Week 3: Closure

Week 3: Closure#

Continue working on the the preparatory exercises and the in-class exercises that you have not yet completed.

Key Concepts#

  • 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

  • Orthogonal and unitary matrices

  • Projections on subspaces

  • The orthogonal complement

If there are still concepts you are unsure about, you should reread the relevant chapters in the textbook or revisit the exercises of the week.

Extra Exercises#

We do not expect you to complete more exercises than those from from the week’s program. The following additional exercises are purely an optional offer for those who want extra practice and challenge.

1: Is this a Projection?#

This exercise continuous from where Exercise IV: A Linear Map that is a Projection? in Week 3: Preparation left off. 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#

Determine the image space \(\operatorname{im}(\mathrm{proj}_Y)\) and the null space/kernel \(\ker(\mathrm{proj}_Y)\).

Answer

We notice that the columns in \(P\) are:

\[\begin{equation*} \text{Col}_1 = \left(\frac{1}{2}, 0, \frac{1}{2}, 0\right)^T, \quad \text{Col}_2 = \left(0, \frac{1}{2}, 0, \frac{1}{2}\right)^T, \end{equation*}\]

with columns 3 and 4 being repetitions of these. Hence,

\[\begin{equation*} \operatorname{im}(\mathrm{proj}_Y) = \operatorname{col}(P) = \operatorname{span}\{(1,0,1,0),\,(0,1,0,1)\}. \end{equation*}\]

The null space can be found by solving \(P\pmb{x}=0\). This will give the conditions:

\[\begin{equation*} \frac{1}{2}(x_1+x_3)=0 \quad \text{og} \quad \frac{1}{2}(x_2+x_4)=0, \end{equation*}\]

so

\[\begin{equation*} x_1+x_3 = 0 \quad \text{og} \quad x_2+x_4 = 0. \end{equation*}\]

Hence

\[\begin{equation*} \ker(\mathrm{proj}_Y) = \{(x_1,x_2,-x_1,-x_2) : x_1,x_2 \in \mathbb{R}\} = \operatorname{span}\{(1,0,-1,0),\,(0,1,0,-1)\}. \end{equation*}\]

Question b#

Show that \(P^2 = P\) and that \(P^* = P\). In fact, this shows that \(\mathrm{proj}_Y\) is an orthogonal projection. Think about why \(P^2 = P\) is a reasonable property of a projection matrix.

Question c#

Onto which subspace \(Y\) does \(\mathrm{proj}_Y\) project?

Hint

What can the matrix “reach”? What is the range of the projection?

Answer

\(Y = \operatorname{col}(P) = \operatorname{span}\{(1,0,1,0),\,(0,1,0,1)\}\).

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