Worked examples · Sheet 3

Computing kernel and image

Theory used: Kernel and image Computing kernel and image Generators of the image and rank Injectivity and kernel Surjectivity and rank Rank-nullity theorem

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Find the kernel and the image of the endomorphism of $R^{2}$

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

and determine whether $f$ is injective or surjective.

A = (1212), det A = 2 - 2 = 0, rk A = 1

f (x, y) = (0, 0) ⟺ x + y = 0 ⟹ ker f = ⟨(1, - 1)⟩

f (x, y) = (x + y) (1, 2) ⟹ Im f = ⟨(1, 2)⟩

y = 2 x

y = -x

The image of

f

is the line

y = 2 x

, the kernel is the line

y = - x

(dashed)

The second row of the matrix is twice the first, so the rank is $1$ . The kernel is the solution set of $f (x, y) = (0, 0)$ , that is, of the single equation $x + y = 0$ ; setting $x = t$ we obtain $y = - t$ , and the kernel is the line spanned by $(1, - 1)$ . Since $ker f \neq = {(0, 0)}$ , $f$ is not injective. Factoring out $x + y$ we obtain $f (x, y) = (x + y) (1, 2)$ , so every image is a multiple of $(1, 2)$ and the image is the line $y = 2 x$ ; it has dimension $1 < 2$ , so $f$ is not surjective. The dimensions satisfy the rank-nullity theorem, $1 + 1 = 2$ .

Exercise 2

Find a basis of the kernel and a basis of the image of

f (x, y, z) = (x + y + 3 z, 2 x - y, x + 2 y + 5 z)

and write the Cartesian equation of the image.

121 1 - 1 2 305 R_{2} \to R_{2} - 2 R_{1} R_{3} \to R_{3} - R_{1} 100 1 - 3 1 3 - 6 2 R_{3} \to 3 R_{3} + R_{2} 100 1 - 3 0 3 - 6 0

rk A = 2, dim ker f = 3 - 2 = 1

{x + y + 3 z = 0 - 3 y - 6 z = 0 ⟹ {y = - 2 z x = - y - 3 z = - z ⟹ ker f = ⟨(- 1, - 2, 1)⟩

Im f = ⟨(1, 2, 1), (1, - 1, 2)⟩

121 1 - 1 2 x y z = x (4 + 1) - y (2 - 1) + z (- 1 - 2) = 5 x - y - 3 z = 0

Row reduction leaves two nonzero rows, so $rk A = 2$ and by the rank-nullity theorem the kernel has dimension $3 - 2 = 1$ . The reduced system has the free variable $z$ : from the second equation $y = - 2 z$ , and substituting into the first $x = - y - 3 z = 2 z - 3 z = - z$ . Setting $z = 1$ we obtain the generator $(- 1, - 2, 1)$ of the kernel.

The pivots of the echelon form are in the first and second columns, therefore the corresponding columns of $A$ , that is $(1, 2, 1)$ and $(1, - 1, 2)$ , form a basis of the image. A vector $(x, y, z)$ belongs to the image if and only if it is linearly dependent on the two basis vectors, that is, if the determinant of the matrix that has the three vectors as columns is zero. Expanding along the third column we obtain the plane $5 x - y - 3 z = 0$ . The third column $(3, 0, 5)$ of $A$ satisfies the equation, since $15 - 0 - 15 = 0$ , in agreement with the fact that it also lies in the image.

Exercise 3

Let $f : R^{4} \to R^{3}$ be

f (x, y, z, w) = (x + y + 2 w, y + z + w, x + 2 y + z + 3 w) .

Find a basis of the kernel and a basis of the image; determine whether $f$ is surjective.

101112011213 R_{3} \to R_{3} - R_{1} 100111011211 R_{3} \to R_{3} - R_{2} 100110010210

rk A = 2, dim ker f = 4 - 2 = 2, dim Im f = 2 < 3

{y = - z - w x = - y - 2 w = z - w ⟹ (x, y, z, w) = z (1, - 1, 1, 0) + w (- 1, - 1, 0, 1)

ker f = ⟨(1, - 1, 1, 0), (- 1, - 1, 0, 1)⟩, Im f = ⟨(1, 0, 1), (1, 1, 2)⟩

The third row of the matrix is the sum of the first two, and indeed the reduction makes it vanish; the rank is $2$ . By the rank-nullity theorem the kernel has dimension $4 - 2 = 2$ , so the reduced system has two free variables, $z$ and $w$ . From the second equation $y = - z - w$ ; substituting into the first we obtain $x = - y - 2 w = z + w - 2 w = z - w$ . Writing the general solution as a combination with coefficients $z$ and $w$ we read off the two vectors of a basis of the kernel.

The pivots are in the first two columns, so $(1, 0, 1)$ and $(1, 1, 2)$ form a basis of the image. The image has dimension $2$ , less than the dimension $3$ of the codomain, therefore $f$ is not surjective; injectivity is not possible either, since a map from a space of dimension $4$ to one of dimension $3$ always has a nonzero kernel.

Exercise 4

Determine whether $f : R^{3} \to R^{2}$ , $f (x, y, z) = (x + y + z, x - y + z)$ , is surjective and find its kernel.

A = (11 1 - 1 11), 11 1 - 1 = - 2 \neq = 0 ⟹ rk A = 2

{x + y + z = 0 x - y + z = 0 ⟹ {y = 0 x = - z ⟹ ker f = ⟨(- 1, 0, 1)⟩

The matrix has a nonzero minor of order $2$ , so the rank is $2$ , equal to the dimension of the codomain; by the surjectivity criterion $f$ is surjective, that is $Im f = R^{2}$ . The rank-nullity theorem gives $dim ker f = 3 - 2 = 1$ . Subtracting the two equations of the homogeneous system side by side we obtain $2 y = 0$ , so $y = 0$ , and the first equation becomes $x = - z$ . Setting $z = 1$ we find the generator $(- 1, 0, 1)$ of the kernel, therefore $f$ is not injective.