Advanced Algebra II — Quiz 2 Solutions

Four problems from the second quiz of Advanced Algebra II (Spring–Summer 2025–26).

Problem 1

Let $a, b, c \in \mathbb{R}^3$ be pairwise orthogonal unit vectors with $a \times b = c$ , where $a = (a_1, a_2, a_3)$ . Define the linear transformation

\mathcal{A}: \mathbb{R}^3 \to \mathbb{R}^3, \quad x \mapsto a \times x.

Find the matrix of $\mathcal{A}$ with respect to the standard basis $\{e_1, e_2, e_3\}$ .
Find the matrix of $\mathcal{A}$ with respect to the basis $\{a, b, c\}$ .

Solution

Part 1. The cross product $a \times x$ in coordinates is

a \times x = \begin{pmatrix} a_2 x_3 - a_3 x_2 \\ a_3 x_1 - a_1 x_3 \\ a_1 x_2 - a_2 x_1 \end{pmatrix}.

Thus the matrix of $\mathcal{A}$ in the standard basis is the skew-symmetric matrix

[\mathcal{A}]_{\{e_i\}} = \begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{pmatrix}.

Part 2. Since $\{a, b, c\}$ is a right-handed orthonormal basis, the cross products among them follow the cyclic relations

a \times a = 0,\quad a \times b = c,\quad a \times c = -b.

Hence

\mathcal{A}(a) = 0,\quad \mathcal{A}(b) = c,\quad \mathcal{A}(c) = -b.

Writing these as coordinate vectors in the basis $\{a, b, c\}$ :

[\mathcal{A}(a)] = \begin{pmatrix}0\\0\\0\end{pmatrix},\quad [\mathcal{A}(b)] = \begin{pmatrix}0\\0\\1\end{pmatrix},\quad [\mathcal{A}(c)] = \begin{pmatrix}0\\-1\\0\end{pmatrix}.

Therefore

[\mathcal{A}]_{\{a,b,c\}} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}.

Problem 2

Let $A = \operatorname{diag}(1, 2, \ldots, n)$ . Define the linear transformation

\mathscr{A}: M_{n \times n}(\mathbb{R}) \to M_{n \times n}(\mathbb{R}), \quad X \mapsto AX - XA.

Find $\operatorname{Ker} \mathscr{A}$ and $\operatorname{Im} \mathscr{A}$ .

Solution

Let $X = (x_{ij})_{n \times n}$ . Compute the $(i,j)$ -entry of $\mathscr{A}(X)$ :

(AX)_{ij} = i \cdot x_{ij},\qquad (XA)_{ij} = x_{ij} \cdot j.

Thus

(\mathscr{A}(X))_{ij} = (i - j)\, x_{ij}.

Kernel. $\mathscr{A}(X) = 0$ means $(i-j)x_{ij} = 0$ for all $i,j$ . When $i \neq j$ , this forces $x_{ij} = 0$ . When $i = j$ , $x_{ii}$ is unrestricted. Hence

\operatorname{Ker} \mathscr{A} = \{\,\text{diagonal matrices}\,\}.

Its dimension is $n$ .

Image. The entries of $\mathscr{A}(X)$ are $(i-j)x_{ij}$ . The diagonal entries ( $i=j$ ) are always $0$ , and every off-diagonal entry can be achieved independently by choosing the corresponding $x_{ij}$ . Therefore

\operatorname{Im} \mathscr{A} = \{\,\text{matrices with zero diagonal}\,\}.

Its dimension is $n^2 - n$ . Note that $\dim \operatorname{Ker} \mathscr{A} + \dim \operatorname{Im} \mathscr{A} = n + (n^2 - n) = n^2$ , confirming the rank-nullity theorem.

Problem 3

Consider the ellipse

E: \frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 1, \qquad (a, b > 0).

Let $\sigma$ be a plane affine transformation with coordinate formula

\begin{cases} x' = a_{11}x + a_{12}y + a_{13}, \\ y' = a_{21}x + a_{22}y + a_{23}, \end{cases} \qquad a_{11}a_{22} - a_{12}a_{21} \neq 0.

Prove that the image $\sigma(E)$ is also an ellipse.

Solution

Write points as column vectors $p = (x, y)^T$ , $p' = (x', y')^T$ . The affine map is

p' = M p + t,

where $M = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ is invertible (since $\det M \neq 0$ ), and $t = (a_{13}, a_{23})^T$ .

Set $p_0 = (x_0, y_0)^T$ and $D = \operatorname{diag}(1/a^2,\, 1/b^2)$ . The ellipse $E$ has equation

(p - p_0)^T D\,(p - p_0) = 1.

Since $M$ is invertible, $p = M^{-1}(p' - t)$ . Substituting:

\bigl(M^{-1}(p' - t) - p_0\bigr)^T D\,\bigl(M^{-1}(p' - t) - p_0\bigr) = 1.

Let $p_0' = \sigma(p_0) = M p_0 + t$ . Then $M^{-1}(p' - t) - p_0 = M^{-1}(p' - p_0')$ , giving

\bigl(M^{-1}(p' - p_0')\bigr)^T D\,\bigl(M^{-1}(p' - p_0')\bigr) = 1,

or equivalently

(p' - p_0')^T \widetilde{D}\,(p' - p_0') = 1, \qquad \widetilde{D} = (M^{-1})^T D\,M^{-1}.

Now $D$ is positive definite (both eigenvalues $1/a^2, 1/b^2 > 0$ ), and $M^{-1}$ is invertible. The congruence $(M^{-1})^T D\,M^{-1}$ preserves positive definiteness, so $\widetilde{D}$ is also positive definite. Its two eigenvalues are real and positive, meaning the quadratic form defines an ellipse (possibly rotated) centered at $p_0'$ .

Thus $\sigma(E)$ is an ellipse.

Problem 4

Let $V$ be a finite-dimensional linear space over a field $F$ , and let $\mathcal{A}: V \to V$ be a linear transformation. Prove that the following three conditions are equivalent:

$\operatorname{Ker} \mathcal{A} = \operatorname{Ker} \mathcal{A}^2$ .
$\operatorname{Im} \mathcal{A} = \operatorname{Im} \mathcal{A}^2$ .
$V = \operatorname{Ker} \mathcal{A} \oplus \operatorname{Im} \mathcal{A}$ .

Solution

Throughout, we will use the elementary facts:

\operatorname{Ker} \mathcal{A} \subseteq \operatorname{Ker} \mathcal{A}^2, \qquad \operatorname{Im} \mathcal{A}^2 \subseteq \operatorname{Im} \mathcal{A},

both of which follow directly from the definitions.

(1) $\Rightarrow$ (3). Assume $\operatorname{Ker} \mathcal{A} = \operatorname{Ker} \mathcal{A}^2$ .

First, we show $\operatorname{Ker} \mathcal{A} \cap \operatorname{Im} \mathcal{A} = \{0\}$ . Take $v \in \operatorname{Ker} \mathcal{A} \cap \operatorname{Im} \mathcal{A}$ . Then $v = \mathcal{A}w$ for some $w$ , and $\mathcal{A}v = 0$ . Hence $\mathcal{A}^2 w = \mathcal{A}v = 0$ , so $w \in \operatorname{Ker} \mathcal{A}^2 = \operatorname{Ker} \mathcal{A}$ . Therefore $v = \mathcal{A}w = 0$ .

By the rank-nullity theorem, $\dim \operatorname{Ker} \mathcal{A} + \dim \operatorname{Im} \mathcal{A} = \dim V$ . Since the intersection is trivial,

\dim(\operatorname{Ker} \mathcal{A} + \operatorname{Im} \mathcal{A}) = \dim \operatorname{Ker} \mathcal{A} + \dim \operatorname{Im} \mathcal{A} = \dim V,

so $\operatorname{Ker} \mathcal{A} + \operatorname{Im} \mathcal{A} = V$ . Together with the trivial intersection, $V = \operatorname{Ker} \mathcal{A} \oplus \operatorname{Im} \mathcal{A}$ .

(2) $\Rightarrow$ (3). Assume $\operatorname{Im} \mathcal{A} = \operatorname{Im} \mathcal{A}^2$ .

For any $v \in V$ , we have $\mathcal{A}v \in \operatorname{Im} \mathcal{A} = \operatorname{Im} \mathcal{A}^2$ , so $\mathcal{A}v = \mathcal{A}^2 w$ for some $w$ . Then

\mathcal{A}(v - \mathcal{A}w) = \mathcal{A}v - \mathcal{A}^2 w = 0,

so $v - \mathcal{A}w \in \operatorname{Ker} \mathcal{A}$ . Writing $v = (v - \mathcal{A}w) + \mathcal{A}w$ , we have expressed an arbitrary $v$ as a sum of an element of $\operatorname{Ker} \mathcal{A}$ and an element of $\operatorname{Im} \mathcal{A}$ . Thus $V = \operatorname{Ker} \mathcal{A} + \operatorname{Im} \mathcal{A}$ .

Again by rank-nullity, $\dim \operatorname{Ker} \mathcal{A} + \dim \operatorname{Im} \mathcal{A} = \dim V$ . With $V = \operatorname{Ker} \mathcal{A} + \operatorname{Im} \mathcal{A}$ , we deduce $\operatorname{Ker} \mathcal{A} \cap \operatorname{Im} \mathcal{A} = \{0\}$ , so it is a direct sum.

(3) $\Rightarrow$ (1). Assume $V = \operatorname{Ker} \mathcal{A} \oplus \operatorname{Im} \mathcal{A}$ .

We always have $\operatorname{Ker} \mathcal{A} \subseteq \operatorname{Ker} \mathcal{A}^2$ . For the reverse, let $v \in \operatorname{Ker} \mathcal{A}^2$ , so $\mathcal{A}^2 v = 0$ . Set $w = \mathcal{A}v$ . Then $w \in \operatorname{Im} \mathcal{A}$ and $\mathcal{A}w = \mathcal{A}^2 v = 0$ , so $w \in \operatorname{Ker} \mathcal{A}$ . Thus $w \in \operatorname{Ker} \mathcal{A} \cap \operatorname{Im} \mathcal{A} = \{0\}$ , giving $w = 0$ , i.e., $\mathcal{A}v = 0$ . Hence $v \in \operatorname{Ker} \mathcal{A}$ , proving $\operatorname{Ker} \mathcal{A}^2 \subseteq \operatorname{Ker} \mathcal{A}$ . Therefore $\operatorname{Ker} \mathcal{A} = \operatorname{Ker} \mathcal{A}^2$ .

(3) $\Rightarrow$ (2). Again assume $V = \operatorname{Ker} \mathcal{A} \oplus \operatorname{Im} \mathcal{A}$ .

We always have $\operatorname{Im} \mathcal{A}^2 \subseteq \operatorname{Im} \mathcal{A}$ . For the reverse, take any $y \in \operatorname{Im} \mathcal{A}$ , so $y = \mathcal{A}x$ for some $x$ . Decompose $x = u + v$ with $u \in \operatorname{Ker} \mathcal{A}$ and $v \in \operatorname{Im} \mathcal{A}$ . Since $v \in \operatorname{Im} \mathcal{A}$ , write $v = \mathcal{A} z$ for some $z$ . Decompose $z = u' + v'$ with $u' \in \operatorname{Ker} \mathcal{A}$ , $v' \in \operatorname{Im} \mathcal{A}$ . Then

y = \mathcal{A}(u + v) = \mathcal{A}v = \mathcal{A}^2 z.

Thus $y \in \operatorname{Im} \mathcal{A}^2$ , proving $\operatorname{Im} \mathcal{A} \subseteq \operatorname{Im} \mathcal{A}^2$ . Therefore $\operatorname{Im} \mathcal{A} = \operatorname{Im} \mathcal{A}^2$ .

All equivalences are proved. The three conditions characterize the situation where the ascending chain $\operatorname{Ker} \mathcal{A} \subseteq \operatorname{Ker} \mathcal{A}^2 \subseteq \cdots$ and the descending chain $\operatorname{Im} \mathcal{A} \supseteq \operatorname{Im} \mathcal{A}^2 \supseteq \cdots$ both stabilize at the first step, which is equivalent to $V$ decomposing as the direct sum of the kernel and image.

Intuition. When $V = \operatorname{Ker} \mathcal{A} \oplus \operatorname{Im} \mathcal{A}$ , the transformation $\mathcal{A}$ acts as an isomorphism on $\operatorname{Im} \mathcal{A}$ (it is injective on this subspace because any non-trivial kernel element would lie in the intersection) and as zero on $\operatorname{Ker} \mathcal{A}$ . This clean decomposition is exactly what guarantees that applying $\mathcal{A}$ twice produces no new kernel elements and no smaller image.