Rank and nullity
A linear transformation crushes some input directions to zero and carries the rest into its outputs. Is there a fixed budget shared between what is lost and what survives?
Where you are 9 in Matrices as linear transformations
Why this exists
A city bike-share system has three stations, North, Market and River, and a depot. Each night a rebalancing truck can run six standard moves, each dropping bikes from the depot at some stations. Siku, who runs the night shift, keeps a count sheet with four lines: the change at North, at Market, at River, and the total change. Running move j once changes the four lines by column j of
A plan says how many times each move is run, a negative count meaning the move is run in reverse, collecting those bikes. So a plan is a vector \mathbf{x} in \mathbb{R}^6 and its effect is A\mathbf{x}.
Which effects can a plan produce? That is the image of A, in \mathbb{R}^4. Which plans change nothing? That is the kernel, in \mathbb{R}^6. Different spaces, yet their dimensions are tied: they always add up to the number of columns.
Day 1 is the theorem and its proof; Day 2, its consequences and the row space.
Definition
Let A be an m \times n matrix. Its rank is the dimension of its image, which is its column space, and by Rank it is also the number of pivot columns. Its nullity is the dimension of its kernel:
For a Linear transformation T: \mathbb{R}^n \to \mathbb{R}^m, \operatorname{rank} T = \dim \operatorname{im} T and \operatorname{nullity} T = \dim \ker T, the rank and nullity of its standard matrix. The row space \operatorname{Row} A is the span of the rows, a subspace of \mathbb{R}^n. Dimension is as in Basis and dimension.
Intuition
The map M(x, y) = (x + 2y,\ x + 2y) sends (2, -1) to \mathbf{0}, so the line through (2, -1) is its kernel. A vector off that line, such as (0, 1), survives: M(0, 1) = (2, 2), whose multiples fill the image line y = x. The plane had two dimensions to spend; one went to the kernel and one to the image.
In general, extend a basis of the kernel to a basis of \mathbb{R}^n: the kernel vectors go to \mathbf{0} and the added ones to a basis of the image, so the n dimensions split into those crushed and those carried through.
Results
The rank-nullity theorem. For every m \times n matrix A,
the number of columns. For a linear T: \mathbb{R}^n \to \mathbb{R}^m, \dim \operatorname{im} T + \dim \ker T = n.
One-to-one and onto from the rank. \mathbf{x} \mapsto A\mathbf{x} is one-to-one exactly when \operatorname{rank} A = n, and onto \mathbb{R}^m exactly when \operatorname{rank} A = m.
What the shape decides. Always \operatorname{rank} A \le \min(m, n).
- If n > m (wide), \operatorname{nullity} A \ge n - m \ge 1: never one-to-one.
- If n < m (tall), \operatorname{rank} A \le n < m: never onto \mathbb{R}^m.
- If n = m (square), one-to-one exactly when onto, exactly when \operatorname{rank} A = n.
The row space. Every vector of \ker A is perpendicular to every row of A, and
Four dimensions from three numbers. If A is m \times n of rank r, then \operatorname{Col} A and \operatorname{Row} A have dimension r, \ker A has dimension n - r, and \ker A^{T} has dimension m - r.
Why it is true
Proof by counting pivots. The pivot columns of A are a basis of \operatorname{Col} A (Image), so the rank is the number of pivot columns. The vectors multiplying the free variables in the solution of A\mathbf{x} = \mathbf{0} are a basis of \ker A (Kernel), one per non-pivot column. Every column is one or the other, so the counts add to n.
Proof by extending a basis. This proof never row reduces, so it works for any linear T: \mathbb{R}^n \to \mathbb{R}^m. Let \mathbf{k}_1, \dots, \mathbf{k}_p be a basis of \ker T (p = 0 if the kernel is \{\mathbf{0}\}). An independent list extends to a basis (Basis and dimension), so add \mathbf{v}_1, \dots, \mathbf{v}_q to get a basis of \mathbb{R}^n, with p + q = n. We show T(\mathbf{v}_1), \dots, T(\mathbf{v}_q) is a basis of \operatorname{im} T, so \dim \operatorname{im} T = q = n - p.
They span. An output is T(\mathbf{x}) with \mathbf{x} = a_1\mathbf{k}_1 + \dots + a_p\mathbf{k}_p + b_1\mathbf{v}_1 + \dots + b_q\mathbf{v}_q. By linearity, and since each T(\mathbf{k}_i) = \mathbf{0},
They are independent. If c_1T(\mathbf{v}_1) + \dots + c_qT(\mathbf{v}_q) = \mathbf{0}, linearity gives T(c_1\mathbf{v}_1 + \dots + c_q\mathbf{v}_q) = \mathbf{0}, so that combination lies in \ker T and equals d_1\mathbf{k}_1 + \dots + d_p\mathbf{k}_p. Moving everything to one side gives a combination of the basis \mathbf{k}_1, \dots, \mathbf{v}_q equal to \mathbf{0}, so every coefficient, each c_i included, is 0.
One-to-one and onto. One-to-one means \ker A = \{\mathbf{0}\} (Kernel): nullity 0, rank n. Onto means \operatorname{Col} A = \mathbb{R}^m, and a subspace of \mathbb{R}^m is all of it exactly when its dimension is m (Basis and dimension).
Shape. \operatorname{Col} A sits in \mathbb{R}^m, so \operatorname{rank} A \le m, and \operatorname{rank} A = n - \operatorname{nullity} A \le n. If n > m, \operatorname{nullity} A = n - \operatorname{rank} A \ge n - m. If n = m, “rank n” and “rank m” are one condition.
Row space. Entry i of A\mathbf{x} is (row i) \cdot\, \mathbf{x}, so a kernel vector is perpendicular to every row, hence to every combination of rows. A vector \mathbf{w} in both \operatorname{Row} A and \ker A has \mathbf{w} \cdot \mathbf{w} = 0, so \mathbf{w} = \mathbf{0}. Put a basis of \operatorname{Row} A and a basis of \ker A together. If a combination of all of them is \mathbf{0}, its row-space part equals minus its kernel part, a vector in both, so both parts are \mathbf{0} and every coefficient is 0. That independent list in \mathbb{R}^n gives
so \dim \operatorname{Row} A \le \dim \operatorname{Col} A for every matrix. The rows of A^{T} are the columns of A (The transpose and symmetric matrices), so the same inequality for A^{T} reads \dim \operatorname{Col} A \le \dim \operatorname{Row} A. They are equal, and the first line is an equality: a second proof of the row-rank result of Rank.
Four dimensions. A^{T} is n \times m of rank r, so its nullity is m - r.
Worked examples
Day 1: Siku’s count sheet
Find the rank and nullity of the count-sheet matrix A of Why this exists, confirm that they add to 6, and say what they mean for the truck.
SolutionRow 4 is the sum of rows 1, 2, 3 (the total line is the sum of the station lines), so subtracting them leaves a zero row, and the rest is already reduced:
The pivots are in columns 1, 3, 6, so \operatorname{rank} A = 3; the image has basis (1, 0, 0, 1), (0, 1, 0, 1), (0, 0, 1, 1), one bike at a single station. The free variables x_2, x_4, x_5 give the kernel basis
so \operatorname{nullity} A = 3, and 3 + 3 = 6. Each kernel vector names a redundant move: move 2 does what two runs of move 1 do, and moves 4 and 5 are likewise combinations of moves 1 and 3.
Day 1: Watching the proof work
Let B = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 0 & 1 & 1 \end{bmatrix}. Find a basis of \ker B, extend it to a basis of \mathbb{R}^3, and check that the images of the added vectors are a basis of \operatorname{im} B.
SolutionSubtracting 2 times row 1 from row 2 gives (0, -3, -3); scaled to (0, 1, 1) it equals row 3, which clears to zero. Subtracting it from row 1:
With x_3 = t free, x_1 = x_2 = -t, so \ker B has basis \mathbf{k} = (-1, -1, 1), nullity 1.
Add \mathbf{e}_1 and \mathbf{e}_2. In a\mathbf{k} + b\mathbf{e}_1 + c\mathbf{e}_2 = (-a + b,\ -a + c,\ a) = \mathbf{0} the third entry forces a = 0, then b = c = 0: three independent vectors in \mathbb{R}^3, a basis.
The images are B\mathbf{k} = \mathbf{0}, B\mathbf{e}_1 = (1, 2, 0), B\mathbf{e}_2 = (1, -1, 1). The last two are independent (not multiples) and span the image: the third column (2, 1, 1) is their sum. So \operatorname{rank} B = 2 and 2 + 1 = 3.
Day 1: A nullity squeezed from both sides
On the east side, Yaku runs five truck moves among four stations, with effects the columns of
Without row reducing, find the rank and nullity of D, and decide whether every pattern of changes can be produced.
SolutionColumn 4 is column 1 plus column 2, and column 5 is column 1 plus column 3, so (1, 1, 0, -1, 0) and (1, 0, 1, 0, -1) are in \ker D. They are independent (slots 4 and 5), so \operatorname{nullity} D \ge 2 and, by the theorem, \operatorname{rank} D \le 3.
Columns 1, 2, 3 are independent: in a(1, 1, 0, 0) + b(0, 1, 1, 0) + c(0, 0, 1, 1) = \mathbf{0} the first entry gives a = 0, the last c = 0, then b = 0. So \operatorname{rank} D \ge 3.
The bounds meet: \operatorname{rank} D = 3, \operatorname{nullity} D = 2. The image is 3-dimensional in \mathbb{R}^4, so some patterns are unreachable. Columns 1, 2, 3 each satisfy y_1 - y_2 + y_3 - y_4 = 0, so every output does, and (1, 0, 0, 0) is never produced.
Day 2: What the shape alone decides
Huaman plans a system with 5 stations and 7 truck moves, and Yaku one with 7 stations and 5 moves. Knowing no entries, what can be said about the two effect matrices? And about a 6 \times 6 matrix?
SolutionHuaman’s matrix is 5 \times 7: rank at most 5, so nullity at least 7 - 5 = 2, and the map is never one-to-one. It is onto exactly when the rank is 5. The possible pairs are (r, 7 - r) for r = 0, \dots, 5.
Yaku’s matrix is 7 \times 5: rank at most 5 < 7, so the image is at most 5-dimensional in \mathbb{R}^7 and some patterns are unreachable, whatever the moves. It is one-to-one exactly when the rank is 5.
A 6 \times 6 matrix with nullity 0 has rank 6: one-to-one and onto. With positive nullity it is neither. The The invertible matrix theorem collects this square case.
Day 2: Transfers between stations
Downtown, the truck moves one bike along a route, from one station to another. Its routes among four stations are 1 \to 2, 2 \to 3, 3 \to 1, 3 \to 4, 4 \to 1, so
Find the rank and nullity, say exactly which patterns are reachable, and find a plan for (-3, 1, 0, 2).
SolutionEach column has one -1 and one 1, so (1, 1, 1, 1) is perpendicular to every column: every output satisfies b_1 + b_2 + b_3 + b_4 = 0. That set is a 3-dimensional subspace H of \mathbb{R}^4, so \operatorname{rank} N \le 3.
Rows 2, 3, 4 have leading entries in columns 1, 2, 4; in a combination equal to \mathbf{0}, entries 1, 2, 4 force the three coefficients to 0 in turn. So \dim \operatorname{Row} N \ge 3, and since row rank equals column rank, \operatorname{rank} N = 3 and \operatorname{nullity} N = 2. A kernel basis is the two loops (1, 1, 1, 0, 0), around 1 \to 2 \to 3 \to 1, and (1, 1, 0, 1, 1), around 1 \to 2 \to 3 \to 4 \to 1.
The image is 3-dimensional inside the 3-dimensional H, so it is H (Basis and dimension): a pattern is reachable exactly when its changes sum to 0. The target does. With x_3 = x_5 = 0 the rows read -x_1 = -3, x_1 - x_2 = 1, x_2 - x_4 = 0, x_4 = 2, so x_1 = 3, x_2 = 2, x_4 = 2. Every other plan adds a combination of the loops.
Day 2: Four subspaces of the count sheet
For Siku’s matrix A, give the dimensions of \operatorname{Col} A, \ker A, \operatorname{Row} A and \ker A^{T}, check a kernel vector against the rows, and find \ker A^{T}.
SolutionA is 4 \times 6 of rank 3 (Example 1), so the dimensions are 3, 6 - 3 = 3, 3 and 4 - 3 = 1. Rows 1, 2, 3 are a basis of the row space, by the reduced form of Example 1.
For \mathbf{k} = (-1, 0, -2, 1, 0, 0), the four rows give -1 + 1 = 0, -2 + 2 = 0, 0, and -1 - 2 + 3 = 0.
A^{T}\mathbf{y} = \mathbf{0} says \mathbf{y} is perpendicular to every column of A. Each column’s last entry is the sum of the first three, so (1, 1, 1, -1) works, and as the dimension is 1, \ker A^{T} = \operatorname{span}\{(1, 1, 1, -1)\}. So the image is the set of \mathbf{b} with b_1 + b_2 + b_3 - b_4 = 0.
Day 2: The rank of a product
Prove that \operatorname{rank}(AB) \le \operatorname{rank} A and \operatorname{rank}(AB) \le \operatorname{rank} B whenever AB is defined, and find the ranks for B of Example 2 and
Let A be m \times n and B be n \times p. Every output A(B\mathbf{x}) is an output of A, so \operatorname{Col}(AB) \subseteq \operatorname{Col} A and \operatorname{rank}(AB) \le \operatorname{rank} A. And B\mathbf{x} = \mathbf{0} gives AB\mathbf{x} = \mathbf{0}, so \ker B \subseteq \ker(AB). Both have p columns, so by the theorem
Now \operatorname{rank} B = 2. In C, row 3 is row 1 plus row 2, which are independent, so \operatorname{rank} C = 2; every row is perpendicular to (1, 2, 0), so \ker C = \operatorname{span}\{(1, 2, 0)\}, and (1, 2, 0) is the first column of B. Multiplying,
of rank 1, below both bounds. The outputs of CB are C applied to the plane \operatorname{Col} B = \operatorname{span}\{(1, 2, 0),\ (1, -1, 1)\}; C crushes (1, 2, 0) and sends (1, -1, 1) to (3, 1, 4), so one of the plane’s two dimensions survives.
Boundaries
- The count is of columns. The n in the theorem is the number of columns, the dimension of the domain. A 4 \times 6 matrix of rank 3 has nullity 3, not 1. The row count enters through \ker A^{T}, of dimension m - r.
- Dimensions, not subspaces. The theorem fixes how big the kernel and image are, not which subspaces they are: \begin{bmatrix} 1 & 0 \end{bmatrix} and \begin{bmatrix} 0 & 1 \end{bmatrix} both have rank 1 and nullity 1, with kernels the y-axis and the x-axis.
- Shape decides only one side. The 7 \times 5 zero matrix is tall yet not one-to-one.
- Linear maps only. f(x, y) = (x^3 - x,\ 0) sends the three lines x = -1, x = 0, x = 1 to \mathbf{0}: not a subspace, no dimension to count.
Where this goes next
For a square matrix, the The invertible matrix theorem and When a matrix or transformation can be undone build on rank n. The perpendicular pair \operatorname{Row} A and \ker A returns in Orthogonal bases and projection onto a subspace as orthogonal complements that split \mathbb{R}^n. In Finding an eigenvector the eigenvectors for \lambda, with \mathbf{0}, fill \ker(A - \lambda I), of dimension n - \operatorname{rank}(A - \lambda I). And \operatorname{rank}(A^{T}A) = \operatorname{rank} A (Exercise 21) is why the normal equations of Least squares: the best approximate solution of Ax = b have one solution when the columns of A are independent.
Exercises
Day 1: rank and nullity of a matrix
- Find the rank and nullity of \begin{bmatrix} 2 & 4 & -2 \\ 1 & 2 & -1 \end{bmatrix}, with a basis of the kernel.
- Find the rank and nullity of \begin{bmatrix} 1 & 0 & 2 & 1 \\ 0 & 1 & -1 & 3 \\ 1 & 1 & 1 & 4 \end{bmatrix}, with a basis of the kernel.
- Find the rank and nullity of \begin{bmatrix} 1 & 2 \\ 3 & 1 \\ 2 & -1 \end{bmatrix}. Is its map one-to-one? Onto \mathbb{R}^3?
- Find the rank and nullity of T: \mathbb{R}^4 \to \mathbb{R}^2, T(\mathbf{x}) = (x_1 + x_2 - x_3,\ x_2 + 2x_4). Is it onto?
- Find the rank, the nullity and a kernel basis of \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 1 \end{bmatrix}.
- Huaman’s three stations get four depot moves with effects the columns of C = \begin{bmatrix} 1 & 0 & 2 & 1 \\ 1 & 1 & 3 & 0 \\ 0 & 2 & 2 & -2 \end{bmatrix} (a negative entry takes bikes away). Find the rank, the nullity and a kernel basis, and say whether every pattern of changes can be produced.
Day 1: counting without solving
- A 4 \times 9 matrix has rank 4. What is its nullity? Is its map onto \mathbb{R}^4? One-to-one?
- A linear T: \mathbb{R}^6 \to \mathbb{R}^4 has a 2-dimensional kernel. What is \dim \operatorname{im} T, and is T onto?
- Which nullities can a 3 \times 4 matrix have? Give a matrix with the smallest one, and describe every matrix with the largest.
Day 1: reasoning
- A linear T: \mathbb{R}^3 \to \mathbb{R}^3 sends every vector of a plane through the origin to \mathbf{0}. What can its image be?
- Prove that if a linear T: \mathbb{R}^n \to \mathbb{R}^m is onto, then n \ge m.
Day 2: what the shape decides
- What can be said, with no entries known, about the map of (a) a 3 \times 8 matrix; (b) an 8 \times 3 matrix; (c) a 6 \times 6 matrix of rank 6; (d) a 4 \times 4 matrix with a nonzero kernel vector?
- Which of the pairs (\operatorname{rank}, \operatorname{nullity}) = (4, 2), (3, 3), (2, 3), (5, 1) can belong to a 4 \times 6 matrix?
- The map of a 6 \times 4 matrix is one-to-one. Find its rank and the dimension of its image. Is it onto \mathbb{R}^6?
- A 5 \times 5 matrix A has A\mathbf{x} = \mathbf{0} only for \mathbf{x} = \mathbf{0}. Explain why A\mathbf{x} = \mathbf{b} has exactly one solution for every \mathbf{b} in \mathbb{R}^5.
Day 2: stations and subspaces
- Yaku’s three stations have transfer routes 1 \to 2, 2 \to 3 and 1 \to 3, each moving one bike. Write the matrix, find its rank and nullity, and decide whether the patterns (2, -5, 3) and (1, 1, 0) are reachable, giving a plan where one exists.
- For the matrix of Exercise 2, give a basis of its row space, check that your kernel vectors are perpendicular to its rows, and confirm \dim \operatorname{Row} + \dim \ker = 4.
- A 5 \times 8 matrix has rank 3. Give the dimensions of its column space, kernel, row space and \ker A^{T}, and say which space each sits in.
- For C of Exercise 6, find \ker C^{T} and use it to describe \operatorname{im} C by one equation.
- For each k, find the rank and nullity of \begin{bmatrix} 1 & 2 & 1 \\ 2 & k & 2 \\ 1 & 2 & k - 1 \end{bmatrix}, with a kernel basis where the nullity is positive.
Day 2: reasoning
- Using \ker(A^{T}A) = \ker A (Exercise 22 of Kernel), prove that \operatorname{rank}(A^{T}A) = \operatorname{rank} A.
- Prove that a 3 \times 3 matrix A with A^2 = 0 has rank at most 1, and give such a matrix of rank exactly 1.
- Refute with a counterexample: “two matrices of the same size and the same rank have the same kernel.”
Answers
- Rank 1, nullity 2, kernel basis \{(-2, 1, 0),\ (1, 0, 1)\}. Row 1 is twice row 2, so the system is x_1 + 2x_2 - x_3 = 0; and 1 + 2 = 3.
- Rank 2, nullity 2, kernel basis \{(-2, 1, 1, 0),\ (-1, -3, 0, 1)\}. Row 3 is row 1 plus row 2, and the first two rows give x_1 = -2x_3 - x_4, x_2 = x_3 - 3x_4.
- Rank 2, nullity 0: the columns are not multiples of each other. One-to-one, but not onto \mathbb{R}^3, since 2 < 3.
- Rank 2, nullity 2. The matrix \begin{bmatrix} 1 & 1 & -1 & 0 \\ 0 & 1 & 0 & 2 \end{bmatrix} has a pivot in each row, so T is onto \mathbb{R}^2.
- Rank 2, nullity 1, kernel basis \{(1, -1, 1)\}. Row 3 is row 1 plus row 2; the rows give x_1 = -x_2 and x_3 = -x_2.
- Rank 2, nullity 2, kernel basis \{(-2, -1, 1, 0),\ (-1, 1, 0, 1)\}. Row 3 is 2(\text{row } 2 - \text{row } 1) and row 2 - row 1 = (0, 1, 1, -1), so x_1 = -2x_3 - x_4, x_2 = -x_3 + x_4. Not every pattern: the image is a plane in \mathbb{R}^3.
- Nullity 9 - 4 = 5. Onto \mathbb{R}^4, since the rank equals the number of rows. Not one-to-one.
- \dim \operatorname{im} T = 6 - 2 = 4, a 4-dimensional subspace of \mathbb{R}^4, so all of it: T is onto.
- The rank is at most 3, so the nullity is 1, 2, 3 or 4. Nullity 1: \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}. Nullity 4 means rank 0: only the zero matrix.
- The nullity is 2 or 3, so the rank is 1 or 0: the image is a line through the origin (kernel exactly the plane) or \{\mathbf{0}\} (the zero map).
- Onto means \operatorname{rank} T = m. By the theorem \operatorname{rank} T = n - \operatorname{nullity} T \le n, so m \le n.
- (a) Nullity at least 5: not one-to-one; onto when the rank is 3. (b) Rank at most 3 < 8: not onto; one-to-one when the rank is 3. (c) Nullity 0: one-to-one and onto. (d) Rank at most 3: neither.
- Only (4, 2) and (3, 3). The sum must be 6, ruling out (2, 3), and the rank is at most 4, ruling out (5, 1).
- Nullity 0, so rank 4 and a 4-dimensional image in \mathbb{R}^6. Not onto.
- Nullity 0 gives rank 5, so \operatorname{Col} A = \mathbb{R}^5 and a solution exists for every \mathbf{b}; it is unique because the map is one-to-one.
- \begin{bmatrix} -1 & 0 & -1 \\ 1 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}, rank 2, nullity 1, kernel \operatorname{span}\{(1, 1, -1)\}, since column 3 is column 1 plus column 2. The image is 2-dimensional inside the plane b_1 + b_2 + b_3 = 0, so it is that plane. (2, -5, 3) sums to 0: plan (-2, 3, 0), route 1 \to 2 reversed twice and 2 \to 3 three times, plus any multiple of (1, 1, -1). (1, 1, 0) sums to 2: not reachable.
- Row basis \{(1, 0, 2, 1),\ (0, 1, -1, 3)\}. With (-2, 1, 1, 0): -2 + 2 = 0, 1 - 1 = 0. With (-1, -3, 0, 1): -1 + 1 = 0, -3 + 3 = 0. Row 3 is their sum, so it is perpendicular too; 2 + 2 = 4.
- \operatorname{Col} A: 3, in \mathbb{R}^5. \ker A: 5, in \mathbb{R}^8. \operatorname{Row} A: 3, in \mathbb{R}^8. \ker A^{T}: 2, in \mathbb{R}^5.
- \ker C^{T} = \operatorname{span}\{(2, -2, 1)\}: 2\,\text{row } 1 - 2\,\text{row } 2 + \text{row } 3 = \mathbf{0} says (2, -2, 1) is perpendicular to every column, and the dimension is 3 - 2 = 1. So \operatorname{im} C is the plane 2b_1 - 2b_2 + b_3 = 0.
- Subtracting multiples of row 1 leaves (0, k - 4, 0) and (0, 0, k - 2). For k \ne 2, 4: rank 3, nullity 0. For k = 4: rank 2, nullity 1, basis \{(-2, 1, 0)\}. For k = 2: rank 2, nullity 1, basis \{(-1, 0, 1)\}.
- A^{T}A and A both have n columns, and equal kernels give equal nullities, so \operatorname{rank}(A^{T}A) = n - \operatorname{nullity} A = \operatorname{rank} A.
- A(A\mathbf{x}) = \mathbf{0} for every \mathbf{x}, so \operatorname{Col} A \subseteq \ker A and r \le 3 - r for r = \operatorname{rank} A; so r \le 1. Example: \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, of rank 1 with zero square.
- \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} and \begin{bmatrix} 1 & -1 \\ 1 & -1 \end{bmatrix} are both 2 \times 2 of rank 1, with kernels \operatorname{span}\{(1, -1)\} and \operatorname{span}\{(1, 1)\}, different lines.