Week 3: Exercises#
The exercises are intended to be done by hand unless otherwise stated (such as when you are asked to plot a graph or run a script).
Exercises – Long Day#
from sympy import*
from dtumathtools import*
init_printing()
def inner(x1, x2):
'''
Computes the standard inner product of two vectors.
'''
return x1.dot(x2, conjugate_convention = 'right')
1: Orthonormal Basis (Coordinates)#
Question a#
Do the vectors
constitute an orthonormal basis in \(\mathbb{R}^3\)?
Hint
The vectors need to be pairwise orthogonal, meaning they must be perpendicular to each other. How do we check for that?
Hint
Two vectors are orthogonal, precisely when their scalar product (dot product) is 0. But more is needed for them to be orthonormal.
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 \(\pmb{x} = [1,2,3]^T\). Calculate the inner products \(\langle \pmb{x}, \pmb{u}_k \rangle\) for \(k=1,2,3\).
Hint
These are just the usual dot products \(\pmb{x} \cdot \pmb{u}_k\).
Question c#
Let’s denote the basis by \(\beta = \pmb{u}_1, \pmb{u}_2, \pmb{u}_3\). State the coordinate vector \({}_{\beta} \pmb{x}\) of \(\pmb{x}\) with respect to \(\beta\). Calculate the norm of both \(\pmb{x}\) and the coordinate vector \({}_{\beta} \pmb{x}\).
u1 = Matrix([1,2,2])/3
u2 = Matrix([2,1,-2])/3
u3 = Matrix([2,-2,1])/3
x = Matrix([1,2,3])
inner(x, u1), inner(x, u2), inner(x, u3)
# Calculated in the previous question. Check:
U = Matrix.hstack(u1, u2, u3) # Change-of-basis matrix from the orthonormal beta basis til the standard basis
e_x = x
beta_x = U.T * e_x # U is unitary, so U^T = U^(-1) is the change-of-basis matrix from the standard basis to the orthonormal beta basis
beta_x
# Another check
inner(x, u1) * u1 + inner(x, u2) * u2 + inner(x, u3) * u3
x.norm(), beta_x.norm()
Since \(U^T\) is unitary, the norm is kept by all vectors.
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 its three columns. Calculate \(U^T \pmb{x}\) and compare the result with that from the previous question.
2: Orthonormal Basis (Construction)#
Create an orthonormal basis for \(\mathbb{R}^3\), within which
is the first basis vector.
Hint
You can, for example, start by guessing on a unit vector that is perpendicular to the given vector.
Hint
What about \(\left(0,0,1\right)\)?
Hint
Can you guess one more? Simply start by finding a vector that is perpendicular to both the given vector and \(\left(0,0,1\right)\).
Hint
What about \(\left(1,-1,0 \right)\)? What are we still missing now?
Hint
\(\left(1,-1,0 \right)\) needs to be normalized.
Answer
A suggestion for an orthonormal basis is
but there are other possibilities! Think about why that is the case.
3: Orthonormalization#
Find the solution set to the homogeneous equation
and justify that it constitutes 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 first need to find a basis for the solution space, meaning 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
How do we form a basis for this solution space?
Hint
The solution space is \(\operatorname{span}\lbrace(-1,1,0),(-1,0,1)\rbrace\), so a basis can be formed by the vectors
Now they just need orthonormalization.
Hint
When two vectors need to be orthonormalized, we must first rotate them relative to each other within the space they span so that they become orthogonal (perpendicular to each other) while still spanning the same space. Secondly, they must be normalized, meaning they must either be lengthened or shortened so that they all have a length of 1.
Hint
Fortunately, we can simply carry this out by following the Gram-Schmidt procedure.
Answer
An orthonormal basis for the solution space of the equation is formed by the vectors
x1, x2, x3 = symbols('x1 x2 x3')
eqn = Eq(x1 + x2 + x3, 0)
A, b = linear_eq_to_matrix(eqn, [x1, x2, x3])
# linsolve((A, b), x1, x2, x3) # alternativt
A.nullspace()
v1, v2 = A.nullspace()
u1 = v1/v1.norm()
w = v2 - inner(v2, u1) * u1
u2 = w/w.norm()
u1, u2
# With Sympy
GramSchmidt([v1, v2], orthonormal = True)
4: Orthogonal Projections#
Let \(\boldsymbol{y}=(2,1,2) \in \mathbb{R}^3\) be given. Then the projection of \(\boldsymbol{x} \in \mathbb{R}^3\) on the line \(Y = \operatorname{span}\{\boldsymbol{y}\}\) is given by:
where \(\boldsymbol{u} = \frac{\boldsymbol{y}}{||\boldsymbol{y}||}\).
Question a#
Let \(\boldsymbol{x} = (1,2,3) \in \mathbb{R}^3\). Calculate \(\operatorname{Proj}_Y(\boldsymbol{x})\), \(\operatorname{Proj}_Y(\boldsymbol{y})\) and \(\operatorname{Proj}_Y(\boldsymbol{u})\).
Question b#
As usual, we consider all vectors as column vectors. Now, find the \(3 \times 3\) matrix \(P = \pmb{u} \pmb{u}^T\) and compute both \(P\boldsymbol{x}\) and \(P\boldsymbol{y}\).
y = Matrix([2,1,2])
x = Matrix([1,2,3])
u = y.normalized()
u, inner(x, u) * u, inner(y, u) * u, inner(u, u) * u
# Note that y (and u) already belong to the line, so y = proj_Y(y)
y == inner(y, u) * u
True
P = u * u.T
P, P * x, P * y
# P is a projection matrix, so P^2 = P
P * P == P
True
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 \(Y = \operatorname{span}\{\pmb{v}_1, \pmb{v}_2\}\) that is spanned by the vectors:
v1 = Matrix([I, 1, 1, 0])
v2 = Matrix([0, I, I, sqrt(2)])
Hint
You can find it using the Gram-Schmidt procedure. Check your result with the Sympy command GramSchmidt([v1,v2], orthonormal=True).
Question b#
Let
Calculate \(\langle \pmb{x}, \pmb{u}_1 \rangle\), \(\langle \pmb{x}, \pmb{u}_2 \rangle\) as well as
What does this linear combination give? Does \(\pmb{x}\) belong to the subspace \(Y\)?
x = 3 * v1 - 2 * v2 # This is how x is constructed in the exercise
x
u1 = v1.normalized()
w = v2 - inner(v2, u1) * u1
u2 = w.normalized()
w.norm(), u1, u2
inner(x, u1), inner(x, u2)
expand(inner(x, u1)), expand(inner(x, u2))
expand(inner(x, u1) * u1 + inner(x, u2) * u2)
\(\pmb{x}\) belongs to \(Y\) since \(\pmb{x}\) can be written as a linear combination of \(\pmb{u}_1\) and \(\pmb{u}_2\).
6: A Python Algorithm#
Question a#
Without running the following code, explain what it does and explain what you expect the output to be. Then run the code in a Jupyter Notebook.
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 will continue the Jupyter Notebook by adding the following code (as before, do not run it yet). Go through the code by hand (think through the for loops). What \(T\) matrix will be the result? Copy-paste the code into an AI tool, such as https://copilot.microsoft.com/ (log in with your DTU account), and ask it to explain the code line by line. Verify the result by running the code in a Jupyter Notebook. Remember that T.shape[0] gives the number of rows in the 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 zeros above the diagonal in the matrix \(T\) so that \(T\) ends up in reduced row-echelon form.
Note
Do not take into account any divisions by zero (for general \(T\) matrices). We will here assume that the computations are possible.
Hint
You will need two for loops, such as:
for col in range(T.shape[0] - 1, -1, -1):
for row in range(col - 1, -1, -1):
Remember that index \(-1\) in Python gives the last element.
Hint
Ask an AI tool, such as https://copilot.microsoft.com/ (log in with your DTU account), for help if you get stuck. You should, though, first attempt to work on the exercise on your own.
Answer
# Backwards 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 kind of algorithm have we 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
This is the Gaussian elimination procedure for solving linear equation systems (review the topic in the Mathematics 1a textbook). The output is the reduced row-echelon form of the input matrix. We have not taken column switches (nor row switches) into account and thus there is a risk of division by zero if 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 \(P_{3}([-1,1])\) equipped with the \(L^{2}\) inner product.
Question a#
Argue that \(\alpha\) 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 will show that \(c_1 = c_2 = c_3 = c_4 = 0\), since that will imply that the polynomials in \(\alpha\) are linearly independent.
Hint
Choose four different \(x\) values and evalute the assumption for those. Then we get four equations in 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 fx \(x = 0, 1, -1, 1/2\). This gives 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.
The students are probably not used to working with linearly independent functions. And they likely don’t know, for example, that the monomial basis is linearly independent. This follows, of course, from “The fundamental theorem of algebra”, taught in Math1a (although it was not proven). The task here can be solved with the following steps:
x = symbols('x', real=True)
c1,c2,c3,c4 = symbols('c1:5', real=True)
v1 = 1
v2 = x
v3 = x**2
v4 = x**3
eq1 = Eq(c1*v1 + c2*v2.subs(x,0) + c3*v3.subs(x,0) + c4*v4.subs(x,0),0)
eq2 = Eq(c1*v1 + c2*v2.subs(x,1) + c3*v3.subs(x,1) + c4*v4.subs(x,1),0)
eq3 = Eq(c1*v1 + c2*v2.subs(x,-1) + c3*v3.subs(x,-1) + c4*v4.subs(x,-1),0)
eq4 = Eq(c1*v1 + c2*v2.subs(x,S(1)/2) + c3*v3.subs(x,S(1)/2) + c4*v4.subs(x,S(1)/2),0)
eq1, eq2, eq3, eq4
A,b = linear_eq_to_matrix([eq1,eq2,eq3,eq4],c1,c2,c3,c4)
linsolve((A,b),c1,c2,c3,c4)
Question b#
Apply the Gram-Schmidt procedure on \(\alpha\) and show that the procedure yields a normalized version of the Legendre polynomials.
Hint
For calculating the norms and the inner products, first determine the integrals by hand. Afterwards, use the integrate command in Sympy with code lines such as x = symbols('x', real=True) and integrate(x**2, (x,-1,1)) to check your work.
Hint
Check your manual calculation of the polynomial \(u_1\) with Sympy:
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
Check your manual calculation of the polynomial \(u_2\) with Sympy:
w2 = v2 - integrate(v2*u1, (x,-1,1)) * u1
u2 = w2/sqrt(integrate(w2**2, (x,-1,1)))
u2
Hint
Check your manual calculation of the polynomial \(u_3\) with Sympy:
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
Check your manual calculation of the polynomial \(u_4\) with Sympy:
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 obtained by running the code as given in the above hints. Compare the result with legendre(0,x), legendre(1,x), legendre(2,x).factor() and legendre(3,x).factor(). They should differ by a scaling factor.
v1_norm = sqrt(integrate(v1**2, (x,-1,1)))
u1 = v1/v1_norm
u1
w2 = v2 - integrate(v2*u1, (x,-1,1)) * u1
u2 = w2/sqrt(integrate(w2**2, (x,-1,1)))
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)))
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)))
u1, u2, u3, u4
# This is a normalized version of the Legendre polynomials:
legendre(0,x), legendre(1,x), legendre(2,x).factor(), legendre(3,x).factor()
Exercises – Short Day#
1: Matrix Multiplications#
Define
Let \(\pmb{a}_1, \pmb{a}_2,\pmb{a}_3,\pmb{a}_4\) denote the columns in \(A\). Let \(\pmb{b}_1, \pmb{b}_2,\pmb{b}_3\) denote the rows in \(A\). We now calculate \(A\pmb{x}\) in two different ways.
Question a#
Method 1: As a linear combination of the columns. Calculate the linear combination
Question b#
Method 2: As “dot product” of the rows in \(A\) with \(x\). Calculate
Note
Since \(\pmb{b}_k\) is a row vector, \((\pmb{b}_k)^T\) is a column vector. Hence, the product \(\pmb{b}_k \pmb{x}\) corresponds to the dot product of \(\pmb{x}\) and \((\pmb{b}_k)^T\).
Question c#
Calculate \(A\pmb{x}\) in Sympy and compare with your calculations from the previous questions.
A = Matrix([[1,2,3,4],[5,6,7,8],[4,4,4,4]])
a1, a2, a3, a4 = A[:,0], A[:,1], A[:,2], A[:,3]
b1, b2, b3 = A[0,:], A[1,:], A[2,:]
x = Matrix([1,2,-1,1])
A*x, x[0]*a1 + x[1]*a2 + x[2]*a3 + x[3]*a4, b1.dot(x), b2.dot(x), b3.dot(x)
2: A Subspace in \(\mathbb{C}^4\) and its Orthogonal Compliment#
Let the following vectors be given in \(\mathbb{C}^4\):
A subspace \(Y\) in \(\mathbb{C}^4\) is determined by \(Y=\operatorname{span}\lbrace\pmb{v}_1,\pmb{v}_2,\pmb{v}_3,\pmb{v}_4\rbrace\).
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 \((\pmb{v}_1,\pmb{v}_2,\pmb{v}_3)\,\) constitutes a basis for \(Y\), and find 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
is the coordinate vector with respect to basis \(v\).
A = Matrix.hstack(v1, v2, v3)
A.rank(), linsolve((A, v4))
Question c#
Provide 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 just need to be orthonormalized, for example with the use of the Gram-Schmidt procedure.
Answer
The vectors
constitute an orthonormal basis for \(Y\).
u1, u2, u3 = GramSchmidt([v1, v2, v3], orthonormal = True)
u1, u2, u3
Question d#
Determine the coordinate vector of \(\pmb{v}_4 \in Y\) with respect to the orthonormal basis for \(Y\).
Hint
This can be calculated with inner products (as we are used to by now) or with a fitting matrix-vector product.
Umat = Matrix.hstack(u1, u2, u3)
simplify(Umat.adjoint() * v4 )
Question e#
Determine the orthogonal complement \(Y^\perp\) in \(\mathbb{C}^4\) to \(Y\).
Hint
First, guess a vector and check if your guess is located in \(Y\). Meaning, check that it is linearly independent to a basis for \(Y\).
Hint
You can, for example, try the guess \((1,0,0,0)\).
Hint
A vector that belongs to the orthogonal complement is a vector that is orthogonal to the basis vectors in a basis for \(Y\).
Hint
Use the Gram-Schmidt procedure - continue from where you left off in the previous question b and just implement the vector you have just guessed. You do not need to normalize the vector in the last step, since it is enough to just orthogonalize. (Why?)
Answer
The orthogonal complement is a 1-dimensional subspace of \(\mathbb{C}^4\) spanned by
An Alternative Approach
It is not necessary to continue with the Gram-Schmidt procedure as described above. Here is another alternative approach that starts directly from the definition of the orthogonal complement: Let \(\boldsymbol=[c1,c2,c3,c4]^T\). Write the three orthogonality conditions as equations in \(c_i\):
Then solve this linear system of equations, and determine/choose a vector that spans the null space (the kernel).
Question f#
Choose a vector \(\pmb{y}\) in \(Y^\perp\) and choose a vector \(\pmb{x}\) in \(Y\). Calculate \(\Vert \pmb{x} \Vert\), \(\Vert \pmb{y} \Vert\) and \(\Vert \pmb{x} + \pmb{y} \Vert\). Check that \(\Vert \pmb{x} \Vert^2 +\Vert \pmb{y} \Vert^2 = \Vert \pmb{x} + \pmb{y} \Vert^2\).
*u_vectors, u_perp = GramSchmidt([v1, v2, v3, Matrix([1,0,0,0])], orthonormal = True)
u_perp
c = symbols('c', real=True)
y = c * u_perp # c=1 or c=2 are ok choices
x = v4 # Could do this more generally
y.norm(), x.norm(), (x+y).norm()
(x+y).norm()**2 == x.norm()**2 + y.norm()**2
True
3: Orthogonal Projection on a Plane#
Let the matrix \(U = [\pmb{u}_1, \pmb{u}_2]\) be given by:
Question a#
Show that \(\pmb{u}_1, \pmb{u}_2\) is an orthonormal basis for \(Y = \operatorname{span}\{\pmb{u}_1, \pmb{u}_2\}\).
Answer
By calculation, it is shown 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. Therefore, this list is linearly independent and thus constitutes a basis for \(Y\). We conclude that \(\pmb{u}_1, \pmb{u}_2\) is an orthonormal basis for \(Y\).
Question b#
Let \(P = U U^* \in \mathbb{R}^{3 \times 3}\). As explained in this section of the textbook, this will give us a projection matrix that describes the orthogonal projection \(\pmb{x} \mapsto P \pmb{x}\), \(\mathbb{R}^3 \to \mathbb{R}^3\) on the plane \(Y = \operatorname{span}\{\pmb{u}_1, \pmb{u}_2\}\). Verify that \(P^2 = P\), \(P \pmb{u}_1 = \pmb{u}_1\), and \(P \pmb{u}_2 = \pmb{u}_2\).
Answer
From this it is easy to show that \(P \pmb{u}_1 = \pmb{u}_1\) and that \(P \pmb{u}_2 = \pmb{u}_2\). 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 its projection \(\operatorname{proj}_Y(\pmb{x})\) af \(\pmb{x}\) down on the plane \(Y\). Illustrate \(\pmb{x}\), \(Y\) and \(\operatorname{proj}_Y(\pmb{x})\) in a plot.
Answer
You can, for example, choose \(\pmb{x} = [1,2,3]^T\). You must then calculate \(\operatorname{proj}_Y(\pmb{x}) = P \pmb{x} = [2,0,2]^T\). 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 \(\pmb{x} - \operatorname{proj}_Y(\pmb{x})\) belongs to \(Y^\perp\).
Hint
With the choice \(\pmb{x} = [1,2,3]^T\) we get \(\pmb{y} := \pmb{x} - \operatorname{proj}_Y(\pmb{x}) = [-1,2,1]^T\). You need to 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 \(\pmb{y} = [-1,2,1]^T \in Y^\perp\).
v1 = Matrix([1,1,-1])
v2 = Matrix([1,0,1])
u1,u2 = GramSchmidt([v1, v2], orthonormal = True)
U = Matrix.hstack(u1, u2)
P = U * U.T
P, P * v1, P * v2
x = Matrix([1,2,3])
P * x
x - P * x
This is the orthogonal projection of \(\pmb{x}\) on \(Y^\perp\). A vector belongs to \(Y^\perp\) if and only if the inner product of \(u_1\) and \(u_2\) is zero.
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
State whether the following propositions are true or false:
\(F\) is unitary
\(F\) is invertible
\(F\) is orthogonal
\(F\) is symmetric
\(F\) is Hermitian
The columns in \(F\) constitute an orthonormal basis for \(\mathbb{C}^4\)
The columns in \(F\) constitute an orthonormal basis for \(\mathbb{R}^4\)
\(-F = F^{-1}\)
Answer
True
True
False
True
False
True
False
False
Note: This 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 discrete Fourier transform and is important in many parts of engineering. See for example 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.
display(F.adjoint() * F == eye(n))
F.adjoint() * F
True
display(F.rank() == n)
F.inv()
True
F.T * F == eye(n)
False
F.is_symmetric()
True
F.adjoint() == F
False
True, False
(True, False)
-F == F.inv()
False
Note: \(F^{-1} = \overline{F}\).