Types of matrices — row, column, square, diagonal, identity, zero, symmetric, skew-symmetric: Exercises

This page is a self-test ladder. Each problem has a fully worked solution hidden inside a collapsible callout — read the problem, try it on paper, then reveal. Every symbol we use ( $A^{T}$ , "diagonal", "order", $diag$ , $\dots$ ) was built in the parent note Types of Matrices; when in doubt, revisit it. We also lean on Matrix Transpose and Matrix Operations.

A quick reminder of the two symmetry words, because half the exercises hinge on them:

The picture below is the mental model for "mirror across the diagonal" that we will use again and again.

Figure — Types of matrices — row, column, square, diagonal, identity, zero, symmetric, skew-symmetric

Level 1 — Recognition

Problem 1.1

Name the type of each matrix (give the most specific correct name): $P = [401], Q = [0000], R = [50 0 - 2], S = [1001]$

Recall Solution 1.1

WHAT we check: shape first, then the zero-pattern.

$P$ has one row, three columns (order $1 \times 3$ ) → row matrix.
$Q$ is square and every entry is $0$ → zero matrix (also technically diagonal, but "zero" is more specific).
$R$ is square, all off-diagonal entries are $0$ , and the diagonal entries $5, - 2$ are not all $1$ → diagonal matrix (not identity).
$S$ is diagonal with all diagonal entries $= 1$ → identity matrix $I_{2}$ . WHY "most specific": identity is a special diagonal, and zero is a special everything — always name the tightest pattern.

Problem 1.2

Which of these are square? Give the order of each. $T = [29], U = [1331], V = [142536]$

Recall Solution 1.2

A matrix is square when rows $=$ columns.

$T$ : order $2 \times 1$ (2 rows, 1 column) → column matrix, not square.
$U$ : order $2 \times 2$ → square (and symmetric, since $a_{12} = 3 = a_{21}$ ).
$V$ : order $2 \times 3$ → not square. Only $U$ is square.

Level 2 — Application

Problem 2.1

Compute $A^{T}$ and state whether $A$ is symmetric, skew-symmetric, or neither. $A = [0 - 7 70]$

Recall Solution 2.1

WHAT: flip rows↔columns. Row $1$ of $A$ is $(0, 7)$ ; it becomes column $1$ of $A^{T}$ . $A^{T} = [07 - 7 0]$ WHY compare to $- A$ : $- A = [07 - 7 0]$ . This equals $A^{T}$ exactly, so $A^{T} = - A$ . Also the diagonal is all zeros — consistent with the skew rule. $A$ is skew-symmetric.

Problem 2.2

Fill in $x, y, z$ so that the matrix is symmetric: $M = 25 z x y 4 - 1 43$

Recall Solution 2.2

Rule used: symmetric means $a_{ij} = a_{j i}$ (mirror across the diagonal — see the figure).

Mirror of $a_{21} = 5$ is $a_{12} = x$ , so $x = 5$ .
Mirror of $a_{31} = z$ is $a_{13} = - 1$ , so $z = - 1$ .
$a_{23} = 4$ and $a_{32} = 4$ already match ✓.
$y$ is on the diagonal, so it mirrors to itself; symmetry places no constraint on it — any value works. Take $y$ free (e.g. $y = 0$ ). Answer: $x = 5, z = - 1, y$ arbitrary.

Problem 2.3

Given $D = diag (3, - 1, 2)$ , compute $D^{2}$ and $D I_{3}$ .

Recall Solution 2.3

WHY diagonal powers are easy: multiplying diagonal matrices just multiplies matching diagonal entries (off-diagonals stay $0$ ). So $D^{k} = diag (d_{1}^{k}, d_{2}^{k}, d_{3}^{k})$ . $D^{2} = diag (3^{2}, (- 1)^{2}, 2^{2}) = diag (9, 1, 4) = 900010004$ And $D I_{3} = D$ because the identity is the multiplicative "1": $D I_{3} = 300 0 - 1 0 002$ .

Level 3 — Analysis

Problem 3.1

Can a matrix be both symmetric and skew-symmetric? If so, describe every such matrix.

Recall Solution 3.1

Set up both conditions at once. Symmetric gives $a_{ij} = a_{j i}$ ; skew gives $a_{ij} = - a_{j i}$ . Combine: $a_{j i} = - a_{j i}$ , so $2 a_{j i} = 0$ , hence $a_{ij} = 0$ for all $i, j$ . Conclusion: the only matrix that is both is the zero matrix $O$ . (This is why the zero matrix quietly satisfies almost every symmetry definition.)

Problem 3.2

A matrix $B$ is skew-symmetric and $3 \times 3$ . Its entries above the diagonal are $b_{12} = a, b_{13} = b, b_{23} = c$ . Write out all $9$ entries, then count how many independent numbers a $3 \times 3$ skew-symmetric matrix has.

Recall Solution 3.2

Diagonal entries are forced to $0$ . Below-diagonal entries are the negatives of the mirror above: $B = 0 - a - b a 0 - c b c 0$ Counting freedom: only the three above-diagonal entries $a, b, c$ are free; everything else is determined by them. So a $3 \times 3$ skew-symmetric matrix has $3$ independent entries. (General rule: $\frac{n ( n - 1 )}{2}$ ; for $n = 3$ that is $\frac{3 \cdot 2}{2} = 3$ ✓.)

Problem 3.3

Is $[1223]$ diagonal? Is it symmetric? Explain the relationship your answer reveals.

Recall Solution 3.3

Diagonal? No — the off-diagonal entry $2 \neq = 0$ .
Symmetric? Yes — $a_{12} = 2 = a_{21}$ . Relationship revealed: every diagonal matrix is symmetric, but not every symmetric matrix is diagonal. Diagonal is a strict subset of symmetric (see the parent's mistake callout).

Level 4 — Synthesis

Problem 4.1

Split $A = [4268]$ into a symmetric part $S = \frac{1}{2} (A + A^{T})$ and a skew part $K = \frac{1}{2} (A - A^{T})$ . Verify $S + K = A$ .

Recall Solution 4.1

WHY this works: $\frac{1}{2} (A + A^{T})$ is always symmetric (adding a matrix to its transpose mirrors it), and $\frac{1}{2} (A - A^{T})$ is always skew (subtracting cancels the diagonal). Their sum telescopes back to $A$ . First $A^{T} = [4628]$ . $S = \frac{1}{2} [88816] = [4448], K = \frac{1}{2} [0 - 4 40] = [0 - 2 20]$ Check: $S + K = [4268] = A$ ✓. Note $S$ is symmetric, $K$ is skew (zero diagonal).

Problem 4.2

Let $A$ be any square matrix. Prove $A + A^{T}$ is symmetric.

Recall Solution 4.1... (4.2)

Tool: two transpose rules — $(X + Y)^{T} = X^{T} + Y^{T}$ and $(X^{T})^{T} = X$ . Let $B = A + A^{T}$ . Then $B^{T} = (A + A^{T})^{T} = A^{T} + (A^{T})^{T} = A^{T} + A = A + A^{T} = B .$ Since $B^{T} = B$ , the matrix $B$ is symmetric. $■$

Level 5 — Mastery

Problem 5.1

Prove that every square matrix $A$ can be written uniquely as symmetric $+$ skew-symmetric.

Recall Solution 5.1

Existence. Set $S = \frac{1}{2} (A + A^{T})$ and $K = \frac{1}{2} (A - A^{T})$ . By Problem 4.2's argument, $S^{T} = S$ (symmetric) and $K^{T} = \frac{1}{2} (A^{T} - A) = - K$ (skew). And $S + K = A$ . So a decomposition exists. Uniqueness. Suppose $A = S^{'} + K^{'}$ with $S^{'}$ symmetric, $K^{'}$ skew. Transpose: $A^{T} = S^{' T} + K^{' T} = S^{'} - K^{'}$ . Add and subtract the two equations: $A + A^{T} = 2 S^{'} \Rightarrow S^{'} = \frac{1}{2} (A + A^{T}) = S, A - A^{T} = 2 K^{'} \Rightarrow K^{'} = \frac{1}{2} (A - A^{T}) = K .$ So the pieces are forced — the decomposition is unique. $■$

Problem 5.2

Prove that for any $n \times n$ skew-symmetric matrix $B$ with $n$ odd, the trace (sum of diagonal entries) is $0$ , and state why this is immediate.

Recall Solution 5.2

Immediate reason: skew-symmetry forces $b_{ii} = 0$ for every $i$ (from $b_{ii} = - b_{ii}$ ). The trace is $b_{11} + b_{22} + \dots + b_{nn} = 0 + 0 + \dots + 0 = 0$ . This holds for all $n$ , odd or even — the "odd" hint is a distractor. The trace of any skew-symmetric matrix is $0$ .

Problem 5.3

Let $D = diag (2, 3, 5)$ . Using the easy-power rule, compute the diagonal of $D^{4}$ .

Recall Solution 5.3

$D^{4} = diag (2^{4}, 3^{4}, 5^{4}) = diag (16, 81, 625)$ . This is why methods like Diagonalization are prized: once a matrix is put in diagonal form, powers (and hence eigenvalue-based analysis) become one-line computations.

Recall Self-test summary

One-line takeaway per level ::: L1 name the tightest pattern; L2 transpose = $(A^{T})_{ij} = A_{j i}$ ; L3 skew diagonals are $0$ ; L4 halve each part; L5 every square matrix = unique symmetric + skew. Only matrix that is both symmetric and skew-symmetric ::: the zero matrix $O$ . Independent entries in a $3 \times 3$ skew matrix ::: $3$ (i.e. $\frac{n ( n - 1 )}{2}$ ).