Symmetric and skew-symmetric matrices

Take any square matrix and flip it across its main diagonal — that flip is called the transpose. Some matrices come back unchanged by the flip; others come back with every entry negated. The first kind are symmetric matrices, the second skew-symmetric matrices, and together they form one of the cleanest classification results in the whole matrices-and-determinants unit: every square matrix, no matter how ordinary or lopsided, can be split uniquely into a symmetric piece and a skew-symmetric piece that add back up to the original. This article states the two defining conditions, proves the decomposition, and works through the arithmetic on real $2\times2$ and $3\times3$ matrices so the definitions become something you can check by eye in seconds during an exam.

The core insight

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

The core idea

Symmetric matrices. A square matrix $A=[a_{ij}]$ is symmetric if it equals its own transpose:

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 $A^T$ is $A$ with rows and columns interchanged. Every off-diagonal entry has a twin reflected across the leading diagonal, and the diagonal entries $a_{ii}$ are unrestricted — they can be any number at all.

Skew-symmetric matrices. A square matrix $A$ is skew-symmetric if

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

Put $i=j$ in this condition and you get $a_{ii}=-a_{ii}$ , so $2a_{ii}=0$ and $a_{ii}=0$ . That is a forced consequence, not an extra rule to memorise: every diagonal entry of a skew-symmetric matrix is necessarily zero. The picture below shows what the two conditions do to a $3\times3$ grid of entries: symmetric mirrors a value across the diagonal, skew-symmetric mirrors it and flips its sign, and the diagonal itself behaves differently in each case.

Figure

A three by three grid showing a symmetric matrix with matching mirrored entries across the diagonal and a skew-symmetric matrix with sign-flipped mirrored entries and a zero diagonal

Neither is the typical case. Most square matrices satisfy neither condition — they are simply "general" square matrices — but the decomposition theorem below says that does not matter: a symmetric part and a skew-symmetric part can always be pulled out of any square $A$ .

Closure properties worth knowing. The sum of two symmetric matrices of the same order is again symmetric, and so is any scalar multiple of a symmetric matrix — both follow directly from $(X+Y)^T=X^T+Y^T$ and $(kX)^T=kX^T$ . The same two rules show the sum of two skew-symmetric matrices, and any scalar multiple of a skew-symmetric matrix, stays skew-symmetric. A frequently tested application goes further: for any matrix $A$ (square or not), the products $AA^T$ and $A^TA$ are always symmetric. The proof needs only the reversal rule $(AB)^T=B^TA^T$ : taking the transpose of $AA^T$ gives $(AA^T)^T=(A^T)^TA^T=AA^T$ , unchanged, so $AA^T$ is symmetric — and the identical argument applied to $A^TA$ gives $(A^TA)^T=A^T(A^T)^T=A^TA$ . This is exactly the kind of "prove that..." claim this unit favours, and it settles in two lines without expanding a single entry.

Derivation AEvery square matrix is the sum of a symmetric matrix and a skew-symmetric matrix

The goal is to show that for any square matrix $A$ of order $2$ or $3$ , the two matrices $P=\tfrac12(A+A^T)$ and $Q=\tfrac12(A-A^T)$ are respectively symmetric and skew-symmetric, that $P+Q$ reconstructs $A$ exactly, and that no other pair of a symmetric matrix and a skew-symmetric matrix can also add up to $A$ — using only the two transpose rules $(A^T)^T=A$ and $(X+Y)^T=X^T+Y^T$ .

Step 1. Define $P=\tfrac12(A+A^T)$ and take its transpose. Since transposing distributes over addition and $(A^T)^T=A$ , $P^T=\tfrac12(A^T+(A^T)^T)=\tfrac12(A^T+A)=\tfrac12(A+A^T)=P$ . So $P^T=P$ : $P$ is symmetric.

Step 2. Define $Q=\tfrac12(A-A^T)$ and take its transpose the same way. $Q^T=\tfrac12(A^T-(A^T)^T)=\tfrac12(A^T-A)=-\tfrac12(A-A^T)=-Q$ . So $Q^T=-Q$ : $Q$ is skew-symmetric.

Step 3. Add $P$ and $Q$ back together: $P+Q=\tfrac12(A+A^T)+\tfrac12(A-A^T)=\tfrac12\big[(A+A^T)+(A-A^T)\big]=\tfrac12(2A)=A$ . The two halves recombine to give back the original matrix exactly, with no leftover terms.

Step 4. Uniqueness. Suppose some other symmetric $P'$ and skew-symmetric $Q'$ also satisfy $P'+Q'=A$ . Then $P-P'=Q'-Q$ . The left side is a difference of two symmetric matrices, hence symmetric; the right side is a difference of two skew-symmetric matrices, hence skew-symmetric. A matrix that is both symmetric and skew-symmetric must satisfy $M^T=M$ and $M^T=-M$ simultaneously, forcing $M=-M$ , i.e. $M=0$ . So $P=P'$ and $Q=Q'$ : the decomposition is unique. Verified: for a general $2\times2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ , $P=\begin{pmatrix}a & \tfrac{b+c}2\\ \tfrac{b+c}2 & d\end{pmatrix}$ is symmetric and $Q=\begin{pmatrix}0 & \tfrac{b-c}2\\ \tfrac{c-b}2 & 0\end{pmatrix}$ is skew-symmetric with zero diagonal, and $P+Q$ reproduces every entry of $A$ .

Worked example

Express $B=\begin{pmatrix}2 & 3\\ 7 & 4\end{pmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix, and then check the result against a $3\times3$ extension.

Step 1 — find $B^T$ . Interchanging rows and columns, $B^T=\begin{pmatrix}2 & 7\\ 3 & 4\end{pmatrix}$ .

Step 2 — compute the symmetric part $P=\tfrac12(B+B^T)$ . $B+B^T=\begin{pmatrix}4 & 10\\ 10 & 8\end{pmatrix}$ , so $P=\begin{pmatrix}2 & 5\\ 5 & 4\end{pmatrix}$ . Checking, $P^T=P$ , and indeed the off-diagonal entries match: symmetric confirmed.

Step 3 — compute the skew-symmetric part $Q=\tfrac12(B-B^T)$ . $B-B^T=\begin{pmatrix}0 & -4\\ 4 & 0\end{pmatrix}$ , so $Q=\begin{pmatrix}0 & -2\\ 2 & 0\end{pmatrix}$ . Checking, $Q^T=-Q$ and the diagonal is zero: skew-symmetric confirmed.

Step 4 — verify the sum. $P+Q=\begin{pmatrix}2+0 & 5-2\\ 5+2 & 4+0\end{pmatrix}=\begin{pmatrix}2 & 3\\ 7 & 4\end{pmatrix}=B$ , exactly recovering the original matrix. Answer: $B=\begin{pmatrix}2 & 5\\ 5 & 4\end{pmatrix}+\begin{pmatrix}0 & -2\\ 2 & 0\end{pmatrix}$ , symmetric part $P$ plus skew-symmetric part $Q$ .

Competitive stretch

For the $3\times3$ matrix $A=\begin{pmatrix}3 & -2 & 4\\ -2 & 5 & 1\\ 4 & 1 & 0\end{pmatrix}$ , since $A^T=A$ already (checked entry by entry: $a_{12}=a_{21}=-2$ , $a_{13}=a_{31}=4$ , $a_{23}=a_{32}=1$ ), $A$ is already symmetric, so its own decomposition is trivial: $P=A$ and $Q$ is the zero matrix, since $A-A^T=0$ . This confirms a general fact worth remembering — a matrix that is already symmetric decomposes into itself plus the zero skew-symmetric matrix, and a purely skew-symmetric matrix decomposes into the zero matrix plus itself. This gives a fast diagnostic during an exam: compute $A-A^T$ first; if it comes out as the zero matrix, $A$ is already symmetric and there is no need to separately compute $\tfrac12(A+A^T)$ , since it will simply equal $A$ .

Common traps

Applying $\tfrac12(A+A^T)$ or $\tfrac12(A-A^T)$ to a non-square matrix — the formula requires $A$ and $A^T$ to have the same order, which only happens for square matrices.
Assuming a zero diagonal is enough to call a matrix skew-symmetric — it is necessary but not sufficient; every off-diagonal pair must also satisfy $a_{ij}=-a_{ji}$ , not just $a_{ii}=0$ .
Sign slips when testing skew-symmetry, such as checking $a_{ij}=a_{ji}$ (the symmetric condition) instead of $a_{ij}=-a_{ji}$ .
Believing every square matrix must be symmetric or skew-symmetric — most are neither, and the whole point of the decomposition theorem is that "neither" still splits cleanly into both.
Forgetting the reversal order in $(AB)^T=B^TA^T$ when proving a product like $AA^T$ is symmetric — writing $A^TA^T$ instead of the correct $(A^T)^TA^T=AA^T$ breaks the whole argument.

Ask the tutor: if you would like to practice this decomposition on a fresh $3\times3$ matrix or see the proof that $AA^T$ is always symmetric, ask the AI tutor about "symmetric and skew-symmetric matrices."

Why does the decomposition formula only work for square matrices?

Because $A+A^T$ and $A-A^T$ must be defined, and matrix addition only makes sense between matrices of the same order. If $A$ is $m\times n$ with $m\neq n$ , then $A^T$ is $n\times m$ , a different shape, so $A+A^T$ is not even a valid operation. The moment $A$ is square, $A^T$ has the identical order and the formula goes through cleanly for every entry.

Do I need to check both AT equals A and AT equals negative A separately, or is there a shortcut?

There is a fast shortcut: compute the diagonal first. If any diagonal entry is nonzero, the matrix cannot be skew-symmetric, so you only need to check $A^T=A$ . If the diagonal is all zeros, check both conditions on the off-diagonal entries, since a zero diagonal alone does not guarantee skew-symmetry. In practice, write out $A^T$ once and compare it to $A$ and to $-A$ side by side.

Is the symmetric plus skew-symmetric decomposition useful for anything beyond this definition, or is it just a classification exercise?

It shows up constantly in "prove that..." questions, where you are asked to show some constructed matrix, such as $A+A^T$ or $AA^T$ , is symmetric or skew-symmetric without expanding every entry. The transpose rules $(AB)^T=B^TA^T$ , $(A+B)^T=A^T+B^T$ and $(kA)^T=kA^T$ let you verify such claims algebraically in a couple of lines, the same toolkit used in the derivation above.

Symmetric and skew-symmetric matrices

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads