Linear Algebra

Kernel

Some inputs to a linear transformation come out as the zero vector. Which ones, what shape does that set have, and what does it say about whether the transformation can tell its inputs apart?

You need

Solving a system by row reduction: one, none or infinitely many solutionsPrecalculus Linear transformationLinear Algebra

Unlocks

Rank and nullityLinear Algebra When a matrix or transformation can be undoneLinear Algebra

Where you are 9 in Matrices as linear transformations

Linear algebra6

Matrices as linear transformations9

▴ 3 above Linear transformation Multiplying a matrix by a vector

Kernel

Rank and nullity When a matrix or transformation can be undone ▾ Show all 9 (1 more)

Why this exists

At a village market, goods change hands by barter. Wara, Asiri and Achik keep a tally of how many baskets of goods each of them holds, and three standing deals run between them and the farms of the valley. Doing deal j once changes the three tallies by the numbers in column j of

A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 1 \\ 1 & 2 & -1 \end{bmatrix},

row 1 for Wara, row 2 for Asiri, row 3 for Achik. A bundle of trades says how many times each deal is done, a negative count meaning the deal runs in reverse, so a bundle is a vector \mathbf{x} in \mathbb{R}^3 and its effect on the tallies is A\mathbf{x}.

Some bundles leave every tally unchanged: goods move, and nobody ends up with more or fewer baskets. They also answer a second question, when two different bundles have the same effect: exactly when their difference is a bundle of the first kind. The set of inputs a linear transformation sends to zero is its kernel, and it measures what the transformation throws away.

Day 1 is the definition, the proof that a kernel is a subspace, and how to find a basis for it. Day 2 is what the kernel says about the transformation: when it is one-to-one, how the kernel looks geometrically, and why a matrix with more columns than rows always has a nonzero kernel vector.

Definition

Let T: \mathbb{R}^n \to \mathbb{R}^m be a Linear transformation. The kernel of T is the set of all inputs that T sends to the zero vector:

\ker T = \{\, \mathbf{x} \in \mathbb{R}^n : T(\mathbf{x}) = \mathbf{0} \,\}.

For an m \times n matrix A, the kernel of A is the kernel of \mathbf{x} \mapsto A\mathbf{x}, which is the solution set of the homogeneous system A\mathbf{x} = \mathbf{0}:

\ker A = \{\, \mathbf{x} \in \mathbb{R}^n : A\mathbf{x} = \mathbf{0} \,\}.

Many books call \ker A the null space of A. The kernel lives in the domain \mathbb{R}^n, not in the codomain.

A transformation T is one-to-one when different inputs always give different outputs: T(\mathbf{u}) = T(\mathbf{v}) only when \mathbf{u} = \mathbf{v}.

Intuition

The kernel always contains \mathbf{0}, since T(\mathbf{0}) = \mathbf{0}. If T(\mathbf{k}) = \mathbf{0}, adding \mathbf{k} to any input leaves the output alone: T(\mathbf{x} + \mathbf{k}) = T(\mathbf{x}) + \mathbf{0}. So the kernel holds the directions in which T cannot see movement.

The map Q that drops each point of the plane straight onto the line y = x squashes the plane along (1, -1). Every vector on the line y = -x lands on the origin, so that line is \ker Q, and each line parallel to it is squashed to one point.

Figure 1The projection Q onto the line y = x flattens the plane along the direction (1, -1). Its kernel is the dashed line of vectors sent to \mathbf{0}. Every point on a line parallel to the kernel, such as x + y = 4, lands at the same image, here (2, 2).

A shear or a rotation collapses no direction: only \mathbf{0} goes to \mathbf{0}, and, by the one-to-one test below, no information is lost.

Results

The kernel is a subspace. For a linear T: \mathbb{R}^n \to \mathbb{R}^m, \ker T is a subspace of \mathbb{R}^n.

A basis from row reduction. Row reduce A to its Reduced row echelon form and write the solutions of A\mathbf{x} = \mathbf{0} in parametric vector form. The vectors that multiply the free variables form a basis of \ker A. So \dim \ker A is the number of free variables, and \ker A = \{\mathbf{0}\} exactly when every column of A has a pivot.

Equal outputs. For a linear T, T(\mathbf{u}) = T(\mathbf{v}) if and only if \mathbf{u} - \mathbf{v} is in \ker T. So if T(\mathbf{p}) = \mathbf{b}, the inputs with output \mathbf{b} are exactly the vectors \mathbf{p} + \mathbf{k} with \mathbf{k} in \ker T.

The one-to-one test. A linear T is one-to-one if and only if \ker T = \{\mathbf{0}\}. For a matrix: \mathbf{x} \mapsto A\mathbf{x} is one-to-one exactly when the columns of A are linearly independent.

Wide matrices. If A is m \times n with n > m, more columns than rows, then \ker A contains a nonzero vector, so \mathbf{x} \mapsto A\mathbf{x} is not one-to-one.

Why it is true

Subspace. First, T(\mathbf{0}) = \mathbf{0}, so \mathbf{0} is in \ker T. If \mathbf{u} and \mathbf{v} are in \ker T, additivity gives T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \mathbf{0} + \mathbf{0} = \mathbf{0}, so \mathbf{u} + \mathbf{v} is in \ker T. For a scalar c, homogeneity gives T(c\mathbf{u}) = c\,T(\mathbf{u}) = c\,\mathbf{0} = \mathbf{0}. The three subspace conditions hold.

Basis. Row operations do not change the solutions of A\mathbf{x} = \mathbf{0} (Solving a system by row reduction: one, none or infinitely many solutions), so \ker A can be read from the reduced form. As Homogeneous systems and solutions in parametric vector form shows, solving for the pivot variables writes every solution as t_1\mathbf{w}_1 + \dots + t_k\mathbf{w}_k, one vector per free variable, so these vectors span \ker A. They are independent: \mathbf{w}_i has a 1 in the slot of the i-th free variable, where every other \mathbf{w}_j has a 0, so in t_1\mathbf{w}_1 + \dots + t_k\mathbf{w}_k = \mathbf{0} that slot forces t_i = 0. With no free variables the only solution is \mathbf{0}.

Equal outputs. By linearity, T(\mathbf{u}) - T(\mathbf{v}) = T(\mathbf{u} - \mathbf{v}), so T(\mathbf{u}) = T(\mathbf{v}) exactly when T(\mathbf{u} - \mathbf{v}) = \mathbf{0}. Applied with \mathbf{v} = \mathbf{p}, the inputs with output \mathbf{b} are the \mathbf{u} with \mathbf{u} - \mathbf{p} in the kernel.

One-to-one. If T is one-to-one, then since T(\mathbf{0}) = \mathbf{0}, no other input can go to \mathbf{0}: the kernel is \{\mathbf{0}\}. If the kernel is \{\mathbf{0}\} and T(\mathbf{u}) = T(\mathbf{v}), then \mathbf{u} - \mathbf{v} is in the kernel, so \mathbf{u} - \mathbf{v} = \mathbf{0} and \mathbf{u} = \mathbf{v}. For a matrix with columns \mathbf{a}_1, \dots, \mathbf{a}_n, A\mathbf{x} = x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n (Multiplying a matrix by a vector), so “only \mathbf{x} = \mathbf{0} solves A\mathbf{x} = \mathbf{0}” is word for word the definition of independent columns.

Wide matrices. Each pivot sits in its own row, so A has at most m pivots. With n > m columns, at least one column has no pivot, which gives a free variable and a nonzero vector in the kernel.

Worked examples

Example 1

Day 1: Testing two bundles

At the market of Why this exists, decide whether the bundles \mathbf{u} = (-2, 2, 2) and \mathbf{w} = (1, 1, 1) leave every tally unchanged.

Solution

A bundle leaves the tallies unchanged exactly when it is in \ker A, so multiply. By rows,

\begin{aligned} A\mathbf{u} &= \begin{bmatrix} -2 - 2 + 4 \\ -4 + 2 + 2 \\ -2 + 4 - 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \\ A\mathbf{w} &= \begin{bmatrix} 2 \\ 4 \\ 2 \end{bmatrix}. \end{aligned}

So \mathbf{u} is in \ker A: reversing deal 1 twice and doing deals 2 and 3 twice each changes no tally. And \mathbf{w} is not: it leaves Wara 2 baskets up, Asiri 4 and Achik 2.

Example 2

Day 1: Every bundle that changes nothing

Find \ker A for the market matrix, and a basis for it.

Solution

The kernel is the solution set of A\mathbf{x} = \mathbf{0}. Row reduce A; the zero right side stays zero under every row operation, so it need not be written. Subtract 2 times row 1 from row 2 and row 1 from row 3, then subtract the new row 2 from row 3:

\begin{aligned} A &\to \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -3 \\ 0 & 3 & -3 \end{bmatrix} \to \begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix} \\ &\to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}, \end{aligned}

where row 2 was divided by 3 and then added to row 1. The pivots are in columns 1 and 2, so x_3 = t is free, and the rows say x_1 = -t and x_2 = t:

\mathbf{x} = t\begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix}.

So \ker A = \operatorname{span}\{(-1, 1, 1)\}, a line through the origin in \mathbb{R}^3, with basis \{(-1, 1, 1)\} and \dim \ker A = 1. The bundle \mathbf{u} of Example 1 is 2(-1, 1, 1). In the market, deal 1 has the effect of deals 2 and 3 together, and every unchanging bundle is a multiple of this one.

Example 3

Day 1: Three free variables

Find a basis for the kernel of

B = \begin{bmatrix} 1 & 2 & 0 & -1 & 3 \\ 2 & 4 & 1 & 0 & 4 \\ -1 & -2 & 1 & 3 & -5 \end{bmatrix}.

Solution

Subtract 2 times row 1 from row 2 and add row 1 to row 3. Both new rows are (0, 0, 1, 2, -2), so subtracting gives a zero row:

\begin{bmatrix} 1 & 2 & 0 & -1 & 3 \\ 0 & 0 & 1 & 2 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}.

This is already reduced. The pivots are in columns 1 and 3, so x_2 = r, x_4 = s, x_5 = t are free and

x_1 = -2r + s - 3t, \qquad x_3 = -2s + 2t.

Collecting the coefficients of each free variable,

\mathbf{x} = r\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} 1 \\ 0 \\ -2 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -3 \\ 0 \\ 2 \\ 0 \\ 1 \end{bmatrix}.

The three vectors span \ker B, and they are independent: in slots 2, 4 and 5 each has a 1 where the other two have 0. So they are a basis, and \ker B is a three-dimensional subspace of \mathbb{R}^5. Check the third: the three rows of B give -3 + 3 = 0, -6 + 2 + 4 = 0 and 3 + 2 - 5 = 0.

Example 4

Day 1: A transformation given by a formula

Find a basis for the kernel of T: \mathbb{R}^3 \to \mathbb{R}, T(x, y, z) = 2x - y + 3z, and describe it geometrically.

Solution

T(\mathbf{x}) = \mathbf{0} is the single equation 2x - y + 3z = 0, a one-row system with its pivot in column 1. So y and z are free, and x = \tfrac12 y - \tfrac32 z:

\mathbf{x} = y\begin{bmatrix} \tfrac12 \\ 1 \\ 0 \end{bmatrix} + z\begin{bmatrix} -\tfrac32 \\ 0 \\ 1 \end{bmatrix}.

Scaling each vector by 2 keeps the pair independent with the same span, and clears fractions: \ker T has basis \{(1, 2, 0),\ (-3, 0, 2)\}. Check: 2 - 2 + 0 = 0 and -6 - 0 + 6 = 0. Geometrically, \ker T is the plane 2x - y + 3z = 0 through the origin. Since (2, -1, 3) \cdot \mathbf{x} = T(\mathbf{x}), it is the plane of vectors perpendicular to (2, -1, 3).

Example 5

Day 2: The kernel of a projection

The matrix

P = \frac13\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{bmatrix}

projects \mathbb{R}^3 onto the plane x + y + z = 0, dropping each point straight onto it. Find \ker P, explain it geometrically, and find two different inputs with the same output.

Solution

The factor \tfrac13 does not change which vectors solve P\mathbf{x} = \mathbf{0}, so row reduce 3P. Swap rows 1 and 2 and scale the new row 1 by -1, then clear column 1:

\begin{aligned} &\begin{bmatrix} 1 & -2 & 1 \\ 2 & -1 & -1 \\ -1 & -1 & 2 \end{bmatrix} \to \begin{bmatrix} 1 & -2 & 1 \\ 0 & 3 & -3 \\ 0 & -3 & 3 \end{bmatrix} \\ &\to \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}. \end{aligned}

So x_3 = t is free, x_1 = x_2 = t, and \ker P = \operatorname{span}\{(1, 1, 1)\}.

Geometrically, (1, 1, 1) is perpendicular to the plane x + y + z = 0, since its dot product with (x, y, z) is x + y + z. Dropping a point straight onto the plane removes its part along that direction, so the vectors landing at the origin are those on the normal line. Space is flattened along the kernel, as Q in the figure flattens the plane along y = -x.

Two inputs whose difference is in the kernel have the same output. Take \mathbf{u} = (2, 0, 1) and \mathbf{v} = \mathbf{u} + (1, 1, 1) = (3, 1, 2):

\begin{aligned} P\mathbf{u} &= \tfrac13(4 - 1,\ -2 - 1,\ -2 + 2) \\ &= (1, -1, 0), \\ P\mathbf{v} &= \tfrac13(6 - 1 - 2,\ -3 + 2 - 2,\ -3 - 1 + 4) \\ &= (1, -1, 0). \end{aligned}

So P is not one-to-one, and every point on the line (2, 0, 1) + t(1, 1, 1) lands at (1, -1, 0).

Example 6

Day 2: A shear and a projection of the plane

Compare the kernels of the shear H(x, y) = (x,\ 2x + y) and the projection Q(x, y) = \big(\tfrac12(x + y),\ \tfrac12(x + y)\big) onto the line y = x. Which is one-to-one?

Solution

The shear. H(x, y) = \mathbf{0} means x = 0 and 2x + y = 0, so y = 0 too. The kernel is \{\mathbf{0}\}, and by the one-to-one test H is one-to-one. Geometrically the shear slides each vertical line x = c along itself by 2c; nothing collapses.

The projection. Q(x, y) = \mathbf{0} means x + y = 0, so \ker Q = \operatorname{span}\{(1, -1)\}, the line y = -x in the figure. Q is not one-to-one, and the Equal outputs result says exactly which inputs share an output: those differing by a multiple of (1, -1). The four dots (4, 0), (3, 1), (1, 3), (0, 4) in the figure differ by such multiples, and each goes to \big(\tfrac12 \cdot 4,\ \tfrac12 \cdot 4\big) = (2, 2).

Example 7

Day 2: Four deals and equal effects

Achik proposes a market with four standing deals, whose effects on the tallies of Wara, Asiri and Achik are the columns of

C = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 2 \\ 1 & 1 & 2 & 0 \end{bmatrix}.

Explain without computing why some nonzero bundle changes nothing, find \ker C, and show that the bundles \mathbf{u} = (2, 0, 1, 3) and \mathbf{v} = (3, 1, 0, 2) have the same effect.

Solution

C is 3 \times 4: four deals but only three tallies. By the Wide matrices result, \ker C contains a nonzero vector. To find it, subtract row 1 from row 3 and then row 2 from row 3, which gives (0, 0, 1, -1); then subtract 2 times row 3 from row 1 and add row 3 to row 2:

\begin{aligned} C &\to \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 1 & -1 \end{bmatrix} \\ &\to \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix}. \end{aligned}

So x_4 = t is free, x_1 = x_2 = -t, x_3 = t, and \ker C = \operatorname{span}\{(-1, -1, 1, 1)\}: deals 3 and 4 together have the same effect as deals 1 and 2 together.

For the two bundles, \mathbf{u} - \mathbf{v} = (-1, -1, 1, 1) is in \ker C, so C\mathbf{u} = C\mathbf{v} without multiplying. Check: both products are (1, 5, 4). The bundles with this effect are exactly \mathbf{v} + t(-1, -1, 1, 1).

Boundaries

The kernel sits in the domain. For an m \times n matrix, \ker A is a subspace of \mathbb{R}^n, while the outputs live in \mathbb{R}^m; the set of outputs is the image, a different subspace.
Only for linear maps. The zero set of a nonlinear function need not be a subspace: f(x, y) = xy is zero on both axes, but (1, 0) + (0, 1) = (1, 1) is not on either. And “only \mathbf{0} goes to \mathbf{0}” no longer means one-to-one: g(x) = x^2 sends only 0 to 0, yet g(1) = g(-1). The proof of the one-to-one test used T(\mathbf{u}) - T(\mathbf{v}) = T(\mathbf{u} - \mathbf{v}), which needs linearity.
Tall matrices can go either way. More rows than columns does not force a trivial kernel: \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix} sends (-2, 1) to \mathbf{0}. Only the wide case is decided by shape alone.
One-to-one is not onto. A trivial kernel does not say every vector of \mathbb{R}^m is an output; that is a question about the image. For square matrices the two meet in When a matrix or transformation can be undone.

Where this goes next

The kernel pairs with the image, the set of all outputs, and Rank and nullity ties their sizes, stated here without proof: the dimension of the kernel (the nullity) plus the rank is the number of columns. A trivial kernel is one of the conditions for a square matrix to be invertible, in When a matrix or transformation can be undone and the The invertible matrix theorem. Finding an eigenvector finds eigenvectors as the kernel of A - \lambda I, by the row reduction of this lesson. And the Equal outputs result is the shape of every solution set in Homogeneous systems and solutions in parametric vector form: one solution plus the kernel.

Exercises

Day 1: testing and finding kernels

Let M = \begin{bmatrix} 2 & 1 & -1 \\ 1 & 3 & 2 \end{bmatrix}. Decide whether (1, -1, 1) and (2, 1, 3) are in \ker M.
Find a basis for the kernel of \begin{bmatrix} 1 & -2 \\ -3 & 6 \end{bmatrix}.
Find a basis for the kernel of \begin{bmatrix} 1 & 1 & 3 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{bmatrix}.
Find the kernel of \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}.
Find a basis for the kernel of \begin{bmatrix} 1 & -1 & 0 & 2 \\ 0 & 0 & 1 & -3 \end{bmatrix}.
Find a basis for the kernel of T: \mathbb{R}^3 \to \mathbb{R}^2, T(x, y, z) = (x + y - z,\ 2x - y + 4z).
Find a basis for the kernel of T: \mathbb{R}^4 \to \mathbb{R}, T(\mathbf{x}) = x_1 - 2x_2 + x_3 + 3x_4, and give its dimension.
Find the kernel of the zero map from \mathbb{R}^2 to \mathbb{R}^3, \mathbf{x} \mapsto \mathbf{0}, and a basis for it.
Find a basis for the kernel of D = \begin{bmatrix} 1 & 2 & 1 & 0 \\ 2 & 4 & 3 & 1 \\ 3 & 6 & 4 & 1 \end{bmatrix}.

Day 1: reasoning

Let T be linear and \mathbf{b} \neq \mathbf{0}. Show that the set of \mathbf{x} with T(\mathbf{x}) = \mathbf{b} is not a subspace.
Show that the set of (x, y) with xy = 0 is not a subspace of \mathbb{R}^2, and explain why this does not contradict the theorem that a kernel is a subspace.
Give a 2 \times 3 matrix whose kernel is exactly \operatorname{span}\{(1, 2, -1)\}, and justify it.
Explain why A and its reduced row echelon form have the same kernel, though they usually have different column spans.

Day 2: one-to-one or not

Is \mathbf{x} \mapsto A\mathbf{x} one-to-one for A = \begin{bmatrix} 2 & -1 \\ 4 & 3 \end{bmatrix}?
Is \mathbf{x} \mapsto A\mathbf{x} one-to-one for A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ 0 & 3 \end{bmatrix}?
Without row reducing, explain why E = \begin{bmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 5 \end{bmatrix} is not one-to-one; then find a basis for \ker E.
Find the kernel of each map and describe it geometrically: (a) the projection (x, y) \mapsto (x, 0) of the plane onto the x-axis; (b) the rotation of the plane by 90^\circ, (x, y) \mapsto (-y, x); (c) the projection (x, y, z) \mapsto (0, 0, z) of \mathbb{R}^3 onto the z-axis.
Find the kernel of T(x, y, z) = (x - y,\ y - z,\ z - x), and give two different inputs with output (1, 0, -1).

Day 2: equal outputs

For the projection P of Example 5, find P(1, 0, 0) and describe every vector with that same image.
In Achik’s market of Example 7, find every bundle with the same effect as (1, 2, 0, 4). Which of them use only whole, nonnegative counts of each deal?
A linear T has T(\mathbf{p}) = \mathbf{b} and \ker T = \operatorname{span}\{\mathbf{k}\} with \mathbf{k} \neq \mathbf{0}. Describe all solutions of T(\mathbf{x}) = \mathbf{b} and of T(\mathbf{x}) = 2\mathbf{b}.

Day 2: reasoning

Let A be m \times n. Prove that \ker(A^T A) = \ker A. (Look at \mathbf{x}^T A^T A\mathbf{x}.)
Explain why a 4 \times 3 matrix can give a one-to-one map but a 3 \times 4 matrix cannot, and give a 4 \times 3 example that does.
Let T: \mathbb{R}^n \to \mathbb{R}^m and S: \mathbb{R}^m \to \mathbb{R}^p be linear. Prove that \ker T is contained in \ker(S \circ T), and give an example where the two are different.

Answers

(1, -1, 1) is in \ker M: M(1, -1, 1) = (2 - 1 - 1,\ 1 - 3 + 2) = (0, 0). (2, 1, 3) is not: M(2, 1, 3) = (2, 11).
Basis \{(2, 1)\}. Row 2 is -3 times row 1, so the system is x_1 - 2x_2 = 0, with x_2 = t free and x_1 = 2t.
Basis \{(-5, 2, 1)\}. The reduced form is \begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \end{bmatrix}, so x_3 = t, x_1 = -5t, x_2 = 2t.
\{\mathbf{0}\}, with the empty set as basis. The matrix reduces to I (its determinant is 6 - 5 = 1), so there are no free variables.
Basis \{(1, 1, 0, 0),\ (-2, 0, 3, 1)\}. The matrix is already reduced; x_2 = s and x_4 = t are free, x_1 = s - 2t and x_3 = 3t.
Basis \{(-1, 2, 1)\}. The matrix \begin{bmatrix} 1 & 1 & -1 \\ 2 & -1 & 4 \end{bmatrix} reduces to \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -2 \end{bmatrix}, so x_1 = -t, x_2 = 2t, x_3 = t.
Basis \{(2, 1, 0, 0),\ (-1, 0, 1, 0),\ (-3, 0, 0, 1)\}, dimension 3: x_1 = 2x_2 - x_3 - 3x_4 with three free variables.
All of \mathbb{R}^2, with basis \{\mathbf{e}_1, \mathbf{e}_2\}: every input goes to \mathbf{0}, and its matrix, the 3 \times 2 zero matrix, has no pivots.
Basis \{(-2, 1, 0, 0),\ (1, 0, -1, 1)\}. The reduced form has rows (1, 2, 0, -1) and (0, 0, 1, 1) and a zero row, so x_1 = -2s + t and x_3 = -t with x_2 = s, x_4 = t.
T(\mathbf{0}) = \mathbf{0} \neq \mathbf{b}, so \mathbf{0} is not in the set, and every subspace contains \mathbf{0}.
(1, 0) and (0, 1) are in the set but their sum (1, 1) is not, since 1 \cdot 1 \neq 0. No contradiction: (x, y) \mapsto xy is not linear (f(2, 2) = 4 \neq 2f(1, 1)).
\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} works. It is reduced with x_3 = t free, x_1 = -t, x_2 = -2t, so the kernel is \operatorname{span}\{(-1, -2, 1)\}, the same line as \operatorname{span}\{(1, 2, -1)\}. (Many other matrices work.)
Each row operation can be undone, so A\mathbf{x} = \mathbf{0} and R\mathbf{x} = \mathbf{0} have the same solutions (Augmented matrices and row operations). But row operations change the columns: \begin{bmatrix} 1 \\ 1 \end{bmatrix} reduces to \begin{bmatrix} 1 \\ 0 \end{bmatrix}.
Yes. Subtracting 2 times row 1 from row 2 gives \begin{bmatrix} 2 & -1 \\ 0 & 5 \end{bmatrix}, a pivot in each column, so \ker A = \{\mathbf{0}\}.
Yes. A\mathbf{x} = \mathbf{0} says x_1 = 0 (row 1) and then 3x_2 = 0 (row 3), so \ker A = \{\mathbf{0}\}. A tall matrix that is one-to-one.
E is 2 \times 4, so by the Wide matrices result it has a nonzero kernel vector. Basis \{(-2, 1, 0, 0),\ (-3, 0, 1, 1)\}: subtracting 2 times row 1 from row 2 gives (0, 0, 1, -1), so x_1 = -2s - 3t, x_3 = t, with x_2 = s, x_4 = t.
(a) The y-axis, \operatorname{span}\{(0, 1)\}, the direction dropped along. (b) \{\mathbf{0}\}: (-y, x) = \mathbf{0} forces x = y = 0. (c) The xy-plane, \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}.
\ker T = \operatorname{span}\{(1, 1, 1)\}: x - y = 0 and y - z = 0 force x = y = z, and the third entry is then 0 as well. Two inputs: (1, 0, 0) and (2, 1, 1) = (1, 0, 0) + (1, 1, 1), both with output (1, 0, -1).
P(1, 0, 0) = (\tfrac23, -\tfrac13, -\tfrac13), the first column of P. The vectors with this image are (1, 0, 0) + t(1, 1, 1), the line through (1, 0, 0) perpendicular to the plane.
Every bundle (1, 2, 0, 4) + t(-1, -1, 1, 1), which is (1 - t,\ 2 - t,\ t,\ 4 + t), all with effect (-3, 10, 3). Whole nonnegative counts need t an integer with 0 \le t \le 1: the bundles (1, 2, 0, 4) and (0, 1, 1, 5).
T(\mathbf{x}) = \mathbf{b}: the line \mathbf{p} + t\mathbf{k}. T(\mathbf{x}) = 2\mathbf{b}: since T(2\mathbf{p}) = 2\mathbf{b}, the parallel line 2\mathbf{p} + t\mathbf{k}.
If A\mathbf{x} = \mathbf{0}, then A^T A\mathbf{x} = A^T\mathbf{0} = \mathbf{0}, so \ker A is contained in \ker(A^T A). Conversely, if A^T A\mathbf{x} = \mathbf{0}, then 0 = \mathbf{x}^T A^T A\mathbf{x} = (A\mathbf{x})^T(A\mathbf{x}) = \lVert A\mathbf{x} \rVert^2, using (A\mathbf{x})^T = \mathbf{x}^T A^T. A vector of length 0 is \mathbf{0}, so A\mathbf{x} = \mathbf{0}.
A 3 \times 4 matrix is wide, so it has at most 3 pivots for 4 columns, a free variable, and a nonzero kernel. A 4 \times 3 matrix can have 3 pivots, one per column: \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} sends \mathbf{x} to (x_1, x_2, x_3, 0), which is \mathbf{0} only for \mathbf{x} = \mathbf{0}.
If T(\mathbf{x}) = \mathbf{0}, then S(T(\mathbf{x})) = S(\mathbf{0}) = \mathbf{0}, since every linear map sends \mathbf{0} to \mathbf{0}. Example: T(x, y) = (x, 0) and S(x, y) = (0, y). Then \ker T is the y-axis, but S(T(x, y)) = (0, 0) for every input, so \ker(S \circ T) is all of \mathbb{R}^2.