# Harrison Chapman

## Linear Algebra I: Homework 11

Due: Friday, November 17, 2017
1. Find the explicit change of basis matrix from the standard basis $E$ of $\mathbb R^3$ to the orthonormal basis $B$

The change of basis matrix which changes between two orthonormal bases is given by the formula $M_{E \to B} = (m^i_j = \langle \vec b_i, \vec e_j \rangle)$

So,

2. Find the explicit change of basis matrix from $B$ to $E$.

Change of basis matrices between orthonormal bases are orthogonal, so

1. Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of $\mathbb R^4$ spanned by the set of vectors,

Applying Gram-Schmidt yields the vectors,

which are all orthogonal. To obtain an orthonormal basis, we divide each vector by its magnitude, and obtain

2. Let

Find a basis of the subspace of $\mathbb R^4$ of all vectors perpendicular to $\vec v$

Vectors perpendicular (orthogonal) to $\vec v$ are all vectors $\vec w$ so that

This equation is the same as,

Which is a matrix equation (there’s only one row, but it’s still a matrix equation):

We’re already in row echelon form, and there are three columns without pivots, with dependent variable,

We play our usual game to find a basis (make all free variables zero except one, repeat, repeat):

3. If $U$ is a subspace of a vector space $W$, prove that $U^{\perp}$ is a subspace of $W$.

We follow the same recipe as we always do to prove something is a subspace:

Let $\vec v$, $\vec w$ be vectors in $U^{\perp}$. Then we just need to show that $a \vec v + b \vec w$ is, too. Let $\vec u$ be a vector in $U$. Then $\langle a \vec v + b \vec w, \vec u \rangle = a\langle \vec v, \vec u \rangle + b \langle w, \vec u \rangle = 0$, meaning that it is indeed in $U^\perp$ (why?). So $U^{\perp}$ is a subspace of $W$.

1. Find a basis for the kernel of the matrix $A$,

To find a basis for the kernel, we first row reduce the matrix $A$, to obtain,

We then read off the solutions to the matrix equation $A\vec x = \vec 0$ (just like finding bases for eigenspaces); there is one pivot and three free variables, so a basis would be,

2. Find a basis for the column space of the matrix $B$,

To find a basis for the column space, we column reduce the matrix $B$, to obtain

We can then take the nonzero columns as a basis for the column space: