Image
A linear transformation sends every input somewhere. Which outputs actually occur, what shape does that set have, and when is every vector of the target space reached?
Where you are 9 in Matrices as linear transformations
Why this exists
In a recording studio, Ekaterine runs a mixing desk with three input channels and three speakers: left, center and right. Channel j with its fader at level 1 sends to the speakers the levels in column j of
row 1 for the left speaker, row 2 for the center, row 3 for the right. Signals add, and a negative setting sends a channel with its polarity flipped, so a setting of the three faders is a vector \mathbf{x} in \mathbb{R}^3 and the speaker levels are B\mathbf{x}.
Irakli wants the same level in all three speakers. Before anyone turns a fader, the whole question can be asked at once: which speaker patterns can this desk produce at all? That set of outputs is the image. It is always a subspace, it can be read from the columns, and when it is all of \mathbb{R}^3 every request can be met.
Day 1 is the definition, the proof that the image of a transformation is the span of its matrix’s columns, and a basis for it. Day 2 is whether every output is reached, and images read geometrically.
Definition
Let T: \mathbb{R}^n \to \mathbb{R}^m be a Linear transformation. The image of T is the set of all its outputs:
So \mathbf{b} is in \operatorname{im} T exactly when T(\mathbf{x}) = \mathbf{b} has at least one solution. The codomain \mathbb{R}^m is where outputs may land; the image is where they do land. Many books say range, as for the functions of Domain and range.
For an m \times n matrix A, \operatorname{im} A is the image of \mathbf{x} \mapsto A\mathbf{x}, and the column space \operatorname{Col} A is the span of its columns, as in Rank.
T is onto \mathbb{R}^m when \operatorname{im} T = \mathbb{R}^m: every \mathbf{b} in \mathbb{R}^m is T(\mathbf{x}) for at least one \mathbf{x}.
Intuition
Read A\mathbf{x} by columns, as in Multiplying a matrix by a vector: A\mathbf{x} = x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n. Making an output means choosing weights for the columns, just as setting the faders chooses how much of each channel to send. The outputs are the combinations of the columns, and nothing else.
For A = \begin{bmatrix} 1 & -2 \\ 2 & -4 \end{bmatrix} the second column is -2 times the first, so A\mathbf{x} = (x_1 - 2x_2)(1, 2). Every output lies on the line y = 2x, and each point (t, 2t) of it is A(t, 0). A whole plane of inputs is squeezed onto that line, and a point such as (3, 1) is never reached.
Geometry often names the image before any arithmetic. A projection onto a line has that line as its image, a projection onto a plane has that plane, and a rotation or a shear of the plane leaves nothing out.
Results
The image is the column space. If T: \mathbb{R}^n \to \mathbb{R}^m is linear with standard matrix A = \begin{bmatrix} T(\mathbf{e}_1) & \cdots & T(\mathbf{e}_n) \end{bmatrix} (as in Finding the matrix of a transformation), then
The image is a subspace of \mathbb{R}^m (Subspaces).
Spanning sets go to spanning sets. If \mathbf{v}_1, \dots, \mathbf{v}_k span \mathbb{R}^n, then \operatorname{im} T = \operatorname{span}\{T(\mathbf{v}_1), \dots, T(\mathbf{v}_k)\}.
A basis from the pivot columns. The pivot columns of A itself form a basis of \operatorname{im} A, so \dim \operatorname{im} A = \operatorname{rank} A. The columns of the reduced form are usually not in \operatorname{im} A at all.
The onto test. For an m \times n matrix A these are equivalent: \mathbf{x} \mapsto A\mathbf{x} is onto \mathbb{R}^m; the columns span \mathbb{R}^m; an echelon form of A has a pivot in every row; \operatorname{rank} A = m. If n < m, the map is never onto.
Projections. If W is a subspace of \mathbb{R}^m and a linear T: \mathbb{R}^m \to \mathbb{R}^m has every output in W and T(\mathbf{w}) = \mathbf{w} for every \mathbf{w} in W, then \operatorname{im} T = W.
Why it is true
Image is column space. Writing \mathbf{x} = x_1\mathbf{e}_1 + \dots + x_n\mathbf{e}_n, linearity gives
a combination of the columns, so \operatorname{im} T \subseteq \operatorname{Col} A. Conversely, c_1 T(\mathbf{e}_1) + \dots + c_n T(\mathbf{e}_n) = T(c_1, \dots, c_n) is an output, so \operatorname{Col} A \subseteq \operatorname{im} T.
Subspace. T(\mathbf{0}) = \mathbf{0} is an output. If T(\mathbf{u}) and T(\mathbf{v}) are outputs, so are T(\mathbf{u}) + T(\mathbf{v}) = T(\mathbf{u} + \mathbf{v}) and c\,T(\mathbf{u}) = T(c\mathbf{u}).
Spanning sets. Each \mathbf{x} is some c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k, so T(\mathbf{x}) = c_1T(\mathbf{v}_1) + \dots + c_kT(\mathbf{v}_k); and every such combination is the output of c_1\mathbf{v}_1 + \dots + c_k\mathbf{v}_k.
Pivot columns. Weights \mathbf{x} give x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n = \mathbf{0} exactly when A\mathbf{x} = \mathbf{0}, and row operations do not change those solutions (Solving a system by row reduction: one, none or infinitely many solutions). So the columns of A obey exactly the linear relations that the columns of its Reduced row echelon form R obey. In R the pivot columns are distinct standard basis vectors, hence independent, and every other column is a combination of them. The same relations hold in A, so its pivot columns are independent and the other columns add nothing to their span: a basis of \operatorname{Col} A. But the columns themselves change: if R has a zero row, every column of R ends in 0, while the columns of A need not.
Onto. By the first result, onto means the columns span \mathbb{R}^m, which Span showed happens exactly when an echelon form has a pivot in every row. Each pivot has its own column, so there are at most n pivots; if n < m, some row has none.
Projections. Every output is in W, and every \mathbf{w} in W is the output T(\mathbf{w}).
Worked examples
Day 1: The patterns the desk can produce
For the desk B of Why this exists, find a basis for \operatorname{im} B and an equation for it. Decide whether the patterns (3, 2, 5) and (1, 1, 1) can be produced, with fader settings for any that can.
Solution\operatorname{im} B is the span of the three columns; row reduction trims them to a basis. Subtracting row 1 from row 3 gives (0, 1, 1), equal to row 2, so subtracting row 2 leaves a zero row; then subtract 2 times row 2 from row 1:
The pivots are in columns 1 and 2, so columns 1 and 2 of B are a basis: \{(1, 0, 1),\ (2, 1, 3)\}, and \dim \operatorname{im} B = 2. Column 3 of the reduced form records the relation \mathbf{c}_3 = -2\mathbf{c}_1 + \mathbf{c}_2, which holds in B: -2(1, 0, 1) + (2, 1, 3) = (0, 1, 1). Channel 3 adds nothing new.
A vector in the image is
whose first two entries add to the third. Conversely, if b_3 = b_1 + b_2, then t = b_2 and s = b_1 - 2b_2 give the first two entries, and the third is s + 3t = b_1 + b_2 = b_3. So \operatorname{im} B is the plane b_3 = b_1 + b_2: the right speaker always gets the sum of the left and center levels. The reduced columns (1, 0, 0) and (0, 1, 0) span the different plane b_3 = 0, and (1, 0, 0) is not even an output, since 0 \neq 1 + 0.
For (3, 2, 5): 5 = 3 + 2, so yes. With t = 2 and s = -1, the faders \mathbf{x} = (-1, 2, 0) work: -(1, 0, 1) + 2(2, 1, 3) = (3, 2, 5). For (1, 1, 1): 1 \neq 1 + 1, so it cannot be produced. Nor can any (c, c, c) with c \neq 0, since c \neq 2c: no setting gives Irakli equal levels except silence.
Day 1: A transformation into space
Let T: \mathbb{R}^2 \to \mathbb{R}^3, T(x, y) = (x + y,\ x - y,\ 2x + y). Show that \operatorname{im} T is a plane through the origin, find its equation, and explain why T is not onto.
SolutionThe standard matrix has columns T(\mathbf{e}_1) = (1, 1, 2) and T(\mathbf{e}_2) = (1, -1, 1), and \operatorname{im} T is their span. Neither is a multiple of the other, so they span a plane through the origin.
For its equation, ask which \mathbf{b} solve T(x, y) = \mathbf{b}. The first two entries give x + y = b_1 and x - y = b_2, so x = \tfrac12(b_1 + b_2) and y = \tfrac12(b_1 - b_2). The third entry must then be
So \mathbf{b} is an output exactly when 3b_1 + b_2 - 2b_3 = 0. Check with the columns: 3 + 1 - 4 = 0 and 3 - 1 - 2 = 0. T is not onto: (0, 0, 1) gives -2 \neq 0. No map from \mathbb{R}^2 to \mathbb{R}^3 is onto, since two columns give at most two pivots for three rows.
Day 1: Four columns, still a plane
Find a basis for the image of
and an equation for it. Is \mathbf{x} \mapsto D\mathbf{x} onto \mathbb{R}^3?
SolutionSubtract 2 times row 1 from row 2 and add row 1 to row 3. Both new rows are (0, 0, 1, -1), so subtracting leaves a zero row; then subtract row 2 from row 1:
The pivots are in columns 1 and 3, so \operatorname{im} D has basis \{(1, 2, -1),\ (1, 3, 0)\}. The reduced form also records \mathbf{d}_2 = 2\mathbf{d}_1 and \mathbf{d}_4 = 3\mathbf{d}_1 - \mathbf{d}_3; check: (3, 6, -3) - (1, 3, 0) = (2, 3, -3). Columns 1 and 2 would not do, being dependent.
A plane through the origin is the set of \mathbf{b} with \mathbf{n} \cdot \mathbf{b} = 0, for a normal \mathbf{n} perpendicular to both basis vectors: n_1 + 2n_2 - n_3 = 0 and n_1 + 3n_2 = 0. Taking n_2 = -1 gives n_1 = 3 and n_3 = 1, so \operatorname{im} D is the plane 3b_1 - b_2 + b_3 = 0. The other columns satisfy it: 6 - 4 - 2 = 0 and 6 - 3 - 3 = 0.
The map is not onto, even with four columns for three rows: row 3 has no pivot. The two free variables make the kernel two-dimensional, and 2 + 2 = 4, the number of columns; Rank and nullity shows this always happens.
Day 2: A headphone feed that reaches everything
Boitumelo’s headphone feed sends the three channels to two earpieces through
Show that \mathbf{x} \mapsto M\mathbf{x} is onto \mathbb{R}^2, and give fader settings that produce any \mathbf{b} = (b_1, b_2).
SolutionSubtracting 2 times row 1 from row 2 gives the echelon form \begin{bmatrix} 1 & 3 & 2 \\ 0 & -1 & -3 \end{bmatrix}, a pivot in each row, so by the onto test every \mathbf{b} is an output.
A formula shows it directly. Carry a general \mathbf{b} through the same step, scale row 2 by -1, then subtract 3 times it from row 1:
With the free fader x_3 = 0, the settings \mathbf{x} = (-5b_1 + 3b_2,\ 2b_1 - b_2,\ 0) give \mathbf{b}. Check: the first entry of M\mathbf{x} is (-5b_1 + 3b_2) + 3(2b_1 - b_2) = b_1, and the second is 2(-5b_1 + 3b_2) + 5(2b_1 - b_2) = b_2. For \mathbf{b} = (1, 1) this gives (-2, 1, 0).
Day 2: A projection and a reflection
Let \mathbf{u} = (1, 2, 2) and L = \operatorname{span}\{\mathbf{u}\}. The projection onto L and the reflection across L have matrices
Find the image of each, from the geometry and then from the columns.
SolutionGeometry for P. Projection onto L is P\mathbf{x} = \dfrac{\mathbf{u} \cdot \mathbf{x}}{\mathbf{u} \cdot \mathbf{u}}\,\mathbf{u} with \mathbf{u} \cdot \mathbf{u} = 9. Every output is a multiple of \mathbf{u}, so it lies in L, and P\mathbf{u} = \mathbf{u}, so P fixes every vector of L. By the projections result, \operatorname{im} P = L.
Columns for P. The columns are \tfrac19\mathbf{u}, \tfrac29\mathbf{u}, \tfrac29\mathbf{u}; only column 1 is a pivot column, so \operatorname{im} P = \operatorname{span}\{(1, 2, 2)\} and \operatorname{rank} P = 1.
Geometry for Q. Reflecting twice returns every vector to where it started: Q(Q\mathbf{b}) = \mathbf{b}. That writes each \mathbf{b} as an output, the output of Q\mathbf{b}, so \operatorname{im} Q = \mathbb{R}^3. In matrices, Q^2 = 4P^2 - 4P + I = I, because projecting twice is projecting once: P^2 = P.
Columns for Q. The factor \tfrac19 does not change the pivots, so reduce 9Q, with rows \mathbf{r}_1 = (-7, 4, 4), \mathbf{r}_2 = (4, -1, 8), \mathbf{r}_3 = (4, 8, -1). Put \mathbf{r}_2 first. Then 4\mathbf{r}_1 + 7\mathbf{r}_2 = (0, 9, 72) and \mathbf{r}_3 - \mathbf{r}_2 = (0, 9, -9), and the difference of these two is (0, 0, 81):
A pivot in every row, so the columns span \mathbb{R}^3, as the geometry said.
Day 2: Rewiring one channel
Ekaterine can rewire channel 3 of the desk so that at level 1 it sends (0, 1, k) to the speakers, leaving channels 1 and 2 alone. For which k can the desk produce every pattern? With k = 3, find fader settings for Irakli’s (1, 1, 1).
SolutionThe new matrix is B_k = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 3 & k \end{bmatrix}. Subtracting row 1 and then row 2 from row 3 leaves (0, 0, k - 1):
Row 3 has a pivot exactly when k \neq 1, so by the onto test the desk reaches every pattern for every k \neq 1. This fits Example 1: the new column lies in the old image b_3 = b_1 + b_2 only when k = 1, the original wiring.
For k = 3, solve x_1 + 2x_2 = 1, x_2 + x_3 = 1, x_1 + 3x_2 + 3x_3 = 1. The first gives x_1 = 1 - 2x_2, and then the third gives x_2 + 3x_3 = 0. Subtracting the second, 2x_3 = -1, so
Check:
Boundaries
- The image sits in the codomain. For an m \times n matrix, \operatorname{im} A is a subspace of \mathbb{R}^m and the kernel a subspace of \mathbb{R}^n; they can be compared only when m = n.
- Only for linear maps. The image of g(x) = (x, |x|) is the V-shaped graph y = |x|, which holds (1, 1) and (-1, 1) but not their sum (0, 2). The subspace proof needed T(\mathbf{u}) + T(\mathbf{v}) = T(\mathbf{u} + \mathbf{v}).
- Columns of A, not of its reduced form. Row operations keep the relations among the columns, which locates the pivot columns, but they change the column space.
- Shape decides onto only one way. Fewer columns than rows is never onto; more columns than rows may or may not be, as D in Example 3 shows.
- Onto is not one-to-one. The headphone matrix M reaches every \mathbf{b}, but a whole line of settings gives each one. For square matrices the two conditions meet, in When a matrix or transformation can be undone.
Where this goes next
The image pairs with the kernel, and Rank and nullity ties their sizes: \dim \operatorname{im} A + \dim \ker A = n. For a square matrix, an image equal to all of \mathbb{R}^n is one of the conditions gathered in the The invertible matrix theorem. When \mathbf{b} is not in the image, A\mathbf{x} = \mathbf{b} has no solution, and Least squares: the best approximate solution of Ax = b looks instead for the output nearest \mathbf{b}, the Orthogonal bases and projection onto a subspace of \mathbf{b} onto \operatorname{im} A.
Exercises
Day 1: finding images
- Describe the image of \begin{bmatrix} 2 & -1 \\ -6 & 3 \end{bmatrix} and decide whether (3, -9) and (3, 9) are in it.
- Let F = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}. Find a basis for \operatorname{im} F and an equation for it.
- Find the image of T: \mathbb{R}^3 \to \mathbb{R}^2, T(x, y, z) = (x + 2y - z,\ 2x + 4y - 2z). Is T onto?
- Find a basis for the image of \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} and one equation that describes it in \mathbb{R}^4.
- Find a basis for the image of E = \begin{bmatrix} 1 & -1 & 2 & 0 \\ 2 & -2 & 5 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} made of columns of E, and explain why columns 1 and 2 would not do.
- Show that T(x, y) = (x - y,\ 2x + y,\ 3x) has the same image as the desk B.
- For F of Exercise 2, decide whether (5, 5, 1) and (1, 0, 1) are outputs, giving an input for each one that is.
Day 1: reasoning
- Show that the image of f(t) = (t, t^2), from \mathbb{R} to \mathbb{R}^2, is not a subspace, and say which step of the subspace proof fails.
- Refute the claim “two matrices with the same reduced row echelon form have the same image”, and say what the two matrices do share.
- Prove that \operatorname{im} A contains every column of A, and that every subspace containing all the columns of A contains \operatorname{im} A.
Day 2: onto or not
- Is \mathbf{x} \mapsto A\mathbf{x} onto \mathbb{R}^2 for A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}?
- Is \mathbf{x} \mapsto A\mathbf{x} onto \mathbb{R}^2 for A = \begin{bmatrix} 1 & -3 & 2 \\ -2 & 6 & -4 \end{bmatrix}? If not, describe its image.
- Is \mathbf{x} \mapsto A\mathbf{x} onto \mathbb{R}^3 for A = \begin{bmatrix} 1 & 0 & 2 & 1 \\ 0 & 1 & -1 & 3 \\ 1 & 1 & 1 & 0 \end{bmatrix}?
- Explain why \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 3 & 1 \end{bmatrix} cannot be onto \mathbb{R}^3, and find an equation for its image.
- For which k is \begin{bmatrix} 1 & k \\ k & 4 \end{bmatrix} onto \mathbb{R}^2? Give the image for each other k.
- For which h is \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & h \\ 0 & 1 & 1 \end{bmatrix} onto \mathbb{R}^3? Where it is not, give a basis and an equation for the image.
Day 2: images in geometry and at the desk
- Find the image of each map and describe it geometrically: (a) (x, y, z) \mapsto (x, 0, z); (b) the shear (x, y) \mapsto (x + 3y,\ y); (c) (x, y) \mapsto \big(\tfrac12(x - y),\ \tfrac12(y - x)\big).
- Irakli mutes channel 2 of the desk B, leaving channels 1 and 3. Which patterns can still be produced?
- Boitumelo adds a fourth channel to the desk B that at level 1 sends (1, 1, 0) to the speakers. Explain why the desk can now produce every pattern, and give one fader setting for (0, 0, 1).
Day 2: reasoning
- Prove that if n < m, no linear T: \mathbb{R}^n \to \mathbb{R}^m is onto.
- A square matrix P has P^2 = P. Prove that \operatorname{im} P = \{\mathbf{b} : P\mathbf{b} = \mathbf{b}\}.
- Let T: \mathbb{R}^n \to \mathbb{R}^m and S: \mathbb{R}^m \to \mathbb{R}^p be linear. Prove that \operatorname{im}(S \circ T) \subseteq \operatorname{im} S, and give an example where they differ.
- Find a 2 \times 2 matrix whose image equals its kernel, and verify it.
Answers
- The line \operatorname{span}\{(1, -3)\}, that is b_2 = -3b_1: column 2 is -\tfrac12 times column 1. (3, -9) is in it, the output of (0, -3). (3, 9) is not: 9 \neq -9.
- Basis \{(1, 2, 1),\ (3, 1, -1)\}; equation 3b_1 - 4b_2 + 5b_3 = 0. The reduced form has rows (1, 0, 1), (0, 1, 1) and a zero row, so columns 1 and 2 are the pivot columns. The normal (3, -4, 5) gives 3 - 8 + 5 = 0 and 9 - 4 - 5 = 0.
- Every output is (x + 2y - z)(1, 2), so the image is \operatorname{span}\{(1, 2)\}, the line b_2 = 2b_1. Not onto: (1, 0) is not on it.
- All three columns, \{(1, 0, 1, 2),\ (0, 1, 1, 0),\ (2, 1, 0, 1)\}: the echelon form has a pivot in each column. The image is b_1 - b_2 + b_3 - b_4 = 0, since (1, -1, 1, -1) has dot product 0 with each column.
- Basis \{(1, 2, 0),\ (2, 5, 1)\}. The reduced form has rows (1, -1, 0, -2), (0, 0, 1, 1) and a zero row, so the pivot columns are 1 and 3. Column 2 is -1 times column 1, so columns 1 and 2 span only a line.
- The columns (1, 2, 3) and (-1, 1, 0) are independent, so the image is a plane, and both satisfy b_3 = b_1 + b_2, so it is that plane, \operatorname{im} B. Directly: (x - y) + (2x + y) = 3x.
- (5, 5, 1) is an output, of (2, 1, 0): 2(1, 2, 1) + (3, 1, -1) = (5, 5, 1). (1, 0, 1) is not: 3 - 0 + 5 = 8 \neq 0.
- The image is the parabola y = x^2. It holds (1, 1) but not 2(1, 1) = (2, 2). The step c\,f(u) = f(cu) fails: 2f(1) = (2, 2) but f(2) = (2, 4).
- \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \begin{bmatrix} 1 \\ 1 \end{bmatrix} both reduce to \begin{bmatrix} 1 \\ 0 \end{bmatrix}, but their images are the x-axis and the line y = x. They share the solutions of A\mathbf{x} = \mathbf{0}, hence the relations among their columns and the pivot positions.
- A\mathbf{e}_j = \mathbf{a}_j, so each column is an output. A subspace holding every column is closed under addition and scaling, so it holds every x_1\mathbf{a}_1 + \dots + x_n\mathbf{a}_n = A\mathbf{x}.
- Yes. Subtracting 2 times row 1 from row 2 gives \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}, a pivot in each row.
- No. Row 2 is -2 times row 1. Every column is a multiple of (1, -2), so the image is the line \operatorname{span}\{(1, -2)\}, b_2 = -2b_1.
- Yes. Subtracting row 1 and then row 2 from row 3 gives (0, 0, 0, -4), so there are pivots in columns 1, 2 and 4, one in every row.
- Two columns give at most two pivots for three rows. The image is 3b_1 - 5b_2 - b_3 = 0: the normal (3, -5, -1) gives 3 - 0 - 3 = 0 and 6 - 5 - 1 = 0 with the columns.
- Onto exactly when k \neq 2 and k \neq -2: subtracting k times row 1 from row 2 leaves (0,\ 4 - k^2). For k = 2 the image is \operatorname{span}\{(1, 2)\}; for k = -2 it is \operatorname{span}\{(1, -2)\}.
- Onto exactly when h \neq 3: the echelon form ends in the row (0, 0, 3 - h). For h = 3, column 3 is the sum of columns 1 and 2; basis \{(1, 2, 0),\ (0, 1, 1)\}, equation 2b_1 - b_2 + b_3 = 0.
- (a) The xz-plane, \operatorname{span}\{(1, 0, 0), (0, 0, 1)\}, onto which the map projects. (b) All of \mathbb{R}^2: \mathbf{b} is the output of (b_1 - 3b_2,\ b_2). (c) The line y = -x, \operatorname{span}\{(1, -1)\}: the map projects onto it.
- The same plane b_3 = b_1 + b_2. The columns (1, 0, 1) and (0, 1, 1) are independent and lie in it, so they span it. Nothing is lost, because (2, 1, 3) = 2(1, 0, 1) + (0, 1, 1).
- (1, 1, 0) is not in the plane b_3 = b_1 + b_2 (0 \neq 2), so the image strictly contains a plane, and the only such subspace of \mathbb{R}^3 is \mathbb{R}^3 (Basis and dimension). One setting: (-\tfrac12, \tfrac12, 0, -\tfrac12), since -\tfrac12(1, 0, 1) + \tfrac12(2, 1, 3) - \tfrac12(1, 1, 0) = (0, 0, 1).
- The standard matrix is m \times n; each pivot has its own column, so there are at most n < m pivots and some row has none. By the onto test, T is not onto.
- If \mathbf{b} = P\mathbf{x}, then P\mathbf{b} = P^2\mathbf{x} = P\mathbf{x} = \mathbf{b}. If P\mathbf{b} = \mathbf{b}, then \mathbf{b} is the output of \mathbf{b}.
- S(T(\mathbf{x})) is S of the vector T(\mathbf{x}), so it is in \operatorname{im} S. Example: T(x, y) = (x, 0) and S the identity on \mathbb{R}^2; \operatorname{im} S = \mathbb{R}^2 but \operatorname{im}(S \circ T) is the x-axis.
- N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, with N(x, y) = (y, 0). The image is the x-axis, and the kernel, where y = 0, is the x-axis too.