Week 3: Preparation

Week 3: Preparation#

Reading Material#

  • Review: We recommend that you review the Mathematics 1a curriculum on vector spaces, spans, and bases

  • Reading: In Chapter 2, read Sections 2.1, 2.3 through 2.6, and 2.10

  • Python: Demo 3

Key Concepts#

Long Day will cover:

  • Vector spaces with inner product and norm

  • Inner product and norm in \(\mathbb{R}^n\) and \(\mathbb{C}^n\)

  • Inner product and norm in \(\mathsf M_{n\times m}(\Bbb R)\) and \(\Bbb R[Z]\)

  • Projections on the line

  • Orthonormal bases

  • The Gram-Schmidt procedure

Short Day will cover:

  • Orthogonal and unitary matrices

  • Projections on subspaces

  • The orthogonal complement

Preparatory Exercises#

I: Absolute Value in \(\mathbb{C}\)#

Calculate the absolute value of the following complex numbers:

  • \(2\)

  • \(3i\)

  • \(-2i\)

  • \(1-2i\)

  • \(-4-4i\)

If \(z=a + i b\), what is then the definition of \(|z|\)? Can \(|z|\) be negative? Is the absolute value \(|z|\) a norm on the vector space \(\mathbb{C}\)?

Answer

The answers to the last two questions are, respectively, no and yes.

II: Inner Product in \(\mathbb{R}^4\)#

Let \(\pmb{u} = (1, -2, 3, 4)\) and \(\pmb{v} = (-1, 0, 2, -3)\) be vectors in \(\mathbb{R}^4\) with the usual inner product defined.

  1. Calculate \(\langle \pmb{u}, \pmb{v} \rangle\).

  2. Compute the length (norm) of each vector.

  3. Can an angle \(\theta\) between the two vectors be determined?

Hint

Use the definition of the inner product:

\[\begin{equation*} \langle \pmb{u}, \pmb{v} \rangle = \pmb{u}_1 v_1 + \pmb{u}_2 v_2 + \pmb{u}_3 v_3 + \pmb{u}_4 v_4, \end{equation*}\]

and of the norm:

\[\begin{equation*} \|\pmb{u}\| = \sqrt{\langle \pmb{u}, \pmb{u} \rangle}. \end{equation*}\]

Answer

  1. Calculate \(\langle \pmb{u}, \pmb{v} \rangle\):

\[\begin{align*} \langle \pmb{u}, \pmb{v} \rangle &= (1)(-1) + (-2)(0) + (3)(2) + (4)(-3) \\ &= -1 + 0 + 6 - 12 = -7. \end{align*}\]

  1. The length (norm) of each vector:

\[\begin{align*} \|\pmb{u}\| &= \sqrt{1^2 + (-2)^2 + 3^2 + 4^2} = \sqrt{1 + 4 + 9 + 16} = \sqrt{30}, \\ \|\pmb{v}\| &= \sqrt{(-1)^2 + 0^2 + 2^2 + (-3)^2} = \sqrt{1 + 0 + 4 + 9} = \sqrt{14}. \end{align*}\]

  1. The angle \(\theta\) can be determined using the formula:

\[\begin{equation*} \cos\theta = \frac{\langle \pmb{u}, \pmb{v} \rangle}{\|\pmb{u}\|\|\pmb{v}\|} = \frac{-7}{\sqrt{30}\sqrt{14}} \approx -0.341. \end{equation*}\]

So the angle is approximately

\[\begin{equation*} \theta \approx \arccos(-0.341) \approx 110^\circ. \end{equation*}\]

III: An Orthonormal Basis in \(\mathbb{R}^5\)?#

Consider these vectors in \(\mathbb{R}^5\):

\[\begin{equation*} \pmb{a}_1 = (1,1,0,0,0), \quad \pmb{a}_2 = (0,1,1,0,0). \end{equation*}\]

Question a#

Verify that the Gram-Schmidt procedure that orthonormalizes vectors produces the following result:

\[\begin{equation*} \pmb{u}_1 = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0, 0, 0\right), \quad \pmb{u}_2 = \left(-\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, 0, 0\right). \end{equation*}\]

Answer

First we let

\[\begin{equation*} \pmb{w}_1 = \pmb{a}_1 = (1,1,0,0,0). \end{equation*}\]

Then we normalize \(\pmb{w}_1\) to obtain the first orthonormal vector:

\[\begin{equation*} \pmb{u}_1 = \frac{\pmb{w}_1}{\|\pmb{w}_1\|} = \frac{(1,1,0,0,0)}{\sqrt{1^2+1^2}} = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,0\right). \end{equation*}\]

Now the next intermediate vector is determined by removing the component along \(\pmb{u}_1\) from \(\pmb{a}_2\):

\[\begin{equation*} \pmb{w}_2 = \pmb{a}_2 - \langle \pmb{a}_2, \pmb{u}_1 \rangle \, \pmb{u}_1. \end{equation*}\]

We calculate the dot product:

\[\begin{equation*} \langle \pmb{a}_2, \pmb{u}_1 \rangle = (0)\left(\frac{1}{\sqrt{2}}\right) + (1)\left(\frac{1}{\sqrt{2}}\right) + (1)(0) = \frac{1}{\sqrt{2}}. \end{equation*}\]

We now have

\[\begin{equation*} \pmb{w}_2 = (0,1,1,0,0) - \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0,0\right) = \left(0-\frac{1}{2},\,1-\frac{1}{2},\,1-0,\,0,\,0\right) = \left(-\frac{1}{2},\,\frac{1}{2},\,1,\,0,\,0\right). \end{equation*}\]

Finally, we normalize \(\pmb{w}_2\) to obtain the next orthonormal vector:

\[\begin{equation*} \pmb{u}_2 = \frac{\pmb{w}_2}{\|\pmb{w}_2\|}. \end{equation*}\]

We compute the norm:

\[\begin{equation*} \|\pmb{w}_2\| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4}+\frac{1}{4}+1} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}. \end{equation*}\]

Hence,

\[\begin{equation*} \pmb{u}_2 = \frac{\left(-\frac{1}{2},\,\frac{1}{2},\,1,\,0,\,0\right)}{\frac{\sqrt{6}}{2}} = \left(-\frac{1}{\sqrt{6}},\,\frac{1}{\sqrt{6}},\,\frac{2}{\sqrt{6}},\,0,\,0\right). \end{equation*}\]

This is the wanted result.

Question b#

Does the resulting list of vectors \(\pmb{u}_1, \pmb{u}_2\) constitute an orthonormal basis for \(\mathbb{R}^5\)?

Answer

No. They are orthonormal but they do not constitute a basis for \(\mathbb{R}^5\) since there are only two vectors in the list. To be a basis for \(\mathbb{R}^5\), five linearly independent vectors are needed.

IV: A Linear Map that is a Projection?#

Consider the linear map \(\mathrm{proj}_Y: \mathbb{R}^4 \to \mathbb{R}^4\) given by:

\[\begin{equation*} \mathrm{proj}_Y(\pmb{x}) = P\pmb{x}, \quad \text{where} \quad P = \begin{bmatrix} 1/2 & 0 & 1/2 & 0 \\ 0 & 1/2 & 0 & 1/2 \\ 1/2 & 0 & 1/2 & 0 \\ 0 & 1/2 & 0 & 1/2 \end{bmatrix}. \end{equation*}\]

Question a#

Show that \(\mathrm{proj}_Y\) is a linear map.

Answer

Since \(\mathrm{proj}_Y(\pmb{x}) = P\pmb{x}\), and since matrix multiplication is linear, we have

\[\begin{equation*} P(\pmb{x}+\pmb{y}) = P\pmb{x} + P\pmb{y},\quad P(c\pmb{x}) = cP\pmb{x}. \end{equation*}\]

Hence, \(\mathrm{proj}_Y\) is linear.

Question b#

Show that the vectors \((1,0,-1,0)\) and \((0,1,0,-1)\) are orthogonal to the rows in \(P\). What does this have to do with the null space/kernel of \(P\)?

Answer

The rows in \(P\) are:

\[\begin{equation*} r_1 = \left(\frac{1}{2}, 0, \frac{1}{2}, 0\right) \quad \text{and} \quad r_2 = \left(0, \frac{1}{2}, 0, \frac{1}{2}\right), \end{equation*}\]

(where \(r_3 = r_1\) and \(r_4 = r_2\)).

We see that

\[\begin{equation*} (1,0,-1,0)\cdot r_1 = 1\cdot\frac{1}{2} + 0\cdot 0 + (-1)\cdot\frac{1}{2} + 0\cdot 0 = \frac{1}{2} - \frac{1}{2} = 0 \end{equation*}\]

and

\[\begin{equation*} (0,1,0,-1)\cdot r_2 = 0\cdot0 + 1\cdot\frac{1}{2} + 0\cdot0 + (-1)\cdot\frac{1}{2} = \frac{1}{2} - \frac{1}{2} = 0. \end{equation*}\]

Since these vectors are orthogonal to all rows in \(P\), they are located within the null space of \(P\).

This exercise will be continued later in this exercise in Week 3: Closure.

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