Algebra / English

Change of basis | Chapter 13, Essence of linear algebra

3Blue1Brown · YouTube · 12:50

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IN THIS VIDEO

01Coordinates are coefficients
02An alternative basis
03Convert coordinates with a matrix

SELECTED MOMENT

Coordinates are coefficients

SUGGESTED PREPARATION

Vector addition and scalar multiplication
Matrix multiplication and inverses

CONCEPT CONNECTIONS

Basis and dimensionExplanation
Why this connection? · 0:32
The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ . Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ . Then $P(-1,2)^T=(-4,1)^T$ : this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ . To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$ , apply $A$ , then convert back with $P^{-1}$ . The resulting matrix is $P^{-1}AP$ . For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$ .
Basis and dimensionExplanation
Why this connection? · 0:32
The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ . Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ . Then $P(-1,2)^T=(-4,1)^T$ : this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ . To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$ , apply $A$ , then convert back with $P^{-1}$ . The resulting matrix is $P^{-1}AP$ . For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$ .

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A vector keeps its geometric meaning when its coordinates change. This visual lesson explains basis vectors, coordinate conversion and how to express the same linear transformation in a different basis.

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Chapters

0:32Coordinates are coefficients 1:40An alternative basis 4:40Convert coordinates with a matrix 7:30Use the inverse for the reverse direction 11:05Represent a transformation in the new basis

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis.

Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ .

Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ . Then $P(-1,2)^T=(-4,1)^T$ : this converts Jennifer's coordinates to standard coordinates.

The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ .

To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$ , apply $A$ , then convert back with $P^{-1}$ . The resulting matrix is $P^{-1}AP$ . For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$ .

Knowledge cards

Coordinates are coefficients

The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis.

An alternative basis

Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ .

Convert coordinates with a matrix

Use the inverse for the reverse direction

The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ .

Represent a transformation in the new basis

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Basis and dimension ExplanationAt 0:32
Why this connection?
The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ . Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ . Then $P(-1,2)^T=(-4,1)^T$ : this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ . To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$ , apply $A$ , then convert back with $P^{-1}$ . The resulting matrix is $P^{-1}AP$ . For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$ .
Basis and dimension ExplanationAt 0:32
Why this connection?
The coordinates $(3,2)$ describe $3\hat{i}+2\hat{j}$ . A coordinate pair makes sense only together with its chosen ordered basis. Jennifer uses $b_1=(2,1)$ and $b_2=(-1,1)$ . These vectors need not be perpendicular. The same vector $(3,2)$ has coordinates $(5/3,1/3)$ in her basis because $(5/3)b_1+(1/3)b_2=(3,2)$ . Place the new basis vectors, expressed in standard coordinates, in the columns of $P=\begin{bmatrix}2&-1\\1&1\end{bmatrix}$ . Then $P(-1,2)^T=(-4,1)^T$ : this converts Jennifer's coordinates to standard coordinates. The reverse conversion is $P^{-1}v$ . Here $P^{-1}=\begin{bmatrix}1/3&1/3\\-1/3&2/3\end{bmatrix}$ , so applying it to $(3,2)^T$ gives $(5/3,1/3)^T$ . To represent a standard-coordinate transformation $A$ in Jennifer's coordinates, first convert with $P$ , apply $A$ , then convert back with $P^{-1}$ . The resulting matrix is $P^{-1}AP$ . For a counterclockwise quarter-turn, the example gives $\begin{bmatrix}1/3&-2/3\\5/3&-1/3\end{bmatrix}$ .

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Change of basis | Chapter 13, Essence of linear algebra

The explanation, unpacked.

Chapters

Learning script

Knowledge cards

Coordinates are coefficients

An alternative basis

Convert coordinates with a matrix

Use the inverse for the reverse direction

Represent a transformation in the new basis

Explore the knowledge in this video

Questions this video answers

How do you convert coordinates from the standard basis to an alternative basis?+

Why does the coordinate pair (3,2) only make sense together with its chosen ordered basis?+

How do you convert coordinates from an alternative basis to the standard basis using a matrix?+

How do you represent a linear transformation in a new basis?+

Do basis vectors need to be perpendicular to form a valid basis?+

How do you find the coordinates of a vector in an alternative basis?+

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