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Linear Transformations¶

For vectors $$x$$ and $$y$$, and scalars $$a$$ and $$b$$, it is sufficient to say that a function, $$F$$, is a linear transformation if

$F(ax + by) = aF(x) + bF(y).$

It can be shown that multiplying an $${m} \times {n}$$ matrix, $$A$$, and an $${n} \times {1}$$ vector, $$v$$, of compatible size is a linear transformation of $$v$$. Therefore from this point forward, a matrix will be synonymous with a linear transformation function.

TRY IT! Let $$x$$ be a vector and let $$F(x)$$ be defined by $$F(x) = Ax$$ where $$A$$ is a rectangular matrix of appropriate size. Show that $$F(x)$$ is a linear transformation.

Proof: Since $$F(x) = Ax$$, then for vectors $$v$$ and $$w$$, and scalars $$a$$ and $$b$$, $$F(av + bw) = A(av + bw)$$ (by definition of $$F$$)$$=$$$$aAv + bAw$$ (by distributive property of matrix multiplication)$$=$$$$aF(v) + bF(w)$$ (by definition of $$F$$).

If $$A$$ is an $${m} \times {n}$$ matrix, then there are two important subpsaces associated with $$A$$, one is $${\mathbb{R}}^n$$, the other is $${\mathbb{R}}^m$$. The domain of $$A$$ is a subspace of $${\mathbb{R}}^n$$. It is the set of all vectors that can be multiplied by $$A$$ on the right. The range of $$A$$ is a subspace of $${\mathbb{R}}^m$$. It is the set of all vectors $$y$$ such that $$y=Ax$$. It can be denoted as $$\mathcal{R}(\mathbf{A})$$, where $$\mathcal{R}(\mathbf{A}) = \{y \in {\mathbb{R}}^m: Ax = y\}$$. Another way to think about the range of $$A$$ is the set of all linear combinations of the columns in $$A$$, where $$x_i$$ is the coefficient of the ith column in $$A$$. The null space of $$A$$, defined as $$\mathcal{N}(\mathbf{A}) = \{x \in {\mathbb{R}}^n: Ax = 0_m\}$$, is the subset of vectors in the domain of $$A, x$$, such that $$Ax = 0_m$$, where $$0_m$$ is the zero vector (i.e., a vector in $${\mathbb{R}}^m$$ with all zeros).

TRY IT! Let $$A = [[1, 0, 0], [0, 1, 0], [0, 0, 0]]$$ and let the domain of $$A$$ be $${\mathbb{R}}^3$$. Characterize the range and nullspace of $$A$$.

Let $$v = [x,y,z]$$ be a vector in $${\mathbb{R}}^3$$. Then $$u = Av$$ is the vector $$u = [x,y,0]$$. Since $$x,y\in {\mathbb{R}}$$, the range of $$A$$ is the $$x$$-$$y$$ plane at $$z = 0$$.

Let $$v = [0,0,z]$$ for $$z\in {\mathbb{R}}$$. Then $$u = Av$$ is the vector $$u = [0, 0, 0]$$. Therefore, the nullspace of $$A$$ is the $$z$$-axis (i.e., the set of vectors $$[0,0,z]$$ $$z\in {\mathbb{R}}$$).

Therefore, this linear transformation “flattens” any $$z$$-component from a vector.