Worked examples · Sheet 1

Linearity checks

Theory used: Linear map First properties Matrix with respect to the standard bases

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine whether the function $f : R^{2} \to R^{3}$

f (x, y) = (x + 2 y, 3 x - y, x)

is linear.

u = (x_{1}, y_{1}), v = (x_{2}, y_{2}), λ \in R

f (u + v) = ((x_{1} + x_{2}) + 2 (y_{1} + y_{2}), 3 (x_{1} + x_{2}) - (y_{1} + y_{2}), x_{1} + x_{2}) = (x_{1} + 2 y_{1}, 3 x_{1} - y_{1}, x_{1}) + (x_{2} + 2 y_{2}, 3 x_{2} - y_{2}, x_{2}) = f (u) + f (v)

f (λ u) = (λ x_{1} + 2 λ y_{1}, 3 λ x_{1} - λ y_{1}, λ x_{1}) = λ f (u)

f (x, y) = 131 2 - 1 0 (x y)

To prove linearity the two conditions with generic vectors and scalars are needed, so we write $u$ and $v$ with literal components. In the sum $f (u + v)$ each component is a polynomial of degree one with no constant term, therefore it splits into the part with index $1$ and the part with index $2$ , and we obtain $f (u) + f (v)$ . In the same way $λ$ factors out of each component of $f (λ u)$ . The two conditions are satisfied, so $f$ is linear. The same conclusion can be read from the expression $f (x, y) = A (x y)$ , since every function of the form $x \mapsto A x$ is linear; the rows of $A$ contain the coefficients of the three components.

Exercise 2

Determine whether the following functions are linear:

g (x, y) = (x - y, x y), h (x, y, z) = (x + z, y - 1) .

h (0, 0, 0) = (0, - 1) \neq = (0, 0)

g (0, 0) = (0, 0), g (2, 2) = (0, 4), 2 g (1, 1) = 2 (0, 1) = (0, 2)

g (2 \cdot (1, 1)) \neq = 2 g (1, 1)

A linear map sends the zero vector to the zero vector. Since $h (0, 0, 0) = (0, - 1)$ , the function $h$ is not linear; the cause is the constant term $- 1$ in the second component.

For $g$ the check on the zero vector is passed, since $g (0, 0) = (0, 0)$ ; this check is necessary and not sufficient, therefore we look for a counterexample to one of the two conditions. The product $x y$ is of degree two, so doubling the vector multiplies the second component by $4$ . Indeed $g (2, 2) = (0, 4)$ , while $2 g (1, 1) = (0, 2)$ , therefore $g$ is not linear.

Exercise 3

Let $T : R_{2} [x] \to R^{2}$ be defined by

T (p) = (p (1), p^{'} (0)) .

Prove that $T$ is linear, write the matrix with respect to the basis $(1, x, x^{2})$ of $R_{2} [x]$ and the standard basis of $R^{2}$ , and compute $T (1 + 2 x - x^{2})$ .

T (p + q) T (λ p) = (p (1) + q (1), p^{'} (0) + q^{'} (0)) = T (p) + T (q) = (λ p (1), λ p^{'} (0)) = λ T (p)

T (1) = (1, 0), T (x) = (1, 1), T (x^{2}) = (1, 0)

M (T) = (101110)

(101110) 12 - 1 = (22) ⟹ T (1 + 2 x - x^{2}) = (2, 2)

The value at a point of the sum of two polynomials is the sum of the values, and the derivative of the sum is the sum of the derivatives; moreover a multiplicative constant can be taken out of both the value and the derivative. Hence the two components of $T$ both satisfy the conditions, and $T$ is linear.

The columns of the matrix are the images of the basis polynomials. For $p = 1$ we have $p (1) = 1$ and $p^{'} = 0$ ; for $p = x$ we have $p (1) = 1$ and $p^{'} = 1$ ; for $p = x^{2}$ we have $p (1) = 1$ and $p^{'} (0) = 2 \cdot 0 = 0$ . The polynomial $1 + 2 x - x^{2}$ has coordinates $(1, 2, - 1)$ with respect to the basis, and the product of the matrix by these coordinates gives $(2, 2)$ . The result is confirmed directly, since $p (1) = 1 + 2 - 1 = 2$ and $p^{'} (x) = 2 - 2 x$ equals $2$ at $x = 0$ .

Exercise 4

Determine for which values of the parameter $k \in R$ the function

f_{k} (x, y) = (x + k^{2} - 1, (k - 1) x y + y)

is linear.

f_{k} (0, 0) = (k^{2} - 1, 0) = (0, 0) ⟺ k = 1 or k = - 1

k = 1 : f_{1} (x, y) = (x, y)

k = - 1 : f_{- 1} (x, y) = (x, - 2 x y + y), f_{- 1} (2, 2) = (2, - 6), 2 f_{- 1} (1, 1) = (2, - 2)

The condition $f_{k} (0, 0) = (0, 0)$ is necessary for linearity, so it is imposed first. The first component of $f_{k} (0, 0)$ is $k^{2} - 1$ , which vanishes only for $k = \pm 1$ . For all other values of $k$ the function is not linear.

The two values found remain to be examined. For $k = 1$ the term $(k - 1) x y$ vanishes and $f_{1}$ is the identity of $R^{2}$ , which is linear. For $k = - 1$ the degree-two term $- 2 x y$ remains, and the counterexample with $(1, 1)$ and $λ = 2$ shows that $f_{- 1} (2, 2) = (2, - 6)$ is different from $2 f_{- 1} (1, 1) = (2, - 2)$ . Therefore $f_{k}$ is linear if and only if $k = 1$ .