Symmetric and skew-symmetric matrices

Take any square matrix and compare it with its own transpose — flip it across the main diagonal — and one of three things happens: it comes back unchanged, it comes back exactly negated, or it comes back as something entirely different. The first two cases are special enough to earn their own names: a matrix that survives the flip unchanged is symmetric, and one that comes back negated is skew-symmetric. These are not exotic curiosities; symmetric matrices show up constantly in applications, from covariance matrices in statistics to $AA^T$ products that appear whenever data is combined with its own transpose, and CBSE Class 12 treats them right after the properties of the transpose operation for exactly this reason. This section builds the definitions from scratch, proves the single most useful fact about them — that every square matrix splits uniquely into a symmetric piece and a skew-symmetric piece — and works through the sums-and-checks style questions that dominate this topic in exams.

The core insight

A square matrix is symmetric when $A^T = A$ and skew-symmetric when $A^T = -A$ , and every square matrix decomposes uniquely as $A = \tfrac12(A+A^T) + \tfrac12(A-A^T)$ , a symmetric part plus a skew-symmetric part.

The core idea

Both definitions are stated entry-by-entry, and it pays to think in terms of entries before jumping to the matrix notation. A square matrix $A = [a_{ij}]$ is symmetric if every entry equals its mirror image across the main diagonal:

A^T = A \quad \Longleftrightarrow \quad a_{ij} = a_{ji} \text{ for all } i, j

where: $a_{ij}$ is the entry in row $i$ , column $j$ , and $a_{ji}$ is the entry obtained by swapping the row and column indices — the entry reflected across the diagonal running from top-left to bottom-right. A square matrix is skew-symmetric (sometimes called anti-symmetric) if every entry is the negative of its mirror image:

A^T = -A \quad \Longleftrightarrow \quad a_{ij} = -a_{ji} \text{ for all } i, j

The diagonal behaves very differently in the two cases. In a symmetric matrix the diagonal entries $a_{ii}$ can be anything at all, because $a_{ii} = a_{ii}$ is automatically true. In a skew-symmetric matrix, setting $i = j$ in the defining relation gives $a_{ii} = -a_{ii}$ , which forces $2a_{ii} = 0$ and hence $a_{ii} = 0$ for every diagonal entry. This single deduction is the most exam-tested fact in the whole topic: a matrix with even one nonzero diagonal entry can never be skew-symmetric, no matter what the off-diagonal entries look like. Because the diagonal of a skew-symmetric matrix is pinned to zero and every entry below the diagonal is forced to be the negative of the entry mirrored above it, a skew-symmetric matrix is completely determined by the entries strictly above the diagonal. A $3\times3$ skew-symmetric matrix has $9$ entries in total, but only $3$ of them are free to choose — the $3$ diagonal entries are zero and the $3$ entries below the diagonal are copies (negated) of the $3$ above it. The figure below shows the mirroring visually: symmetric matrices reflect with the same sign, skew-symmetric matrices reflect with a sign flip and a forced-zero diagonal.

Figure

Two three by three grids side by side showing a symmetric matrix reflecting with equal values across the diagonal and a skew symmetric matrix reflecting with negated values and a zero diagonal

Derivation AEvery square matrix splits into a symmetric part and a skew-symmetric part

Prove that for any square matrix $A$ , defining $P = \tfrac12(A+A^T)$ and $S = \tfrac12(A-A^T)$ produces a symmetric matrix $P$ , a skew-symmetric matrix $S$ , and a sum $P + S$ that reconstructs $A$ exactly, and that this is the only such pair of matrices with these properties.

Step 1. Check that $P = \tfrac12(A + A^T)$ is symmetric. Taking its transpose gives $P^T = \tfrac12(A^T + (A^T)^T) = \tfrac12(A^T + A) = P$ , using $(A^T)^T = A$ and the fact that transposing a sum transposes each term separately. Since $P^T = P$ , $P$ is symmetric by definition.

Step 2. Check that $S = \tfrac12(A - A^T)$ is skew-symmetric. Its transpose is $S^T = \tfrac12(A^T - (A^T)^T) = \tfrac12(A^T - A) = -\tfrac12(A - A^T) = -S$ . Since $S^T = -S$ , $S$ is skew-symmetric.

Step 3. Confirm the pieces reconstruct $A$ . Adding them gives $P + S = \tfrac12(A+A^T) + \tfrac12(A-A^T) = \tfrac12(2A) = A$ , so the factor of one-half is exactly what makes the two halves add back to $A$ instead of $2A$ .

Step 4. Prove uniqueness. Suppose $A = P_1 + S_1 = P_2 + S_2$ with $P_1, P_2$ symmetric and $S_1, S_2$ skew-symmetric. Rearranging gives $P_1 - P_2 = S_2 - S_1$ ; the left side is symmetric and the right side is skew-symmetric, but the only matrix that is both is the zero matrix (from $M = M^T$ and $M = -M^T$ together), so $P_1 = P_2$ and $S_1 = S_2$ — the decomposition is forced, not just one option among many.

A = \underbrace{\tfrac12(A+A^T)}_{P,\ \text{symmetric}} \ + \ \underbrace{\tfrac12(A-A^T)}_{S,\ \text{skew-symmetric}}

where: $P$ is called the symmetric part of $A$ and $S$ the skew-symmetric part; both are built purely from $A$ and its transpose, so the decomposition exists for every square matrix and is completely mechanical to compute.

Worked example

Express $A = \begin{bmatrix} 2 & 3 \\ 7 & 4 \end{bmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix, and separately check whether $P = \begin{bmatrix} 3 & -2 \\ -2 & 5 \end{bmatrix}$ and $Q = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$ are symmetric, skew-symmetric, or neither.

Step 1. Classify $P$ and $Q$ first, since the same transpose check drives everything else. $P^T = \begin{bmatrix} 3 & -2 \\ -2 & 5 \end{bmatrix} = P$ , so $P$ is symmetric. $Q^T = \begin{bmatrix} 0 & -4 \\ 4 & 0 \end{bmatrix} = -Q$ , and its diagonal is already zero, so $Q$ is skew-symmetric.

Step 2. Now decompose $A = \begin{bmatrix} 2 & 3 \\ 7 & 4 \end{bmatrix}$ . Its transpose is $A^T = \begin{bmatrix} 2 & 7 \\ 3 & 4 \end{bmatrix}$ .

Step 3. Compute the symmetric part. $A + A^T = \begin{bmatrix} 4 & 10 \\ 10 & 8 \end{bmatrix}$ , so $\tfrac12(A+A^T) = \begin{bmatrix} 2 & 5 \\ 5 & 4 \end{bmatrix}$ . Its transpose is itself, confirming it is symmetric.

Step 4. Compute the skew-symmetric part. $A - A^T = \begin{bmatrix} 0 & -4 \\ 4 & 0 \end{bmatrix}$ , so $\tfrac12(A-A^T) = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$ . Its transpose is $\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ , the negative of itself, confirming it is skew-symmetric with a zero diagonal as required.

Step 5. Add the two parts back to verify: $\begin{bmatrix} 2 & 5 \\ 5 & 4 \end{bmatrix} + \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 7 & 4 \end{bmatrix} = A$ . The reconstruction matches exactly, so $A = \begin{bmatrix} 2 & 5 \\ 5 & 4 \end{bmatrix} + \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$ , a symmetric matrix plus a skew-symmetric matrix.

Competitive stretch

For a general $2\times2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the same formula gives the symmetric part $\begin{bmatrix} a & \tfrac{b+c}{2} \\ \tfrac{b+c}{2} & d \end{bmatrix}$ and the skew-symmetric part $\begin{bmatrix} 0 & \tfrac{b-c}{2} \\ \tfrac{c-b}{2} & 0 \end{bmatrix}$ in one line, without redoing the transpose arithmetic each time — useful for checking a computed decomposition quickly, since only $b$ and $c$ ever change between the two parts.

Common traps

Forgetting that skew-symmetric matrices must have an all-zero diagonal — any matrix with a nonzero diagonal entry is automatically disqualified, no matter what the rest of it looks like.
Checking only one or two entries instead of every pair when classifying a matrix — a single mismatched pair is enough to make it neither symmetric nor skew-symmetric.
Getting the decomposition signs backwards — the symmetric part uses $A + A^T$ and the skew-symmetric part uses $A - A^T$ (not $A^T - A$ , which would just flip the sign of $S$ ).
Forgetting the factor of $\tfrac12$ — without halving, $A + A^T$ and $A - A^T$ do not add back to $A$ , they add back to $2A$ .
Trying to apply either definition to a non-square matrix — $A^T$ must have the same order as $A$ for the comparison $A^T = A$ or $A^T = -A$ to even make sense, which only happens when $A$ is square.

Ask the tutor: Want to try classifying a trickier $3\times3$ matrix or walk through the general $2\times2$ decomposition formula step by step? Ask the AI tutor.

Why must every diagonal entry of a skew-symmetric matrix be zero?

The defining relation $a_{ij} = -a_{ji}$ must hold for every pair of indices, including the case $i = j$ . Substituting $i = j$ gives $a_{ii} = -a_{ii}$ , and the only number equal to its own negative is zero, so $2a_{ii} = 0$ forces $a_{ii} = 0$ . This applies to every diagonal entry independently, so the whole diagonal is pinned to zero — it is not an extra assumption, it falls straight out of the definition.

Is the symmetric plus skew-symmetric decomposition always unique, or could there be another way to split A?

It is unique. If $A = P_1 + S_1 = P_2 + S_2$ with $P_1, P_2$ symmetric and $S_1, S_2$ skew-symmetric, then $P_1 - P_2 = S_2 - S_1$ . The left side is symmetric (a difference of symmetric matrices is symmetric) and the right side is skew-symmetric, but the only matrix that is both symmetric and skew-symmetric is the zero matrix, since $M = M^T$ and $M = -M^T$ together force $M = -M$ . So $P_1 = P_2$ and $S_1 = S_2$ , and the formula $\tfrac12(A+A^T)$ , $\tfrac12(A-A^T)$ is the only decomposition that works.

Does a matrix have to be square before we can even ask if it is symmetric?

Yes. Symmetric and skew-symmetric are defined by comparing $A$ to $A^T$ , and $A^T$ only has the same order (same number of rows and columns) as $A$ when $A$ is square — for an $m \times n$ matrix with $m \neq n$ , $A^T$ is $n \times m$ , a different shape entirely, so $A^T = A$ or $A^T = -A$ cannot even be written down. Always confirm the matrix is square before attempting the classification.

Symmetric and skew-symmetric matrices

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads