The Central Idea

In the ordinary coordinate plane, we use the standard basis

\[ \mathbf{i} = \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad \mathbf{j} = \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]

A vector

\[ \mathbf{v} = \begin{bmatrix} x\\ y \end{bmatrix} \]

means

\[ \boxed{ \mathbf{v} = x\mathbf{i}+y\mathbf{j} }. \]

The numbers \(x\) and \(y\) are coordinates relative to that basis.

The Vector Is Not the Coordinate List

It is useful to distinguish the vector itself from the numbers used to describe it.

A vector can remain the same while its coordinate representation changes.

Coordinates depend on the chosen basis.

What Is a Basis?

A basis is a collection of vectors that provides enough independent directions to build every vector in the space uniquely.

In the plane, a basis consists of two nonparallel vectors.

If

\[ B= \{\mathbf{b}_1,\mathbf{b}_2\}, \]

then every plane vector can be written uniquely as

\[ \boxed{ \mathbf{v} = c_1\mathbf{b}_1+c_2\mathbf{b}_2 }. \]

Why the Basis Vectors Must Be Independent

If one basis candidate is a scalar multiple of the other, the two vectors provide only one direction.

They cannot build every vector in the plane.

Therefore, a plane basis requires two nonparallel vectors.

The Standard Basis

The usual basis of \(\mathbb{R}^2\) is

\[ \boxed{ E= \{\mathbf{i},\mathbf{j}\} }. \]

Because

\[ \mathbf{i} = \begin{bmatrix} 1\\ 0 \end{bmatrix} \]

and

\[ \mathbf{j} = \begin{bmatrix} 0\\ 1 \end{bmatrix}, \]

ordinary coordinate pairs are already standard-basis coordinates.

A Nonstandard Basis

Consider

\[ \mathbf{b}_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]

These vectors are not parallel, so they form a basis of the plane.

We may write

\[ B= \{\mathbf{b}_1,\mathbf{b}_2\}. \]

Coordinates Relative to a Basis

Suppose

\[ \mathbf{v} = c_1\mathbf{b}_1+c_2\mathbf{b}_2. \]

The coordinate vector of \(\mathbf{v}\) relative to basis \(B\) is

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} c_1\\ c_2 \end{bmatrix} }. \]

These are not necessarily the ordinary \(x\)- and \(y\)-coordinates.

Example: Coordinates in a Nonstandard Basis

Let

\[ \mathbf{b}_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]

Find the \(B\)-coordinates of

\[ \mathbf{v} = \begin{bmatrix} 6\\ 2 \end{bmatrix}. \]

Set Up the Basis Equation

We want scalars \(c_1\) and \(c_2\) such that

\[ c_1 \begin{bmatrix} 1\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} 6\\ 2 \end{bmatrix}. \]

Matching components gives

\[ c_1+c_2=6 \]

and

\[ c_1-c_2=2. \]

Solve for the Basis Coordinates

Add the equations:

\[ 2c_1=8. \]

Therefore,

\[ c_1=4. \]

Then

\[ c_2=2. \]

Thus,

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 2 \end{bmatrix} }. \]

Same Vector, Different Coordinates

The vector is

\[ \mathbf{v} = \begin{bmatrix} 6\\ 2 \end{bmatrix} \]

in standard coordinates.

Relative to basis \(B\), it is represented by

\[ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 2 \end{bmatrix}. \]

The vector did not change; only the coordinate description changed.

Basis Coordinates as Instructions

The coordinate vector

\[ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 2 \end{bmatrix} \]

means:

\[ \boxed{ \mathbf{v} = 4\mathbf{b}_1+2\mathbf{b}_2 }. \]

Coordinates tell us how much of each basis direction is needed to build the vector.

The Basis Matrix

Place the basis vectors into the columns of a matrix:

\[ P_B = \begin{bmatrix} \vert&\vert\\ \mathbf{b}_1&\mathbf{b}_2\\ \vert&\vert \end{bmatrix}. \]

For

\[ \mathbf{b}_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix}, \]

we have

\[ \boxed{ P_B = \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} }. \]

From Basis Coordinates to Standard Coordinates

If

\[ [\mathbf{v}]_B = \begin{bmatrix} c_1\\ c_2 \end{bmatrix}, \]

then

\[ \boxed{ \mathbf{v} = P_B[\mathbf{v}]_B }. \]

The basis matrix converts \(B\)-coordinates into standard coordinates.

Example: Convert to Standard Coordinates

Let

\[ P_B = \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \]

and

\[ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 2 \end{bmatrix}. \]

Then

\[ \mathbf{v} = \begin{bmatrix} 1&1\\ 1&-1 \end{bmatrix} \begin{bmatrix} 4\\ 2 \end{bmatrix}. \]

Therefore,

\[ \boxed{ \mathbf{v} = \begin{bmatrix} 6\\ 2 \end{bmatrix} }. \]

From Standard Coordinates to Basis Coordinates

Starting from

\[ \mathbf{v} = P_B[\mathbf{v}]_B, \]

multiply by \(P_B^{-1}\):

\[ P_B^{-1}\mathbf{v} = [\mathbf{v}]_B. \]

Thus,

\[ \boxed{ [\mathbf{v}]_B = P_B^{-1}\mathbf{v} }. \]

Why a Basis Matrix Must Be Invertible

A valid basis must give every vector exactly one coordinate representation.

Therefore, the basis matrix must be invertible.

If the columns were dependent, then the matrix would collapse information and could not provide unique coordinates.

Basis and Invertibility

In \(\mathbb{R}^2\), two vectors form a basis exactly when the matrix having those vectors as columns is invertible.

Thus,

\[ \boxed{ \{\mathbf{b}_1,\mathbf{b}_2\} \text{ is a basis} \iff P_B \text{ is invertible} }. \]

Example: A Pair That Is Not a Basis

Consider

\[ \mathbf{b}_1 = \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 2\\ 4 \end{bmatrix}. \]

Since

\[ \mathbf{b}_2=2\mathbf{b}_1, \]

the vectors are parallel.

They do not form a basis of the plane.

Change of Coordinates Is Not a Geometric Transformation

This distinction is essential.

A geometric transformation changes the vector itself:

\[ \mathbf{v}\mapsto A\mathbf{v}. \]

A change of basis can leave the geometric vector unchanged while changing only the coordinate numbers used to describe it.

Active versus Passive Viewpoints

One useful distinction is:

The same matrix ideas can appear in both settings, so interpretation matters.

Changing from One Basis to Another

Suppose \(B\) and \(C\) are two bases.

A vector can have coordinate forms

\[ [\mathbf{v}]_B \]

and

\[ [\mathbf{v}]_C. \]

To convert between them, we may pass through standard coordinates.

From \(B\)-Coordinates to \(C\)-Coordinates

First convert \(B\)-coordinates to the standard vector:

\[ \mathbf{v} = P_B[\mathbf{v}]_B. \]

Then convert the standard vector to \(C\)-coordinates:

\[ [\mathbf{v}]_C = P_C^{-1}\mathbf{v}. \]

Combining these steps gives

\[ \boxed{ [\mathbf{v}]_C = P_C^{-1}P_B[\mathbf{v}]_B }. \]

The Change-of-Basis Matrix

The matrix

\[ \boxed{ P_{C\leftarrow B} = P_C^{-1}P_B } \]

converts \(B\)-coordinates directly into \(C\)-coordinates:

\[ \boxed{ [\mathbf{v}]_C = P_{C\leftarrow B}[\mathbf{v}]_B }. \]

Reading the Arrow Notation

The notation

\[ P_{C\leftarrow B} \]

may be read as:

\[ \boxed{ \text{“coordinates in }B \text{ converted to coordinates in }C\text{”} }. \]

The arrow points toward the destination basis.

The Reverse Change of Basis

To convert from \(C\)-coordinates back to \(B\)-coordinates,

\[ \boxed{ P_{B\leftarrow C} = P_B^{-1}P_C }. \]

The two change-of-basis matrices are inverses:

\[ \boxed{ P_{B\leftarrow C} = \left(P_{C\leftarrow B}\right)^{-1} }. \]

Example with the Standard Basis as Destination

If \(C=E\) is the standard basis, then

\[ P_E=I. \]

Therefore,

\[ P_{E\leftarrow B} = I^{-1}P_B = P_B. \]

This confirms that the basis matrix itself converts \(B\)-coordinates into standard coordinates.

Example with the Standard Basis as Source

If we convert from standard coordinates into basis \(B\), then

\[ P_{B\leftarrow E} = P_B^{-1}I. \]

Therefore,

\[ \boxed{ P_{B\leftarrow E} = P_B^{-1} }. \]

Three-Dimensional Bases

In \(\mathbb{R}^3\), a basis consists of three independent vectors:

\[ B= \{\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3\}. \]

Every vector can then be written uniquely as

\[ \boxed{ \mathbf{v} = c_1\mathbf{b}_1 + c_2\mathbf{b}_2 + c_3\mathbf{b}_3 }. \]

The Three-Dimensional Basis Matrix

Place the basis vectors into columns:

\[ \boxed{ P_B = \begin{bmatrix} \vert&\vert&\vert\\ \mathbf{b}_1&\mathbf{b}_2&\mathbf{b}_3\\ \vert&\vert&\vert \end{bmatrix} }. \]

Then

\[ \mathbf{v} = P_B[\mathbf{v}]_B \]

and

\[ [\mathbf{v}]_B = P_B^{-1}\mathbf{v}. \]

The Same Principle in Higher Dimensions

In an \(n\)-dimensional vector space, a basis contains \(n\) independent vectors.

Every vector is represented by \(n\) coordinates relative to that basis.

The basis matrix collects the basis vectors as columns and translates between basis coordinates and a reference coordinate system.

Why Different Bases Are Useful

The standard basis is convenient, but another basis can reveal structure that is hidden in ordinary coordinates.

A useful basis may:

Example: Basis Aligned with Diagonals

The vectors

\[ \mathbf{b}_1 = \begin{bmatrix} 1\\ 1 \end{bmatrix} \]

and

\[ \mathbf{b}_2 = \begin{bmatrix} 1\\ -1 \end{bmatrix} \]

align with the diagonal lines

\[ y=x \]

and

\[ y=-x. \]

Problems involving those directions may become simpler in this basis.

Coordinates Depend on Scale as Well as Direction

Two bases may point along the same directions but use differently sized basis vectors.

For example,

\[ \left\{ \begin{bmatrix} 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \end{bmatrix} \right\} \]

and

\[ \left\{ \begin{bmatrix} 2\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 3 \end{bmatrix} \right\} \]

use the same axis directions but different coordinate units.

Example: Rescaled Basis Coordinates

Let

\[ \mathbf{b}_1 = \begin{bmatrix} 2\\ 0 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 0\\ 3 \end{bmatrix}. \]

For

\[ \mathbf{v} = \begin{bmatrix} 8\\ 9 \end{bmatrix}, \]

we have

\[ \mathbf{v} = 4\mathbf{b}_1+3\mathbf{b}_2. \]

Therefore,

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 3 \end{bmatrix} }. \]

Coordinate Axes versus Basis Vectors

In the standard basis, the basis vectors align with the familiar coordinate axes.

In a nonstandard basis, the coordinate directions may be tilted, stretched, or otherwise different.

The coordinate grid is conceptually built from the chosen basis directions.

Oblique Bases

Basis vectors do not need to be perpendicular.

For example,

\[ \mathbf{b}_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 1\\ 1 \end{bmatrix} \]

form a valid basis even though they meet at an angle other than \(90^\circ\).

Such a basis is called oblique.

Orthonormal Bases

A basis is orthonormal when its basis vectors are mutually perpendicular and each has magnitude \(1\).

The standard basis is orthonormal.

Orthonormal bases are especially convenient because lengths, angles, and coordinate projections behave cleanly.

Coordinate Extraction in an Orthonormal Basis

If

\[ B= \{\mathbf{b}_1,\mathbf{b}_2\} \]

is orthonormal, then the coordinates of \(\mathbf{v}\) are

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} \mathbf{v}\cdot\mathbf{b}_1\\ \mathbf{v}\cdot\mathbf{b}_2 \end{bmatrix} }. \]

Each coordinate is simply the projection onto one unit basis direction.

Why the Dot Product Formula Works for Orthonormal Bases

If

\[ \mathbf{v} = c_1\mathbf{b}_1+c_2\mathbf{b}_2, \]

take the dot product with \(\mathbf{b}_1\):

\[ \mathbf{v}\cdot\mathbf{b}_1 = c_1(\mathbf{b}_1\cdot\mathbf{b}_1) + c_2(\mathbf{b}_2\cdot\mathbf{b}_1). \]

Orthonormality gives

\[ \mathbf{b}_1\cdot\mathbf{b}_1=1 \]

and

\[ \mathbf{b}_2\cdot\mathbf{b}_1=0. \]

Therefore,

\[ \boxed{ c_1=\mathbf{v}\cdot\mathbf{b}_1 }. \]

Matrices Also Depend on the Chosen Basis

Just as vectors have different coordinate representations in different bases, a linear transformation can have different matrix representations.

The transformation itself may be unchanged while the matrix used to describe it changes.

A Transformation in Standard Coordinates

Suppose a linear transformation \(T\) is represented in the standard basis by matrix

\[ A. \]

Then

\[ \mathbf{v} \mapsto A\mathbf{v}. \]

We may ask for the matrix that represents the same transformation when both input and output coordinates are written in basis \(B\).

Transformation Matrix in a New Basis

Start with \(B\)-coordinates:

\[ [\mathbf{v}]_B. \]

Convert to standard coordinates:

\[ \mathbf{v} = P_B[\mathbf{v}]_B. \]

Apply the transformation:

\[ A\mathbf{v} = AP_B[\mathbf{v}]_B. \]

Then convert the result back to \(B\)-coordinates:

\[ P_B^{-1}AP_B[\mathbf{v}]_B. \]

The Similarity Transformation Formula

Therefore, the matrix representing \(T\) in basis \(B\) is

\[ \boxed{ [T]_B = P_B^{-1}AP_B }. \]

This formula changes the coordinate description of the transformation without changing the underlying linear transformation.

Why the Order in \(P_B^{-1}AP_B\) Matters

Read from right to left:

  1. \(P_B\) converts \(B\)-coordinates into standard coordinates.
  2. \(A\) applies the transformation in standard coordinates.
  3. \(P_B^{-1}\) converts the result back into \(B\)-coordinates.

Matrix composition records the coordinate workflow exactly.

Why a New Basis Can Simplify a Matrix

A complicated transformation matrix in the standard basis can sometimes become much simpler in a basis aligned with important transformation directions.

In especially favorable cases, the new matrix becomes diagonal.

This possibility leads directly toward eigenvectors and eigenvalues.

Basis Vectors as a Coordinate Framework

A basis gives a reference framework for describing vectors.

Choosing a basis is similar to choosing:

Changing that framework changes coordinates but need not change the vector.

Common Error: Assuming Coordinates Are Intrinsic

The coordinate list

\[ \begin{bmatrix} x\\ y \end{bmatrix} \]

has meaning only after the basis is known.

The same pair of numbers can describe different geometric vectors in different bases.

Common Error: Confusing \(\mathbf{v}\) with \([\mathbf{v}]_B\)

The notation

\[ \mathbf{v} \]

refers to the vector itself.

The notation

\[ [\mathbf{v}]_B \]

refers to the coordinate vector relative to basis \(B\).

They may contain different numerical entries.

Common Error: Placing Basis Vectors as Rows

In the standard column-vector convention, the basis matrix is formed by placing basis vectors as columns:

\[ \boxed{ P_B = \begin{bmatrix} \vert&\vert\\ \mathbf{b}_1&\mathbf{b}_2\\ \vert&\vert \end{bmatrix} }. \]

Replacing those columns with rows changes the conversion rule.

Common Error: Using \(P_B\) in the Wrong Direction

The basis matrix converts from \(B\)-coordinates to standard coordinates:

\[ \mathbf{v} = P_B[\mathbf{v}]_B. \]

The inverse converts from standard coordinates to \(B\)-coordinates:

\[ [\mathbf{v}]_B = P_B^{-1}\mathbf{v}. \]

Common Error: Using Dependent Vectors as a Basis

A basis must provide independent directions.

If one candidate basis vector is a scalar multiple or linear combination of the others in a way that creates dependence, the basis matrix is not invertible.

Unique coordinates then fail.

Common Error: Treating Change of Basis as Moving the Vector

A passive change of basis can leave the actual vector fixed.

Only its coordinate representation changes.

This differs from an active transformation such as a rotation or scaling, which changes the vector itself.

Common Error: Reversing Change-of-Basis Order

To convert \(B\)-coordinates to \(C\)-coordinates,

\[ \boxed{ [\mathbf{v}]_C = P_C^{-1}P_B[\mathbf{v}]_B }. \]

Reversing the product generally gives the wrong conversion.

Common Error: Forgetting That Matrices Also Change Representation

A transformation matrix is not independent of basis.

The same linear transformation can have different matrices in different bases.

What stays the same is the underlying transformation.

A Reliable Procedure for Finding Basis Coordinates

  1. Write the vector as a linear combination of the basis vectors.
  2. Introduce one unknown coefficient for each basis vector.
  3. Match standard-coordinate components.
  4. Solve the resulting system.
  5. Place the coefficients into the coordinate vector in basis order.
  6. Verify by reconstructing the original vector.

A Reliable Procedure for Converting to Standard Coordinates

  1. Place the basis vectors into the columns of \(P_B\).
  2. Write the basis-coordinate vector \([\mathbf{v}]_B\).
  3. Multiply \(P_B[\mathbf{v}]_B\).
  4. Interpret the result as the standard coordinate vector \(\mathbf{v}\).

A Reliable Procedure for Converting from Standard Coordinates

  1. Construct the basis matrix \(P_B\).
  2. Verify that it is invertible.
  3. Compute or otherwise determine \(P_B^{-1}\).
  4. Multiply \(P_B^{-1}\mathbf{v}\).
  5. Interpret the result as \([\mathbf{v}]_B\).
  6. Check by multiplying back by \(P_B\).

A Reliable Procedure for Changing from Basis \(B\) to Basis \(C\)

  1. Construct \(P_B\) and \(P_C\).
  2. Convert \(B\)-coordinates to standard coordinates with \(P_B\).
  3. Convert standard coordinates to \(C\)-coordinates with \(P_C^{-1}\).
  4. Combine the two steps as \(P_C^{-1}P_B\).
  5. Apply the resulting change-of-basis matrix.
  6. Verify by reconstructing the same vector in both bases.

A Reliable Procedure for Rewriting a Transformation in a New Basis

  1. Identify the standard matrix \(A\).
  2. Construct the basis matrix \(P_B\).
  3. Compute \(P_B^{-1}\).
  4. Form \(P_B^{-1}AP_B\).
  5. Interpret the result as the matrix of the same transformation in basis \(B\).
  6. Check the transformation on a simple basis-coordinate vector if needed.

Example: Full Basis-Coordinate Workflow

Let

\[ \mathbf{b}_1= \begin{bmatrix} 2\\ 1 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 1\\ -1 \end{bmatrix} \]

and let

\[ \mathbf{v} = \begin{bmatrix} 7\\ 1 \end{bmatrix}. \]

Find \([\mathbf{v}]_B\).

Set Up the Full Basis Problem

Write

\[ c_1 \begin{bmatrix} 2\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1\\ -1 \end{bmatrix} = \begin{bmatrix} 7\\ 1 \end{bmatrix}. \]

Matching components gives

\[ 2c_1+c_2=7 \]

and

\[ c_1-c_2=1. \]

Complete the Full Basis Problem

From

\[ c_1-c_2=1, \]

we have

\[ c_2=c_1-1. \]

Substitute:

\[ 2c_1+(c_1-1)=7. \]

Therefore,

\[ 3c_1=8, \qquad c_1=\frac83. \]

Then

\[ c_2=\frac53. \]

Thus,

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} \frac83\\ \frac53 \end{bmatrix} }. \]

Verify the Full Basis Problem

Reconstruct:

\[ \frac83 \begin{bmatrix} 2\\ 1 \end{bmatrix} + \frac53 \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]

This gives

\[ \begin{bmatrix} \frac{16}{3}\\ \frac83 \end{bmatrix} + \begin{bmatrix} \frac53\\ -\frac53 \end{bmatrix} = \begin{bmatrix} 7\\ 1 \end{bmatrix}. \]

The coordinate representation is verified.

Practice 1: Standard-Basis Interpretation

Write

\[ \begin{bmatrix} 5\\ -2 \end{bmatrix} \]

as a linear combination of \(\mathbf{i}\) and \(\mathbf{j}\).

Practice 2: Find Basis Coordinates

Let

\[ \mathbf{b}_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}. \]

Find the \(B\)-coordinates of

\[ \mathbf{v} = \begin{bmatrix} 8\\ 4 \end{bmatrix}. \]

Practice 3: Convert to Standard Coordinates

Let

\[ P_B= \begin{bmatrix} 2&1\\ 0&3 \end{bmatrix} \]

and

\[ [\mathbf{v}]_B = \begin{bmatrix} 4\\ 2 \end{bmatrix}. \]

Find \(\mathbf{v}\) in standard coordinates.

Practice 4: Determine Whether Vectors Form a Basis

Determine whether

\[ \begin{bmatrix} 1\\ 2 \end{bmatrix} \]

and

\[ \begin{bmatrix} 3\\ 6 \end{bmatrix} \]

form a basis of \(\mathbb{R}^2\).

Practice 5: Rescaled Basis

Let

\[ \mathbf{b}_1= \begin{bmatrix} 2\\ 0 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 0\\ 4 \end{bmatrix}. \]

Find the \(B\)-coordinates of

\[ \begin{bmatrix} 10\\ 12 \end{bmatrix}. \]

Practice 6: Read a Basis Matrix

The basis matrix is

\[ P_B= \begin{bmatrix} 3&1\\ 2&4 \end{bmatrix}. \]

State the two basis vectors.

Practice 7: Orthonormal Coordinates

Let

\[ \mathbf{b}_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 0\\ -1 \end{bmatrix}. \]

Find the \(B\)-coordinates of

\[ \mathbf{v} = \begin{bmatrix} 3\\ 5 \end{bmatrix}. \]

Practice 8: Interpret a Coordinate Vector

Suppose

\[ [\mathbf{v}]_B = \begin{bmatrix} -2\\ 5 \end{bmatrix}. \]

Write \(\mathbf{v}\) as a linear combination of \(\mathbf{b}_1\) and \(\mathbf{b}_2\).

Practice 9: Transformation Matrix in a New Basis

If \(A\) is the standard matrix of a transformation and \(P_B\) is a basis matrix, write the formula for the matrix of the same transformation in basis \(B\).

Practice 10: Identify the Error

A student says that changing from the standard basis to another basis must physically rotate or move the vector.

Explain the error.

Practice Answers

Practice 1

\[ \boxed{ \begin{bmatrix} 5\\ -2 \end{bmatrix} = 5\mathbf{i}-2\mathbf{j} }. \]

Practice 2

Solve

\[ c_1+c_2=8 \]

and

\[ c_1-c_2=4. \]

This gives

\[ c_1=6, \qquad c_2=2. \]

Therefore,

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} 6\\ 2 \end{bmatrix} }. \]

Practice 3

\[ \mathbf{v} = \begin{bmatrix} 2&1\\ 0&3 \end{bmatrix} \begin{bmatrix} 4\\ 2 \end{bmatrix} = \begin{bmatrix} 10\\ 6 \end{bmatrix}. \]

Thus,

\[ \boxed{ \mathbf{v} = \begin{bmatrix} 10\\ 6 \end{bmatrix} }. \]

Practice 4

No. Since

\[ \begin{bmatrix} 3\\ 6 \end{bmatrix} = 3 \begin{bmatrix} 1\\ 2 \end{bmatrix}, \]

the vectors are dependent and parallel.

Practice 5

\[ \begin{bmatrix} 10\\ 12 \end{bmatrix} = 5 \begin{bmatrix} 2\\ 0 \end{bmatrix} + 3 \begin{bmatrix} 0\\ 4 \end{bmatrix}. \]

Therefore,

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} 5\\ 3 \end{bmatrix} }. \]

Practice 6

The basis vectors are the columns:

\[ \boxed{ \mathbf{b}_1= \begin{bmatrix} 3\\ 2 \end{bmatrix}, \qquad \mathbf{b}_2= \begin{bmatrix} 1\\ 4 \end{bmatrix} }. \]

Practice 7

Since

\[ \mathbf{v} = 3\mathbf{b}_1-5\mathbf{b}_2, \]

we have

\[ \boxed{ [\mathbf{v}]_B = \begin{bmatrix} 3\\ -5 \end{bmatrix} }. \]

Practice 8

\[ \boxed{ \mathbf{v} = -2\mathbf{b}_1+5\mathbf{b}_2 }. \]

Practice 9

\[ \boxed{ [T]_B = P_B^{-1}AP_B }. \]

Practice 10

A passive change of basis does not need to move the vector at all.

The vector may remain fixed while only the coordinate system used to describe it changes.

What to Remember

Where We Go Next

Changing basis can reveal directions in which a transformation becomes much simpler. Some especially important vectors keep their direction under a transformation and are only scaled.

The next lesson, Eigenvalues, Eigenvectors, and Spatial Effects, develops these special directions, their scale factors, and the geometric structure they reveal inside matrix transformations.