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

This page hunts down every kind of question the topic Types of Matrices can throw at you. Before any worked example, we lay out a scenario matrix: a checklist of every distinct case. Then each worked example is tagged with the cell it covers, so by the end there are no surprises left.

Everything here rests on one operation you must already picture clearly — the transpose, written $A^T$. Its rule is simple: the entry sitting in row $i$, column $j$ of $A$ moves to row $j$, column $i$ of $A^T$. In symbols, $(A^T{)}_{ij}=A_{ji}$. Picture it as flipping the whole grid over its main diagonal (the top-left to bottom-right staircase), like folding a piece of paper along that crease. If that fold picture is fuzzy, revisit Matrix Transpose first.

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The scenario matrix

Every exam question on matrix types is really one of these cells. We will hit all of them.

Cell	Case class	What makes it tricky	Example
A	Classify by shape (row / column / square)	order $m\times n$ vs $n\times m$	Ex 1
B	Recognise diagonal vs identity	zeros on diagonal allowed for diagonal, not identity	Ex 2
C	Degenerate: zero matrix & the $1\times 1$ matrix	is $[0]$ symmetric? skew? both?	Ex 3
D	Verify symmetry with mixed signs	negative and positive mirror pairs	Ex 4
E	Verify skew-symmetry, catch a bad diagonal	non-zero diagonal ⇒ instantly fails	Ex 5
F	Split any square matrix into symmetric + skew parts	the universal decomposition	Ex 6
G	Solve for unknowns to force a type (exam twist)	set up equations from $a_{ij}=a_{ji}$	Ex 7
H	Word problem — symmetry from a real situation	distances / mutual relations are symmetric	Ex 8
I	Limiting / boundary: sum of symmetric + skew, and $A+A^T$ always symmetric	prove a general fact, not a number	Ex 9

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Example 1 — Cell A: classify by shape

Forecast: Guess the order (rows × columns) of each and one type-name for each before reading on.

1. Count $P$'s rows and columns. One row, three columns ⇒ order $1\times 3$. Why this step? Shape is decided purely by counting; "row matrix" is defined as exactly one row.
2. $P$ is a row matrix. It is not square (rows $\ne$ columns), so it has no determinant.
3. Count $Q$. Three rows, one column ⇒ order $3\times 1$ ⇒ column matrix.
4. Count $R$. Two rows, two columns ⇒ order $2\times 2$ ⇒ square matrix. Being square is the entry ticket for determinants, eigenvalues, and inverses (see Determinants, Inverse of a Matrix).

Verify: Total entries must match order. $P$: $1\times 3=3$ entries ✓. $Q$: $3\times 1=3$ ✓. $R$: $2\times 2=4$ ✓.

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Example 2 — Cell B: diagonal vs identity

Forecast: Which one is the identity? Is $D_2$ still "diagonal" even though a diagonal entry is $0$?

1. Check the off-diagonal entries ($a_{12},a_{21}$) of each. All are $0$, so all three are diagonal matrices by definition ($a_{ij}=0$ whenever $i\ne j$). Why this step? "Diagonal" cares only about the off-diagonal being zero — it says nothing about what the diagonal itself holds.
2. Look at the diagonal of $D_2$. It contains a $0$ (top-left). That is perfectly allowed — a diagonal matrix may have zeros on its diagonal. Look at the figure: $D_2$ still lives entirely on the staircase.
3. Identity test: diagonal and every diagonal entry equals $1$. Only $D_3$ passes ⇒ $D_3=I_2$.
4. Scalar test: diagonal with all diagonal entries equal. $D_3$ (all $1$) qualifies; $D_1$ (entries $2,5$) and $D_2$ (entries $0,3$) do not.

Verify: Identity must satisfy $AI=A$. Take $A=\left[\begin{array}{cc}7& 8\\ 9& 1\end{array}\right]$: $A{D}_3=A$ ✓ (checked in VERIFY). And $D_1\ne I$ because $2\ne 1$.

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Example 3 — Cell C: the degenerate cases

Forecast: Can a matrix be symmetric and skew-symmetric at the same time? Guess before step 1.

1. Transpose $O$. Flipping an all-zero grid changes nothing: $O^T=O$. So $O$ is symmetric. Why this step? Symmetry is exactly the equation $A^T=A$; we test it directly.
2. Skew test on $O$: need $O^T=-O$. But $-O=O$ (negating zero gives zero), and $O^T=O$, so $O^T=-O$ holds too. So $O$ is also skew-symmetric. Why this step? The zero matrix is the unique case satisfying both conditions — because $A^T=A$ and $A^T=-A$ together force $A=-A$, i.e. $A=O$.
3. Transpose $M=[7]$. A $1\times 1$ matrix is its own transpose (nothing to flip): $M^T=[7]=M$. So $M$ is symmetric.
4. Skew test on $M$: need $M^T=-M$, i.e. $7=-7$. False. So $[7]$ is not skew-symmetric. (For $1\times 1$ skew, the single entry sits on the diagonal, and diagonal entries of a skew matrix must be $0$.)

Verify: Only skew $1\times 1$ matrix is $[0]$, since $a_{11}=-a_{11}\Rightarrow a_{11}=0$. ✓

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Example 4 — Cell D: verify symmetry with mixed signs

Forecast: Symmetry means each entry equals its mirror across the diagonal — even the negatives. Guess yes/no.

1. List the mirror pairs $(a_{ij},a_{ji})$ for $i<j$.
   • $a_{12}=-2$ and $a_{21}=-2$: equal ✓ (both negative — the sign must match, not flip).
   • $a_{13}=4$ and $a_{31}=4$: equal ✓
   • $a_{23}=-6$ and $a_{32}=-6$: equal ✓ Why this step? Symmetry is $a_{ij}=a_{ji}$; we only need the entries above the diagonal to match their reflections below.
2. Diagonal entries always satisfy $a_{ii}=a_{ii}$, so they never block symmetry.
3. All pairs match ⇒ $A$ is symmetric, $A^T=A$.

Verify: Compute $A^T$ and compare entrywise to $A$ (done in VERIFY, returns equal). ✓

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Example 5 — Cell E: catch the bad diagonal

Forecast: One of them fails at the very first glance. Which, and why?

1. Scan the diagonal of $C$ first. $c_{11}=2\ne 0$. A skew-symmetric matrix must have all diagonal entries $0$ (because $a_{ii}=-a_{ii}\Rightarrow a_{ii}=0$). So $C$ is rejected instantly — no transpose needed. Why this step? The zero-diagonal rule is the fastest disqualifier; use it before doing arithmetic.
2. Diagonal of $B$: all zeros ✓. Passes the gate.
3. Check off-diagonal pairs of $B$ for $a_{ij}=-a_{ji}$:
   • $b_{12}=3,b_{21}=-3$: $3=-(-3)$ ✓
   • $b_{13}=-1,b_{31}=1$: $-1=-(1)$ ✓
   • $b_{23}=5,b_{32}=-5$: $5=-(-5)$ ✓
4. So $B$ is skew-symmetric, $B^T=-B$.

Verify: $B^T=-B$ checked entrywise; $C$ fails since $C^T\ne -C$. ✓

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Example 6 — Cell F: split any square matrix

Forecast: Every square matrix splits this way, uniquely. Guess the formula for $S$ and $K$ from $A$ and $A^T$.

1. Compute the transpose. $A^T=\left[\begin{array}{cc}4& 6\\ 2& 8\end{array}\right]$. Why this step? The whole decomposition is built from $A$ and its mirror $A^T$.
2. Symmetric part $S=\frac{1}{2}(A+A^T)$. $S=\frac{1}{2}\left[\begin{array}{cc}8& 8\\ 8& 16\end{array}\right]=\left[\begin{array}{cc}4& 4\\ 4& 8\end{array}\right]$ Why this step? $A+A^T$ is always symmetric (its transpose is $A^T+A$, the same thing), so half of it is the symmetric piece.
3. Skew-symmetric part $K=\frac{1}{2}(A-A^T)$. $K=\frac{1}{2}\left[\begin{array}{cc}0& -4\\ 4& 0\end{array}\right]=\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]$ Why this step? $A-A^T$ is always skew (its transpose is $A^T-A=-(A-A^T)$), giving the skew piece — note its zero diagonal, as required.
4. Reassemble: $S+K=\left[\begin{array}{cc}4& 4\\ 4& 8\end{array}\right]+\left[\begin{array}{cc}0& -2\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}4& 2\\ 6& 8\end{array}\right]=A$ ✓

Verify: $S^T=S$, $K^T=-K$, and $S+K=A$ (all three in VERIFY). ✓

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Example 7 — Cell G: solve for unknowns (exam twist)

Forecast: Symmetry forces one equation. Guess how many unknowns it can pin down.

1. Write the symmetry condition for the off-diagonal pair: $a_{12}=a_{21}$, i.e. $x+2=2y-1.$ Why this step? Symmetry only constrains mirror pairs; the diagonal ($3,7$) is free.
2. One equation, two unknowns ⇒ infinitely many solutions along a line: $x=2y-3$. Why this step? We must report this honestly — a single symmetry equation cannot fix both variables.
3. Pick a sample solution: let $y=2\Rightarrow x=2(2)-3=1$. Then $a_{12}=1+2=3$ and $a_{21}=2(2)-1=3$ ✓.

Verify: With $x=1,y=2$: $A=\left[\begin{array}{cc}3& 3\\ 3& 7\end{array}\right]$, and $A^T=A$. ✓

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Example 8 — Cell H: word problem (symmetry from reality)

Forecast: Distance from X to Y equals distance from Y to X. What does that force?

1. Fill the matrix (order towns X, Y, Z): $A=\left[\begin{array}{ccc}0& 12& 8\\ 12& 0& 15\\ 8& 15& 0\end{array}\right]$ Why this step? Each entry is a physical distance; the diagonal is $0$ because a town is zero km from itself.
2. Observe the physics: distance is mutual — X→Y = Y→X. So $a_{ij}=a_{ji}$ for all pairs. Why this step? This is exactly the symmetry condition, arising naturally from a real property.
3. Therefore $A$ is symmetric. (It is not skew: skew would need $a_{ij}=-a_{ji}$, meaning negative distances — physically impossible.)

Verify: $A^T=A$ entrywise ✓. Note the zero diagonal here is a coincidence with skew's requirement but $A$ is not skew because the off-diagonals are positive, not sign-flipped.

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Example 9 — Cell I: prove a general fact (limiting case)

Forecast: No numbers needed for the proof — only the transpose rules. Guess which two rules you'll use.

1. Transpose rule for sums: $(P+Q{)}^T=P^T+Q^T$. Transpose of a transpose: $(A^T{)}^T=A$ (folding twice returns the original). Why this step? These two facts are the only machinery required.
2. Apply them: let $S=A+A^T$. Then $S^T=(A+A^T{)}^T=A^T+(A^T{)}^T=A^T+A=S.$ Since $S^T=S$, $S$ is symmetric — for every $A$, no exceptions.
3. Concrete test: $A^T=\left[\begin{array}{cc}1& -3\\ 5& 2\end{array}\right]$, so $A+A^T=\left[\begin{array}{cc}2& 2\\ 2& 4\end{array}\right],$ which is visibly symmetric.

Verify: $(A+A^T{)}^T=A+A^T$ for the numeric case ✓.

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Recall

Recall When can a matrix be both symmetric and skew-symmetric?

Only the zero matrix ::: because $A^T=A$ and $A^T=-A$ give $A=-A\Rightarrow A=O$.

Recall Fastest way to reject a skew-symmetric candidate?

Look at the diagonal — any non-zero diagonal entry disqualifies it instantly.

Recall Formula to split

$A$ into symmetric + skew parts? $A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$.

Recall Why is a distance matrix symmetric but not skew?

Distance is mutual so $a_{ij}=a_{ji}$ (symmetric); distances are non-negative, never sign-flipped (not skew).