The Lumenary · Early undergraduate

Linear algebra

Name: ______________________

Date: ______________________

You will learn to

Solve linear systems by Gaussian elimination and describe every solution set as a translate of a null space
Prove that all bases of a finite-dimensional vector space have the same size (the exchange lemma), and prove and apply the rank–nullity theorem
Represent linear maps by matrices, prove that composition corresponds to matrix multiplication, and change basis
Prove that a square matrix is invertible if and only if its determinant is non-zero, if and only if its columns are linearly independent
Compute eigenvalues and eigenvectors, prove that eigenvectors for distinct eigenvalues are independent, and decide whether a matrix is diagonalisable

1.
What is the rank of the matrix $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 0 & 1 \end{pmatrix}$ ?
2.
Which of these are subspaces of the given vector space? (Choose all that apply.)
- A. $\{(t, 2t) : t \in \R\} \subseteq \R^2$
- B. $\{(x, y, z) : x + y + z = 1\} \subseteq \R^3$
- C. $\{x \in \R^n : Ax = 0\}$ for a fixed $m \times n$ matrix $A$
- D. $\{(x, y) : xy = 0\} \subseteq \R^2$
- E.The polynomials of degree exactly $2$ , inside all polynomials
3.
A linear map $T : \R^7 \to \R^4$ is surjective. What is $\dim \ker T$ ?
4.
$A$ is a $3 \times 3$ matrix with $\det A = 5$ . What is $\det(2A)$ ?
5.
Which of these real matrices are diagonalisable over $\R$ ? (Choose all that apply.)
- A. $\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$
- B. $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
- C. $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
- D. $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$
- E. $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$
6.
In the proof that eigenvectors $v_1, \dots, v_m$ with distinct eigenvalues are independent, we reach $c_i(\lambda_i - \lambda_m) = 0$ for $i < m$ . Which fact lets us conclude $c_i = 0$ ?
- A. $\lambda_i \neq \lambda_m$ , and in a field a product of two non-zero elements is non-zero
- B. $v_i \neq 0$
- C.The induction hypothesis
- D.Linearity of $T$
7.
For a linear map $T : V \to V$ we proved: if $V$ is finite-dimensional, then $T$ injective $\Rightarrow$ $T$ surjective. Which example shows that the finite-dimensionality hypothesis cannot be dropped?
- A.On the space of all real sequences, $T(a_0, a_1, a_2, \dots) = (0, a_0, a_1, \dots)$
- B.On all polynomials, $T = d/dx$
- C.On $\R^2$ , $T(x, y) = (x, 0)$
- D.On $\R^2$ , rotation by $90^\circ$
8.
Let f : V →ₗ[K] W with [FiniteDimensional K V]. Which Lean statement correctly formalises the rank–nullity theorem?
- A.finrank K (LinearMap.range f) + finrank K (LinearMap.ker f) = finrank K V
- B.finrank K (LinearMap.range f) + finrank K (LinearMap.ker f) = finrank K W
- C.finrank K (LinearMap.range f) ≤ finrank K V
- D.finrank K (LinearMap.ker f) = finrank K V - finrank K W
9.
Write rank(rows) that returns the rank of a matrix (a list of rows of integers) by Gaussian elimination with exact Fraction arithmetic. Then use it to check rank–nullity on the photosynthesis matrix: its null space should be one-dimensional.
Computational — Practice

Answer key — Linear algebra

1.
$R_2 - 2R_1 = (0, 0, 0)$ and $R_3 - R_1 = (0, -2, -2)$ . So the echelon form has the two pivot rows $(1, 2, 3)$ and $(0, -2, -2)$ and one zero row: the rank is $2$ . By rank–nullity the null space has dimension $3 - 2 = 1$ ; indeed it is spanned by $(-1, -1, 1)$ .
2.
The line through the origin and the null space are subspaces. The plane $x + y + z = 1$ misses $0$ ; the union of the axes is not closed under addition; the polynomials of degree exactly $2$ miss $0$ and are not closed under addition.
3.
$\operatorname{im} T = \R^4$ has dimension $4$ , so $\dim \ker T = 7 - 4 = 3$ .
4.
$\det(2A) = 2^3 \det A = 8 \cdot 5 = 40$ . Geometrically: doubling all lengths in space multiplies volumes by $8$ . In general $\det(cA) = c^n \det A$ for $n \times n$ matrices — the determinant is not linear in the whole matrix.
5.
Diagonalisable over $\R$ : $\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ (distinct eigenvalues), $3I$ (already diagonal) and $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ (eigenvalues $3, 1$ ). The shear has a single eigenvalue with a one-dimensional eigenspace, and the rotation has no real eigenvalues.
6.
Because the eigenvalues are distinct, $\lambda_i - \lambda_m \neq 0$ , and a field has no zero divisors, so $c_i = 0$ . If two eigenvalues coincided the argument would fail — and so would the theorem: $e_1$ and $2e_1$ are both eigenvectors of $I$ for the eigenvalue $1$ , and they are dependent.
7.
The right shift on sequences is injective but not surjective. In the proof, rank–nullity gave $\dim \operatorname{im} T = \dim V$ , and then 'a subspace of full dimension is everything' — both steps need $\dim V < \infty$ .
8.
Rank–nullity: $\dim \operatorname{im} f + \dim \ker f = \dim V$ , i.e. finrank K (LinearMap.range f) + finrank K (LinearMap.ker f) = finrank K V. The codomain version fails for the zero map $\R \to \R^5$ , and the subtraction version assumes surjectivity.

from fractions import Fraction

def rank(rows):
    m = [[Fraction(x) for x in row] for row in rows]
    r = 0
    for c in range(len(m[0]) if m else 0):
        p = next((i for i in range(r, len(m)) if m[i][c] != 0), None)
        if p is None:
            continue
        m[r], m[p] = m[p], m[r]
        for i in range(r + 1, len(m)):
            f = m[i][c] / m[r][c]
            m[i] = [a - f * b for a, b in zip(m[i], m[r])]
        r += 1
    return r

The photosynthesis matrix has rank $3$ , so by rank–nullity its null space has dimension $4 - 3 = 1$ : all balancings are multiples of $(6, 6, 1, 6)$ . Exact fractions matter — with floating point, a value like $10^{-17}$ left over from rounding would wrongly count as a pivot.