linearly independent, while \(\pmb{v}_4=2\pmb{v}_1-\pmb{v}_2+\pmb{v}_3\), so

\[\begin{equation*} {}_v\pmb{v}_4=\begin{bmatrix} 2 \\ -1 \newline 1 \end{bmatrix} \end{equation*}\]

constitutes the coordinate vector with respect to the \(v\) basis.

Question c#

State an orthonormal basis for \(Y\).

Hint

The vectors \(\pmb{v}_1\), \(\pmb{v}_2\), and \(\pmb{v}_3\) constitute a basis for \(Y\), so they must be orthonormalized using Gram-Schmidt.

Answer

\[\begin{equation*} (\pmb{q}_1,\pmb{q}_2,\pmb{q}_3)=\left( \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{bmatrix},\begin{bmatrix} \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \\ \frac{1}{2} \end{bmatrix},\begin{bmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ \frac{1}{2} \end{bmatrix}\right) \end{equation*}\]

constitutes an orthonormal basis for \(Y\).

Question d#

Find the coordinate vector of \(\pmb{v}_4 \in Y\) with respect to the orthonormal basis for \(Y\).

Hint

This can be computed using inner products (as usual) or via a fitting matrix-vector product.

Question e#

Determine the orthogonal complement \(Y^\perp\) in \(\mathbb{C}^4\) to \(Y\).

Hint

First, try guessing a vector and check whether it is located within \(Y\). That is, check whether it is linearly independent to the basis for \(Y\).

Hint

You could for example guess on \((1,0,0,0)\).

Hint

To be within the orthogonal compliment, we must find a vector that is orthogonal to the basis vectors in \(Y\).

Hint

Use Gram-Shmidt by continuing from Question b and just add the vector you have guessed above. You do not need to normalize the vector in the final step. (Why not?)

Answer

The orthogonal compliment is a 1-dimensional subspace of \(\mathbb{C}^4\) that is spanned by

\[\begin{equation*} \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix}. \end{equation*}\]

Question f#

Choose a vector \(\pmb{y}\) in \(Y^\perp\) and choose a vector \(\pmb{x}\) in \(Y\). Compute \(\Vert \pmb{x} \Vert\), \(\Vert \pmb{y} \Vert\) and \(\Vert \pmb{x} + \pmb{y} \Vert\). Check that \(\Vert \pmb{x} \Vert^2 +\Vert \pmb{y} \Vert^2 = \Vert \pmb{x} + \pmb{y} \Vert^2\).

3: Orthogonal Projection on a Plane#

Let a matrix \(U = [\pmb{u}_1, \pmb{u}_2]\) be given by:

\[\begin{equation*} U = \left[\begin{matrix}\frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\\ \frac{\sqrt{3}}{3} & 0\\- \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2}\end{matrix}\right]. \end{equation*}\]

Question a#

Show that \(\pmb{u}_1, \pmb{u}_2\) is an orthonormal basis for \(Y = \mathrm{span}\{\pmb{u}_1, \pmb{u}_2\}\).

Answer

A calculation will show that \(\langle \pmb{u}_1, \pmb{u}_2 \rangle = 0\) and \(\Vert \pmb{u}_1 \Vert = \Vert \pmb{u}_2 \Vert = 1\). This shows that \(\pmb{u}_1, \pmb{u}_2\) is an orthonormal list of vectors. This list is thus linearly independent and hence a basis for \(Y\). We conclude that \(\pmb{u}_1, \pmb{u}_2\) is an orthonormal basis for \(Y\).

Question b#

Let \(P = U U^* \in \mathbb{R}^{3 \times 3}\). This will give us a projection matrix that describes the orthogonal projection \(\pmb{x} \mapsto P \pmb{x}\), \(\mathbb{R}^3 \to \mathbb{R}^3\) onto the plane \(Y = \mathrm{span}\{\pmb{u}_1, \pmb{u}_2\}\). Verify that \(P^2 = P\), \(P \pmb{u}_1 = \pmb{u}_1\), and \(P \pmb{u}_2 = \pmb{u}_2\).

Answer

\[\begin{equation*} P = \left[\begin{matrix}\frac{5}{6} & \frac{1}{3} & \frac{1}{6} \\ \frac{1}{3} & \frac{1}{3} & - \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{3} & \frac{5}{6} \end{matrix}\right] \end{equation*}\]

From this it is easy to show that \(P \pmb{u}_1 = \pmb{u}_1\), and \(P \pmb{u}_2 = \pmb{u}_2\). In fact it applies that \(P (c_1 \pmb{u}_1 + c_2 \pmb{u}_2) = c_1 \pmb{u}_1 + c_2 \pmb{u}_2\) for all choices of \(c_1, c_2 \in \mathbb{R}\).

Question c#

Choose a vector \(\pmb{x} \in \mathbb{R}^3\) that does not belong to \(Y\), and find the projection \(\mathrm{proj}_Y(\pmb{x})\) of \(\pmb{x}\) down onto the plane \(Y\). Illustrate \(\pmb{x}\), \(Y\), and \(\mathrm{proj}_Y(\pmb{x})\) in a plot.

Answer

One can for instance choose \(\pmb{x} = [1,2,3]^T\). One must then compute \(\mathrm{proj}_Y(\pmb{x}) = P \pmb{x} = [2,0,2]^T\). Note that \([2,0,2]^T = 2\sqrt{2} \pmb{u}_2 \in Y\).

U = Matrix([[sqrt(3)/3, sqrt(2)/2], [sqrt(3)/3, 0], [-sqrt(3)/3, sqrt(2)/2]])
P = U * U.T
x = Matrix([1,2,3])
t1, t2 = symbols('t1 t2')
r = t1 * U[:,0] + t2 * U[:,1]
p = dtuplot.plot3d_parametric_surface(r[0], r[1], r[2], (t1, -5, 5), (t2, -5, 5), show = False, rendering_kw = {"alpha": 0.5})  
p.extend(dtuplot.scatter((x, P*x), show = False))
p.xlim = [-5,5]
p.ylim = [-5,5]
p.zlim = [-5,5]

p.show()

Question d#

Show that \(\pmb{x} - \mathrm{proj}_Y(\pmb{x})\) belongs to \(Y^\perp\).

Hint

With the choice \(\pmb{x} = [1,2,3]^T\) we achieve \(\pmb{y} := \pmb{x} - \mathrm{proj}_Y(\pmb{x}) = [-1,2,1]^T\). You must show that the \(\pmb{y}\) vector is orthogonal to all vectors in \(Y\).

Hint

An arbitrary vector in \(Y\) can be written as \(c_1 \pmb{u}_1 + c_2 \pmb{u}_2\), where \(c_1, c_2 \in \mathbb{R}\).

Hint

Show that \(\langle \pmb{y}, \pmb{u}_1 \rangle = 0\) and \(\langle \pmb{y}, \pmb{u}_2 \rangle = 0\)

Answer

Since \(\langle c_1 \pmb{u}_1 + c_2 \pmb{u}_2, \pmb{y} \rangle = c_1 \langle \pmb{u}_1 , \pmb{y} \rangle + c_2 \langle \pmb{u}_2 , \pmb{y} \rangle = 0 + 0 = 0\) we see that \(\pmb{y} = [-1,2,1]^T \in Y^\perp\).

4: Unitary Matrices#

Let a matrix \(F\) be given by:

n = 4
F = 1/sqrt(n) * Matrix(n, n, lambda k,j: exp(2*pi*I*k*j/n))
F
\[\begin{split}\displaystyle \left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & \frac{i}{2} & - \frac{1}{2} & - \frac{i}{2}\\\frac{1}{2} & - \frac{1}{2} & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & - \frac{i}{2} & - \frac{1}{2} & \frac{i}{2}\end{matrix}\right]\end{split}\]

Determine whether the following propositions are true or false:

  1. \(F\) is unitary

  2. \(F\) is invertible

  3. \(F\) is orthogonal

  4. \(F\) is symmetric

  5. \(F\) is Hermitian

  6. The columns of \(F\) constitute an orthonormal basis for \(\mathbb{C}^4\)

  7. The columns of \(F\) constitute an orthonormal basis for \(\mathbb{R}^4\)

  8. \(-F = F^{-1}\)

Answer

  1. Yes

  2. Yes

  3. No

  4. Yes

  5. No

  6. Yes

  7. No

  8. No

Remark: The matrix is called the Fourier matrix. It fulfills for all \(n\) that: \(\overline{F} = F^* = F^{-1}\) and \(F = F^T\). The map \(\pmb{x} \mapsto F^* \pmb{x}\) is called the descrete Fourier transform and is important in many parts of engineering science. See for instance https://bootcamp.uxdesign.cc/the-most-important-algorithm-of-all-time-9ff1659ff3ef - Fast Fourier transform is the same map as \(\pmb{x} \mapsto F^* \pmb{x}\), just implemented in a smart way.

Contents