Recall that a basis for a vector space is a linearly independent, spanning set.
Example 1¶
The set $\left\{ \ihat,\ \jhat \right\}$ is the standard basis for $\mathbb{R}^2$
Example 2¶
The set $E = \left\{\,\begin{bmatrix} 1 \\ 2 \end{bmatrix},\ \begin{bmatrix} 2 \\ 1 \end{bmatrix}\,\right\}$ is a basis for $\mathbb{R}^2$.
We know that they are linearly independent because we can reduce the $2\times 2$ matrix $A$ below to the identity.
$$ A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \xrightarrow{R_2-2R_1} \begin{bmatrix} 1 &\ \ 2 \\ 0 & -3 \end{bmatrix} \xrightarrow{-R_2/3} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \xrightarrow{R_1-2R_2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
This means that the matrix $A$ is a non-singular matrix so every vector $\mathbf{v}\in\mathbb{R}^2$ can be written
$$ \mathbf{v} = A\mathbf{x} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}x_1 + \begin{bmatrix} 2 \\ 1 \end{bmatrix}x_2. $$
We call the scalars $x_1$ and $x_2$ the coordinates of the vector $\mathbf{v}$ with respect to the basis $E = \left\{\,\begin{bmatrix} 1 \\ 2 \end{bmatrix},\ \begin{bmatrix} 2 \\ 1 \end{bmatrix}\,\right\}$.
We denote a vector in $\mathbb{R}^2$ with respect to this basis
$$ \mathbf{v} = \mathbf{v}_E = \begin{bmatrix} \mathbf{v} \end{bmatrix}_E = \begin{bmatrix}\ x_1\ \\ \ x_2\ \end{bmatrix}_E $$
Example 3¶
In fact there are infinitely many pairs of linearly independent vectors in $\mathbb{R}^2$
$$ U = \left\{\,\mathbf{u}_1,\ \mathbf{u}_2\,\right\} $$
and every such pair makes up a basis for $\mathbb{R}^2$. For a vector $\mathbf{x}\in\mathbb{R}^2$ we can note the coefficients $x_1$, $x_2$ of each basis vector that comprises the linear combination
$$ \mathbf{x} = x_1\mathbf{u}_1 + x_2\mathbf{u}_2 $$
In this case the coordinates of vector $\mathbf{x}$ are the scalars $x_1$ and $x_2$ with respect to the basis $U$.
Example 4¶
Consider the vector $\mathbf{v}$ whose tail sits at the point $(0,0)$ on the Cartesian plane, and whose tip sits at the point $(3,3)$. One describes this vector using a list of two numbers. These numbers can be the coefficients of $\mathbf{\hat{\imath}}$ and $\mathbf{\hat{\jmath}}$.
$$ \mathbf{v} = 3\ihat + 3\jhat $$
We describe vector $\mathbf{v}$ in standard coordinates
$$ \mathbf{v} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} $$
One can also describe the vector using the basis from Example 2,
$$ E = \left\{\,\begin{bmatrix} 1 \\ 2 \end{bmatrix},\ \begin{bmatrix} 2 \\ 1 \end{bmatrix}\,\right\} $$
Either solving the linear system of equations
$$ A\mathbf{v}_E = \begin{bmatrix} 3 \\ 3 \end{bmatrix} $$
or observing that
$$ \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 1\begin{bmatrix} 1 \\ 2 \end{bmatrix} + 1\begin{bmatrix} 2 \\ 1 \end{bmatrix} $$
one determines that the coefficients of each vector in basis $E$ are 1 and 1. It is possible to describe vector $\mathbf{v}$ using a different list of two numbers, the coefficients of our new basis vectors. This new list of numbers is the coordinates of the vector with respect to our new basis. We would also like to refer to vector $\mathbf{v}$ using the list of coordinates. To clarify that the list refers to coordinates of basis $E$ instead of the standard basis decorate the square brackets with a subscript indicating the set of basis vectors.
$$ \mathbf{v}_E = \begin{bmatrix} 1 \\ 1 \end{bmatrix}_E $$
This set of coordinates indicates that the coefficients of the ordered basis $E$ are 1 for the first basis vector, and 1 for the second basis vector
$$ \mathbf{v} = 1\begin{bmatrix} 1 \\ 2 \end{bmatrix} + 1\begin{bmatrix} 2 \\ 1 \end{bmatrix} $$
One can describe vectors in $\mathbb{R}^2$ using the basis in Example 3,
$$ U = \left\{\,\mathbf{u}_1,\ \mathbf{u}_2\,\right\} $$
In this case we can solve the system of equations
$$ \left[\,\mathbf{u}_1,\ \mathbf{u}_2\,\right]\mathbf{v}_U = U\mathbf{v}_U = \begin{bmatrix} 3 \\ 3 \end{bmatrix} $$
Notice that to simplify our notation we used the same name of the set of vectors and the $2\times 2$ matrix formed by the two vectors. Since the $2\times 2$ matrix $U$ is non-singular (its columns are linearly independent), the system has a unique solution
$$ \mathbf{v} = v_1\mathbf{u}_1 + v_2\mathbf{u}_2 $$
As in the example of vector space $\mathbb{R}^2$, we would like to describe our vector with a list of coefficients or coordinates with respect to basis U,
$$ \mathbf{v}_U = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}_U $$
Notice that the vector in this example has not changed.
A vector is an element of a vector space. It is a mathematical object, which is a fancy way of saying it exists only in our imaginations.
The coordinates or list of numbers used to describe a vector exist only when a basis has been chosen.
The coordinates only describe how to refer to a vector using a chosen basis. The list of numbers is a numerical representation of a vector.
It seems odd that we need to take the time to distinguish between a vector and the column of numbers we use to refer to that vector. When we discuss real numbers or scalars we rarely distinguish the number one from the Arabic numeral 1 that we often use to represent it. However the decimal system does not represent numbers uniquely.
$$ \begin{align*} \text{One} &= 1 = 0.999\bar{9} \\ \\ \text{Two} &= 2 = 1.999\bar{9} \\ \\ 2.5 &= \dfrac{5}{2} = 2.4999\bar{9} \end{align*} $$
The ratio of the circumference of a circle and its diameter is typically represented by the Greek small letter $\pi$. When we perform algebraic computations with an unknown number, we typically use a single letter to represent it.
Numbers are mathematical objects and exist only in our imaginations. Numerals are the symbols we write down to represent them.
The unit of a field of numbers (scalars) is unambiguous and commonly denoted with the symbol $1$. If this seems like a lot of vocabulary to describe a simple and little used distinction remember that vectors are not like a field of scalars. They are more complicated and can represent a lot more information than a scalar quantity.
Choosing a basis so that we can represent our vectors as a list of coordinates is called
- choosing an orientation,
- choosing a coordinate system,
- choosing a frame of reference.
- choosing an ordered basis
The vector $\mathbf{v}$ is the same vector without regard to what orientation or basis is chosen to describe vector $\mathbf{v}$.
$$ \mathbf{v} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}_E = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}_U $$
Video Lecture 1 - Change of Basis¶
Change of BasisTake a break from studying linear algebra after watching this video. Consider watching it one or two more times before you continue reading this section.
One may represent the same vector using a ridiculously large collection of different lists of coordinates that depend on the choice of orientation or basis. One may also construct linear transformations from one coordinate system to another.
The transition matrices are non-singular matrices. When you multiply a vector on the left by a transition matrix the result is the same vector, with a different coordinate system.
Every non-singular matrix can be thought of as a transition matrix from one coordinate system to another. This adds a new interpretation for a non-singular matrix.
A non-singular matrix can represent¶
- a transformation on a vector space with respect to the same basis
- a transformation from a vector with respect to one basis representation to the same vector with respect to a different basis representation
- a transformation from a vector with respect to one basis representation to another vector with respect to another basis (This combines 1. and 2.)
Example 5¶
Consider the basis for $\mathbb{R}^2$ given by
$$ U = \left\{\,\mathbf{u}_1,\ \mathbf{u}_2\,\right\} = \left\{\, {\color{royalblue}\begin{bmatrix} -2\ \\ \ 0\ \end{bmatrix}},\ {\color{crimson}\begin{bmatrix}\ 2\ \\ \ 5\ \end{bmatrix}} \,\right\} $$
Create the transition matrix $U$ from this basis to the standard basis, and create the transition matrix from the standard basis to this new basis $U$.
To give the transition matrix from this new basis $U$ to the standard basis the same name as the set of vectors
$$ U = \begin{bmatrix} {\color{royalblue}-2}\ & \ {\color{crimson}2}\ \\ \ {\color{royalblue}0}\ &\ {\color{crimson}5}\ \end{bmatrix} $$
This works because the two new basis vectors are written with respect to the standard basis!
Using the new basis corresponds to using Jennifer's language instead of the standard language, however Jennifer's basis vectors are written using the standard basis. Any vector in $\mathbb{R}^2$ can be written uniquely as a list of coordinates of Jennifer's basis vectors $U$.
$$ \mathbf{v} = u_1\mathbf{u}_1 + u_2\mathbf{u}_2 = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_U $$
Since Jennifer's basis vectors can be written with respect to the standard basis
$$ \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = u_1\begin{bmatrix} -2\ \\ \ 0\ \end{bmatrix} + u_2\begin{bmatrix}\ 2\ \\ \ 5\ \end{bmatrix} = U\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_U = U\mathbf{v}_U $$
This makes the matrix with columns that are Jennifer's basis vectors written in the standard coordinate system the transition matrix from Jennifer's basis $U$ to the standard basis. As matrix $U$ is a non-singular matrix, it is invertible and
$$ \begin{align*} \mathbf{v} &= v_1\ihat + v_2\jhat = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = U\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_U = U\mathbf{v}_U \\ \\ U^{-1}\mathbf{v} &= U^{-1}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = U^{-1}U\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_U = U^{-1}U\mathbf{v}_U \\ \\ U^{-1}\mathbf{v} &= U^{-1}\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_U = \mathbf{v}_U \\ \\ \begin{bmatrix} U\,|\,I \end{bmatrix} &= \left[ \begin{array}{cc|cc} -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ 0\ &\ 5\ &\ \ 0\ &\ \ 1\ \end{array} \right] \xrightarrow{R_2/5} \left[ \begin{array}{cc|cc} -2\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ 0\ &\ 1\ &\ \ 0\ &\ \ \frac{1}{5}\ \end{array} \right] \xrightarrow{R_1-2R_2} \\ \\ \left[ \begin{array}{cc|cc} -2\ &\ \ 0\ &\ \ 1\ & -\frac{2}{5}\ \\ \ 0\ &\ 1\ &\ \ 0\ &\ \ \frac{1}{5}\ \end{array} \right] &\xrightarrow{-R_1/2} \left[ \begin{array}{cc|cc}\ \ 1\ & \ 0\ & -\frac{1}{2}\ &\ \ \frac{1}{5}\ \\ \ 0\ &\ 1\ &\ \ 0\ &\ \ \frac{1}{5}\ \end{array} \right] = \begin{bmatrix} I\,|\,U^{-1} \end{bmatrix} \\ \\ U^{-1} &= \begin{bmatrix} -\frac{1}{2}\ &\ \ \frac{1}{5}\ \\ \ \ 0\ &\ \ \frac{1}{5}\ \end{bmatrix} \end{align*} $$
Matrix $U^{-1}$ is the matrix one multiplies by vector $\mathbf{v}$ in standard coordinates to obtain the same vector in Jennifer's language $\mathbf{v}_U$. Notice that the first column of matrix $U^{-1}$ are the coordinates of $\ihat$ in Jennifer's language, and the second column are the coordinates of $\jhat$ in Jennifer's language.
Example 6¶
Consider the bases of vector space $P_2$, ${\color{crimson} U = \left\{ 1, x, x^2 - 2 \right\} }$, and ${\color{royalblue} V = \left\{ 2, x-1, x^2\right\}}$. Create the transition matrix from basis ${\color{crimson} U}$ to basis ${\color{royalblue} V}$ and from basis ${\color{royalblue} V}$ to basis ${\color{crimson} U}$. By using
$$S = {\color{royalblue} V^{-1}}{\color{crimson} U},\text{ and }{S^{-1} = \color{crimson} U^{-1}}{\color{royalblue} V}$$
In order to obtain the transition matrix $S$ from basis ${\color{crimson} U}$ to basis ${\color{royalblue} V}$, we need both
- the transition matrix ${\color{crimson} U}$ from basis ${\color{crimson} U}$ to the standard basis, and
- the transition matrix ${\color{royalblue} V}$ from basis ${\color{royalblue} V}$ to the standard basis.
Together, $S = {\color{royalblue} V^{-1}}{\color{crimson} U}$
$${\color{crimson}U = \begin{bmatrix}\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix}},\qquad {\color{royalblue} V = \begin{bmatrix}\ \ 2\ & -1\ &\ \ 0\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix} }$$
The transition matrix from basis ${\color{crimson}U}$ to basis ${\color{royalblue}V}$ can be obtained using the same method for inverting a non-singular matrix.
$$ \left[ \begin{array}{ccc|ccc}\ \ 2\ & -1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \end{array} \right] \xrightarrow{R_1+R_2} \left[ \begin{array}{ccc|ccc}\ \ 2\ &\ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 1\ & -2\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \end{array} \right] \xrightarrow{R_1/2} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1/2\ &\ \ 1/2\ & -1\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \end{array} \right] $$
$$S = \begin{bmatrix}\ 1/2\ &\ 1/2\ & -1\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix}$$
$S$ is the transition matrix from basis ${\color{crimson}U}$ to basis ${\color{royalblue}V}$.
Notice that $S$ is upper triangular. Hence its determinant is easy to compute, $\det(S) = \dfrac{1}{2}$. The cofactor matrix is equally simple to compute
$$\mathrm{adj}(S) = \begin{bmatrix}\ \ 1\ &\ \ 0\ &\ \ 0\ \\ -1/2\ &\ 1/2\ &\ \ 0\ \\ \ \ 1\ &\ \ 0\ &\ 1/2\ \end{bmatrix}^T = \begin{bmatrix}\ \ 1\ & -1/2\ &\ \ 1\ \\ \ \ 0\ &\ 1/2\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ 1/2\ \end{bmatrix}$$
$$S^{-1} = \frac{1}{\det(S)}\mathrm{adj}(S) = 2\begin{bmatrix}\ \ 1\ & -1/2\ &\ \ 1\ \\ \ \ 0\ &\ 1/2\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ 1/2\ \end{bmatrix} = \begin{bmatrix}\ \ 2\ & -1\ &\ \ 2\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix}$$
$S^{-1}$ is the transition matrix from basis ${\color{royalblue}V}$ to basis ${\color{crimson}U}$.
Example 7¶
Using the same bases ${\color{crimson} U}$ and ${\color{royalblue} V}$ in Example 6, compute $S$ and $S^{-1}$ directly by computing the basis vectors in ${\color{crimson} U}$ in ${\color{royalblue} V}$-coordinates for $S$, and computing the basis vectors in ${\color{royalblue} V}$ in ${\color{crimson} U}$-coordinates for $S^{-1}$.
$$\begin{align*} {\color{crimson}1} &= \frac{1}{2}\left({\color{royalblue}2}\right) & {\color{royalblue}2} &= 2\left( {\color{crimson}1} \right) \\ {\color{crimson}x} &= \frac{1}{2}\left({\color{royalblue}2}\right) + 1\left({\color{royalblue}x-1}\right) & {\color{royalblue}x-1} &= (-1)\left( {\color{crimson}1} \right) + 1\left( {\color{crimson}x} \right) \\ {\color{crimson}x^2-2} &= -1\left({\color{royalblue}2}\right) + 1\left({\color{royalblue}x^2}\right) & {\color{royalblue}x^2} &= 2\left( {\color{crimson}1} \right) + 1\left( {\color{crimson}x^2-2} \right) \\ \\ S &= \begin{bmatrix}\ 1/2\ &\ 1/2\ & -1\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix} & S^{-1} &= \begin{bmatrix}\ \ 2\ & -1\ &\ \ 2\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ \end{bmatrix} \end{align*}$$
Exercise 1¶
Consider bases
$$ U = \left\{\,\begin{bmatrix} 1 \\ 4 \end{bmatrix},\ \begin{bmatrix} -1\ \\ \ 1\ \end{bmatrix}\,\right\} $$
and
$$ V = \left\{\,\begin{bmatrix} 4 \\ 3 \end{bmatrix},\ \begin{bmatrix} 1 \\ 3 \end{bmatrix}\,\right\} $$
Create the transition matrix from basis $V$ to basis $U$.
Follow Along
We need the transition matrices $U$ and $V$ from each basis to the standard basis.$$ U = \begin{bmatrix}\ 1\ & -1\ \\ \ 4\ &\ 1\ \end{bmatrix},\ V = \begin{bmatrix}\ 4\ &\ 1\ \\ \ 3\ &\ 3\ \end{bmatrix} $$
The transition from basis $V$ to basis $U$ is a composition of first the transition from basis $V$ to the standard basis, and second from the standard basis to basis $U$.
$$ \begin{align*} S &= U^{-1}V = \dfrac{1}{1+4}\begin{bmatrix}\ 1\ & -4\ \\ \ 1\ &\ 1\ \end{bmatrix}^T\begin{bmatrix}\ 4\ &\ 1\ \\ \ 3\ &\ 3\ \end{bmatrix} \\ \\ &= \dfrac{1}{5}\begin{bmatrix}\ 1\ &\ 1\ \\ -4\ &\ 1\ \end{bmatrix}\begin{bmatrix}\ 4\ &\ 1\ \\ \ 3\ &\ 3\ \end{bmatrix} \\ \\ &= \dfrac{1}{5}\begin{bmatrix}\ 7\ &\ 4\ \\ -13\ & -1\ \end{bmatrix} = \begin{bmatrix}\ \frac{7}{5}\ &\ \frac{4}{5}\ \\ -\frac{13}{5}\ & -\frac{1}{5}\ \end{bmatrix} \end{align*} $$
Exercise 2¶
Use the bases defined in Exercise 1 for $\mathbb{R}^2$ and determine the coordinates with respect to basis $U$.
(a) $\mathbf{x} = \begin{bmatrix}\ \ 5\ \\ \ \ 10\ \end{bmatrix}_V = $
(b) $\mathbf{y} = \begin{bmatrix}\ \ 10\ \\ \ \ 5\ \end{bmatrix}_V = $
(c) $\mathbf{z} = \begin{bmatrix}\ \ 1\ \\ -1\ \end{bmatrix} = $
(d) $\mathbf{a} = \begin{bmatrix} -2\ \\ \ \ 7\ \end{bmatrix} = $
(e) $\mathbf{b} = \begin{bmatrix} -1\ \\ \ \ 1\ \end{bmatrix}_U = $
Follow Along
Identify the basis used to describe each vector and then determine the representation of the vector with respect to basis $U$.(a) $\begin{bmatrix}\ \ 5\ \\ \ \ 10\ \end{bmatrix}_V$ describes the vector in $V$-coordinates. One uses the transition matrix $S$ from basis $V$ to basis $U$ to determine the coordinates of the vector in $U$-coordinates. $$ S\begin{bmatrix}\ \ 5\ \\ \ \ 10\ \end{bmatrix}_V = \dfrac{1}{5}\begin{bmatrix}\ 7\ &\ 4\ \\ -13\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ 5\ \\ \ \ 10\ \end{bmatrix}_V = \begin{bmatrix}\ 7\ &\ 4\ \\ -13\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ 1\ \\ \ \ 2\ \end{bmatrix}_V = \begin{bmatrix}\ \ 15\ \\ -15\ \end{bmatrix}_U = \mathbf{x}_U $$
(b) $\begin{bmatrix}\ \ 10\ \\ \ \ 5\ \end{bmatrix}_V$ describes the vector in $V$-coordinates. One uses the transition matrix $S$ from basis $V$ to basis $U$ to determine the coordinates of the vector in $U$-coordinates. $$ S\begin{bmatrix}\ \ 10\ \\ \ \ 5\ \end{bmatrix}_V = \dfrac{1}{5}\begin{bmatrix}\ 7\ &\ 4\ \\ -13\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ 10\ \\ \ \ 5\ \end{bmatrix}_V = \begin{bmatrix}\ 7\ &\ 4\ \\ -13\ & -1\ \end{bmatrix}\begin{bmatrix}\ \ 2\ \\ \ \ 1\ \end{bmatrix}_V = \begin{bmatrix}\ \ 18\ \\ -27\ \end{bmatrix}_U = \mathbf{y}_U $$
(c) $\begin{bmatrix}\ \ 1\ \\ -1\ \end{bmatrix}$ describes the vector in standard coordinates. One uses the transition matrix $U^{-1}$ from the standard basis to basis $U$ to determine the coordinates of the vector in $U$-coordinates. $$ \mathbf{z} = U^{-1}\begin{bmatrix}\ \ 1\ \\ -1\ \end{bmatrix} = \dfrac{1}{5}\begin{bmatrix}\ 1\ &\ 1\ \\ -4\ &\ 1\ \end{bmatrix}\begin{bmatrix}\ \ 1\ \\ -1\ \end{bmatrix} = \begin{bmatrix}\ \ 0 \ \\ -1\ \end{bmatrix}_U = \mathbf{z}_U $$
(d) $\begin{bmatrix} -2\ \\ \ \ 7\ \end{bmatrix}$ describes the vector in standard coordinates. One uses the transition matrix $U^{-1}$ from the standard basis to basis $U$ to determine the coordinates of the vector in $U$-coordinates. $$ \mathbf{a} = U^{-1}\begin{bmatrix} -2\ \\ \ \ 7\ \end{bmatrix} = \dfrac{1}{5}\begin{bmatrix}\ 1\ &\ 1\ \\ -4\ &\ 1\ \end{bmatrix}\begin{bmatrix} -2\ \\ \ \ 7\ \end{bmatrix} = \begin{bmatrix}\ \ 1 \ \\ \ \ 3\ \end{bmatrix}_U = \mathbf{a}_U $$
(e) $\mathbf{b} = \begin{bmatrix} -1\ \\ \ \ 1\ \end{bmatrix}_U$ describes vector $b$ in $U$-coordinates already $$ \mathbf{b}_U = \begin{bmatrix} -1\ \\ \ \ 1\ \end{bmatrix}_U $$
Exercise 3¶
Consider bases
$$ U = \left\{\,\begin{bmatrix}\ 1\ \\ \ 4\ \\ \ \ 2\ \end{bmatrix},\ \begin{bmatrix} -1\ \\ \ \ 1\ \\ \ \ 4\ \end{bmatrix},\ \begin{bmatrix}\ 2\ \\ \ 4\ \\ \ 1\ \end{bmatrix}\,\right\} $$
and
$$ V = \left\{\,\begin{bmatrix}\ 2\ \\ \ 3\ \\ \ 1\ \end{bmatrix},\ \begin{bmatrix}\ 1\ \\ \ 3\ \\ \ 0\ \end{bmatrix},\ \begin{bmatrix}\ 0\ \\ \ 2\ \\ \ 1\ \end{bmatrix}\,\right\} $$
Create the transition matrix from basis $U$ to basis $V$.
Follow Along
We need the transition matrix from basis $U$ to basis $V$. That is $T = V^{-1}U$ since$$ U = \begin{bmatrix}\ \ 1\ & -1\ &\ \ 2\ \\ \ \ 4\ &\ \ 1\ &\ \ 4\ \\ \ \ 2\ &\ \ 4\ &\ \ 1\ \end{bmatrix}, \\ V = \begin{bmatrix}\ 2\ &\ 1\ &\ 0\ \\ \ 3\ &\ 3\ &\ 2\ \\ \ 1\ &\ 0\ &\ 1\ \end{bmatrix}, $$
The transition from basis $U$ to basis $V$ is a composition of first the transition from basis $U$ to the standard basis, and second from the standard basis to basis $V$.
$$ \begin{align*} T &= V^{-1}U \\ \\ \begin{bmatrix}\,V\,|\,U\,\end{bmatrix} &= \left[ \begin{array}{ccc|ccc}\ \ 2\ &\ \ 1\ &\ \ 0 &\ \ 1\ & -1\ &\ \ 2\ \\ \ \ 3\ &\ \ 3\ &\ \ 2 &\ \ 4\ &\ \ 1\ &\ \ 4\ \\ \ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \end{array} \right] \xrightarrow{R_1\leftrightarrow R_3} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \\ \ \ 3\ &\ \ 3\ &\ \ 2 &\ \ 4\ &\ \ 1\ &\ \ 4\ \\ \ \ 2\ &\ \ 1\ &\ \ 0 &\ \ 1\ & -1\ &\ \ 2\ \end{array} \right] \xrightarrow{\substack{ R_2-3R_1 \\ R_3-2R_1 }} \\ \\ &\left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \\ \ \ 0\ &\ \ 3\ & -1 & -2\ & -11\ &\ \ 1\ \\ \ \ 0\ &\ \ 1\ & -2 & -3\ & -9\ &\ \ 0\ \end{array} \right] \xrightarrow{R_2\leftrightarrow R_3} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \\ \ \ 0\ &\ \ 1\ & -2 & -3\ & -9\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -1 & -2\ & -11\ &\ \ 1\ \end{array} \right] \xrightarrow{R_3-3R_2} \\ \\ &\left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \\ \ \ 0\ &\ \ 1\ & -2 & -3\ & -9\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 5 &\ \ 7\ &\ \,16\ &\ \ 1\ \end{array} \right] \xrightarrow{R_3/5} \ \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1 &\ \ 2\ &\ \ 4\ &\ \ 1\ \\ \ \ 0\ &\ \ 1\ & -2 & -3\ & -9\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 1 &\ \ \frac{7}{5}\ &\ \,\frac{16}{5}\ &\ \ \frac{1}{5}\ \end{array} \right] \xrightarrow{\substack{R_1-R_3 \\ R_2+2R_3}} \\ \\ &\left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 0 &\ \ \frac{3}{5}\ &\ \ \frac{4}{5}\ &\ \ \frac{4}{5}\ \\ \ \ 0\ &\ \ 1\ &\ \ 0 & -\frac{1}{5}\ & -\frac{13}{5}\ &\ \ \frac{2}{5}\ \\ \ \ 0\ &\ \ 0\ &\ \ 1 &\ \ \frac{7}{5}\ &\ \,\frac{16}{5}\ &\ \ \frac{1}{5}\ \end{array} \right] \\ \\ T &= \begin{bmatrix}\ \ \frac{3}{5}\ &\ \ \frac{4}{5}\ &\ \ \frac{4}{5}\ \\ -\frac{1}{5}\ & -\frac{13}{5}\ &\ \ \frac{2}{5}\ \\ \ \ \frac{7}{5}\ &\ \,\frac{16}{5}\ &\ \ \frac{1}{5}\ \end{bmatrix} \end{align*} $$
Exercise 4¶
Use the bases defined in Exercise 3 for $\mathbb{R}^3$ and determine the coordinates with respect to basis $V$.
(a) $\mathbf{x} = \begin{bmatrix}\ \ 6\ \\ \ \ 8\ \\ -7\ \end{bmatrix} = $
(b) $\mathbf{y}_U = \begin{bmatrix}\ \ 5\ \\ -10\ \\ \ \ 5\ \end{bmatrix}_U = $
(c) $\mathbf{z}_U = \begin{bmatrix} -5\ \\ \ \ 0\ \\ \ \,10\ \end{bmatrix}_U = $
(d) $\mathbf{a}_U = \begin{bmatrix}\ \,10\ \\ -5\ \\ \ \,10\ \end{bmatrix}_U = $
(e) $\mathbf{b} = \begin{bmatrix}\ 9\ \\ \ 0\ \\ \ 6\ \end{bmatrix} = $
Follow Along
(a) $\mathbf{x} = \begin{bmatrix}\ \ 6\ \\ \ \ 8\ \\ -7\ \end{bmatrix}$ is in standard coordinates. I either need to compute $V^{-1}$, or solve $V\mathbf{x}_V = \mathbf{x}$ for vector $\mathbf{x}$ in $V$-coordinates. We will need $V^{-1}$ again in (e), so let's compute $V^{-1}$. $$ \begin{align*} \left[ \begin{array}{c|c} V & I \end{array} \right] = \left[ \begin{array}{ccc|ccc}\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \\ \ \ 3\ &\ \ 3\ &\ \ 2\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\ \ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \end{array} \right] &\xrightarrow{R_1\leftrightarrow R_3} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ \ 3\ &\ \ 3\ &\ \ 2\ &\ \ 0\ &\ \ 1\ &\ \ 0\ \\\ \ 2\ &\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ \end{array} \right] \xrightarrow{\substack{R_2-3R_1 \\ R_3-2R_1}} \\ \\ \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ \ 0\ &\ \ 3\ & -1\ &\ \ 0\ &\ \ 1\ & -3\ \\ \ 0\ &\ \ 1\ & -2\ &\ \ 1\ &\ \ 0\ & -2\ \end{array} \right] &\xrightarrow{R_2\leftrightarrow R_3} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ 0\ &\ \ 1\ & -2\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 0\ &\ \ 3\ & -1\ &\ \ 0\ &\ \ 1\ & -3\ \end{array} \right] \xrightarrow{R_3-3R_2} \\ \\ \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ 0\ &\ \ 1\ & -2\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 0\ &\ \ 0\ &\ \ 5\ & -3\ &\ \ 1\ &\ \ 3\ \end{array} \right] &\xrightarrow{R_3/5} \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ 0\ &\ \ 1\ & -2\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ & -3/5\ &\ 1/5\ &\ 3/5\ \end{array} \right] \xrightarrow{\substack{R_1-R_3 \\ R_2+2R_3}} \\ \\ \left[ \begin{array}{ccc|ccc}\ \ 1\ &\ \ 0\ &\ \ 0\ &\ 3/5\ & -1/5\ &\ 2/5\ \\ \ \ 0\ &\ \ 1\ &\ \ 0\ & -1/5\ &\ 2/5\ & -4/5\ \\ \ \ 0\ &\ \ 0\ &\ \ 1\ & -3/5\ &\ 1/5\ &\ 3/5\ \end{array} \right] &\qquad \\ \\ V^{-1} &= \frac{1}{5}\begin{bmatrix}\ \ 3\ & -1\ &\ \ 2\ \\ -1\ &\ \ 2\ & -4\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{bmatrix} \end{align*}$$
$$\mathbf{x}_V = V^{-1}\mathbf{x} = \frac{1}{5}\begin{bmatrix}\ \ 3\ & -1\ &\ \ 2\ \\ -1\ &\ \ 2\ & -4\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{bmatrix}\begin{bmatrix}\ \ 6\ \\ \ \ 8\ \\ -7\ \end{bmatrix} = \frac{1}{5}\begin{bmatrix} -4\ \\ \ \ 38\ \\ -31\ \end{bmatrix}_V$$
(b) Use $T$ from Exercise 3 to obtain $$\mathbf{y}_V = T\mathbf{y}_U = \frac{1}{5}\begin{bmatrix}\ \ 3\ &\ \ 4\ &\ \ 4\ \\ -1\ & -13\ &\ \ 2\ \\ \ \ 7\ &\ \ 16\ &\ \ 1\ \end{bmatrix}\begin{bmatrix}\ \ 5\ \\ -10\ \\ \ \ 5\ \end{bmatrix}_U = \begin{bmatrix} -1\ \\ \ \ 27\ \\ -24\ \end{bmatrix}_V$$
(c) Use $T$ from Exercise 3 to get the result $$\mathbf{z}_V = T\mathbf{z}_U = \frac{1}{5}\begin{bmatrix}\ \ 3\ &\ \ 4\ &\ \ 4\ \\ -1\ & -13\ &\ \ 2\ \\ \ \ 7\ &\ \ 16\ &\ \ 1\ \end{bmatrix}\begin{bmatrix} -5\ \\ \ \ 0\ \\ \ \ 10\ \end{bmatrix}_U = \begin{bmatrix}\ \ 5\ \\ \ \ 5\ \\ -5\ \end{bmatrix}_V$$
(d) Use $T$ from Exercise 3 to evaluate $$\mathbf{a}_V = T\mathbf{a}_U = \frac{1}{5}\begin{bmatrix}\ \ 3\ &\ \ 4\ &\ \ 4\ \\ -1\ & -13\ &\ \ 2\ \\ \ \ 7\ &\ \ 16\ &\ \ 1\ \end{bmatrix}\begin{bmatrix}\ \ 10\ \\ -5\ \\ \ \ 10\ \end{bmatrix}_U = \begin{bmatrix}\ \ 10\ \\ \ \ 15\ \\ \ \ 0\ \end{bmatrix}_V$$
(e) Use $V^{-1}$ to compute $\mathbf{b}_V$ $$\mathbf{b}_V = V^{-1}\mathbf{b} = \frac{1}{5}\begin{bmatrix}\ \ 3\ & -1\ &\ \ 2\ \\ -1\ &\ \ 2\ & -4\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{bmatrix}\begin{bmatrix}\ \ 9\ \\ \ \ 0\ \\ \ \ 6\ \end{bmatrix} = \frac{1}{5}\begin{bmatrix}\ \ 39\ \\ -33\ \\ -9\ \end{bmatrix}_V$$
