Shift of Basis

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Related pages:

Shift of Basis for Vectors
Shift of Basis (in general).

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Representation and Reconstruction of a Presentant with respect to a Background

Representation: $\, [ \, p_{resentant} \, ]_{B_{ackground}} \, \mapsto \, \left< \, r_{epresentant} \, \right>_{B_{ackground}}$

Reconstruction: $\, \left( \, \left< \, r_{epresentant} \, \right>_{B_{ackground}} \, \right)_{B_{ackground}} \mapsto \,\, p_{resentant}$

/////// In Swedish:

Representation och Rekonstruktion av en Presentant med avseende på en Bakgrund

Representation: $\, [ \, p_{resentant} \, ]_{B_{akgrund}} \, \mapsto \, \left< \, r_{epresentant} \, \right>_{B_{akgrund}}$

Rekonstruktion: $\, \left( \, \left< \, r_{epresentant} \, \right>_{B_{akgrund}} \, \right)_{B_{akgrund}} \mapsto \,\, p_{resentant}$

/////// Back to English:

Representation and Reconstruction of a Linear Transformation
with respect to a Basis for its Domain and and a Basis for its Codomain
:

Representation: $[ \,\, a \, l_{inearMap} \, ]_{a \, B_{asisForItsCodomain}}^{a \, B_{asisForItsDomain}} \, \mapsto \, \left< \, a \, m_{atrix} \, \right>_{a \, B_{asisForItsCodomain}}^{a \, B_{asisForItsDomain}}$

Reconstruction: $\left( \, \left< \, t_{he} \, m_{atrix} \, \right>_{t_{he} \, B_{asisForItsCodomain}}^{t_{he} \, B_{asisForItsDomain}} \, \right)_{t_{he} \, B_{asisForItsCodomain}}^{t_{he} \, B_{asisForItsDomain}} \, \mapsto \, t_{he} \, l_{inearMap}$

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Let $\, X \,$ and $\, Y \,$ be two linear spaces (= vector spaces).
Given a linear map $\; F: X \rightarrow Y \;$, we can express its action on a vector $\, x \in X \,$ as

$\, Y \ni F(x) \leftarrow x \in X$.

This reversal of the usual direction in the visual representation of $\, F \,$ is due to advantages in connecting smoothly with matrix algebra, and it is demonstrated in our section on Linear Transformations.

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The Basis-Shift formula for Linear Transformations:

Let $\, U \,$ and $\, U' \,$ be two different bases for $\, X$,
and let $\, V \,$ and $\, V' \,$ be two different bases for $\, Y$.

Hence (from the basis shift formula for vectors) we can write:

$U \triangleq \begin{bmatrix} | & & | \\ u_1 & \cdots & u_m \\ | & & | \end{bmatrix}, \; U' \triangleq \begin{bmatrix} | & & | \\ u'_1 & \cdots & u'_m \\ | & & | \end{bmatrix} \,$,

$V \triangleq \begin{bmatrix} \, | & & | \\ v_1 & \cdots & v_n \\ | & & | \end{bmatrix}, \; V' \triangleq \begin{bmatrix} \, | & & | \\ v'_1 & \cdots & v'_n \\ | & & | \end{bmatrix}$.

From the discussion above we have:

$[F(x)]_V \equiv [\, F \,]_V^U \, [\, x \,]_U \,$

$[F(x)]_{V'} \equiv [\, F \,]_{V'}^{U'} \, [\, x \,]_{U'} \,$.

Moreover

$[\, x \,]_{U'} \equiv {[B_{asisShift}]}_{U'}^U [x]_U \,$, and

$[F(x)]_{V'} \equiv {[B_{asisShift}]}_{V'}^V [F(x)]_V \,$.

Hence we get

$[F(x)]_{V'} \equiv [B_{asisShift}]_{V'}^V [\, F \,]_V^U [B_{asisShift}]_U^{U'} [\, x \,]_{U'} \,$,

and therefore we arrive at

the basis-shift formula for linear transformations:

$[\, F \,]_{V'}^{U'} \equiv [B_{asisShift}]_{V'}^V [\, F \,]_V^U [B_{asisShift}]_U^{U'} \,$.

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Change of basis for the domain
and the codomain of a linear transformation
:

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$[\;\;]_U^{U'} \triangleq {[\left( \;\; \right)_{U'}]}_U$

$[\;\;]_U \equiv [\;\;]_U^{U'} [\;\;]_{U'} \,$

$[F(x)]_V \equiv [\, F \,]_V^U \, [\, x \,]_U \,$

$[\, F(x) \,]_{V'} \equiv [\, F \,]_{V'}^{U'} [ \, x \, ]_{U'} \equiv [\;\;]_{V'}^V [\, F \,]_V^U [\;\;]_U^{U'} [ \, x \, ]_{U'} \,$.

$[\, F \,]_{V'}^{U'} \equiv [\;\;]_{V'}^V [\, F \,]_V^U [\;\;]_U^{U'} \,$.

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Numbers and Vectors are naturally related to each other:

Hence Numbers and Vectors are naturally equivalent.

Numbers and Vectors are naturally related to Music:

Check out our section on Category Theory

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