Linear Maps

Linear Maps

Definition: A Linear Map or Linear Transformation from the vector spaces $V$ to $W$ over the field $\mathbb{F}$ is a function $T : V \to W$ with the property that $T(au + bv) = aT(u) + bT(v)$ for all vectors $u, v \in V$ and for all scalars $a, b \in \mathbb{F}$. The Set of All Linear Maps from $V$ to $W$ is denoted $\mathcal L (V, W)$.

An equivalent definition of a linear map $T$ from $V$ to $W$ is a function $T : V \to W$ such that for all vectors $u, v \in V$, $T(u + v) + T(u) + T(v)$ (Additivity) and for all $a \in \mathbb{F}$, $T(au) = aT(u)$ (Homogeneity).

We will now look at some examples of linear maps and verify that they in fact a linear maps by showing that $T(au + bv) = aT(u) + bT(v)$ for all $u, v \in V$ and $a, b \in \mathbb{F}$.

The Zero Map

The zero map denoted $0 \in \mathcal L (V, W)$ defined by $0(v) = 0$ maps every vector $v \in V$ to the zero vector $0_W \in W$.

Let $u, v \in V$ and $a, b \in \mathbb{F}$.

  • $0(au + bv) = 0_W$
  • $a 0(u) + b 0(v) = 0_W + 0_W = 0_W$.

Therefore $0(au + bv) = a0(u) + b0(v)$, so we have verified that $0$ is a linear map.

The Identity Map

The identity map denoted $I \in \mathcal L (V, V)$ defined by $I(v) = v$ maps every vector $v \in V$ back to itself.

Let $u, v \in V$ and $a, b \in \mathbb{F}$.

  • $I(au + bv) = au + bv$
  • $aI(u) + bI(v) = au + bv$.

Therefore $I(au + bv) = aI(u) + bI(v)$, so we have verified that $I$ is a linear map.

The Differentiation of Polynomials Map

The differentiation of polynomials map $T \in \mathcal L (\wp (\mathbb{R}), \wp (\mathbb{R}))$ defined by $T(p(x)) = p'(x)$ takes every vector $p(x) \in \wp (\mathbb{R})$ and maps it to its derivative $p'(x)$.

Let $p(x), q(x) \in \wp ( \mathbb{R} )$ and $a, b \in \mathbb{F}$.

  • $T(a p(x) + b q(x)) = (ap(x) + bq(x))' = ap'(x) + bq'(x)$
  • $aT(p(x)) + bT(q(x)) = ap'(x) + bp'(x)$.

Therefore $T(ap(x) + bq(x)) = aT(p(x)) + bT(q(x))$, and so we have verified that $T$ is a linear map.

The Definite Integration of Polynomials Map

The integration of polynomials map from $c$ to $d$ ($a, b \in \mathbb{R}$) denoted $T \in \mathcal L (\wp (\mathbb{R}), \mathbb{R} )$ defined by $T(p(x)) = \int_c^d p(x) \: dx$ maps vector $p(x) \in \wp (\mathbb{R})$ to a number in $\mathbb{R}$.

Let $p(x), q(x) \in \wp ( \mathbb{R} )$ and $a, b \in \mathbb{F}$.

  • $T(a p(x) + b q(x)) = \int_c^d ap(x) + bq(x) \: dx = a \int_c^d p(x) \: dx + b \int_c^d q(x) \: dx$.
  • $aT(p(x)) + bT(q(x)) = a \int_c^d p(x) \: dx + b \int_c^d q(x) \: dx$.

Therefore $T(ap(x) + bq(x)) = aT(p(x)) + bT(q(x))$, and so we have verified that $T$ is a linear map.

The Left Shift Operator

The left shift operator $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ defined by $T((x_1, x_2, ...)) = (x_2, x_3, ...)$ takes every sequence in $\mathbb{F}^{\infty}$ and shifts every term one position to the left.

Let $(x_1, x_2, ...), (y_1, y_2, ...) \in \mathbb{F}^{\infty}$ and $a, b \in \mathbb{F}$.

  • $T(a(x_1, x_2, ...) + b(y_1, y_2, ...)) = T((ax_1 + by_1, ax_2 + by_2, ...)) = (ax_2 + by_2, ax_3 + by_3, ...)$.
  • $aT((x_1, x_2, ...)) + bT((y_1, y_2, ...)) = a(x_2, x_3, ...) + b(y_2, y_3, ...) = (ax_2, ax_3, ...) + (by_2, by_3, ...) = (ax_2 + by_2, ax_3 + by_3, ...)$.

Therefore $T(a(x_1, x_2, ...) + b(y_1, y_2, ...)) = aT((x_1, x_2, ...)) + bT((y_1, y_2, ...))$, and so we have verified that $T$ is a linear map.

The Right Shift Operator

The right shift operator $T \in \mathcal L (\mathbb{F}^{\infty}, \mathbb{F}^{\infty})$ defined by $T((x_1, x_2, ...)) = (0, x_1, ...)$ takes every sequence in $\mathbb{F}^{\infty}$ and shifts every term one position to the right while replacing the first term with zero.

Let $(x_1, x_2, ...), (y_1, y_2, ...) \in \mathbb{F}^{\infty}$ and $a, b \in \mathbb{F}$.

  • $T(a(x_1, x_2, ...) + b(y_1, y_2, ...)) = T((ax_1 + by_1, ax_2 + by_2, ...)) = (0, ax_1 + by_1, ...)$
  • $aT((x_1, x_2, ...)) + bT((y_1, y_2, ...)) = a(0, x_1, ...) + b(0, y_1 ...) = (0, ax_1, ...) + (0, by_1, ...) = (0, ax_1 + by_1, ...)$.

Therefore $T(a(x_1, x_2, ...) + b(y_1, y_2, ...)) = aT((x_1, x_2, ...)) + bT((y_1, y_2, ...))$, and so we have verified that $T$ is a linear map.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License