Symmetric vs Skew-Symmetric Matrices: Tests and Decomposition

Two entries sit across the main diagonal of a matrix. Do they match, or are they negatives of each other?

That small question is the starting point for symmetric and skew-symmetric matrices. Once you know which entries to compare, the definitions become easy to recognise. The more interesting part is that even a matrix that belongs to neither type can be split into one part of each type.

Think of a complicated surface with two hidden layers: one carries its shared balance, and the other carries the differences between opposite positions. Adding those layers restores the whole surface. Matrix decomposition does something similar, with exact numbers.

We will use real entries throughout this lesson. The numerical examples and practice questions below are original, so you can work through them without needing a separate exercise book.

The quick difference: same entries or opposite entries?

Here is the essential comparison. A^T means the transpose of A; some textbooks write it as A'.

Test	Symmetric matrix	Skew-symmetric matrix
Shape	Square	Square
Transpose condition	A^T = A	A^T = -A
Across the main diagonal	Matching entries	Entries are negatives
Main diagonal	Any real entries	Every entry is zero

These definitions and the zero-diagonal result are given in NCERT’s Matrices chapter, section 3.6.

The main diagonal runs from the top-left corner to the bottom-right corner. For a 3 by 3 matrix, compare these three pairs:

Position (1, 2) with position (2, 1)
Position (1, 3) with position (3, 1)
Position (2, 3) with position (3, 2)

For symmetry, each pair must match. For skew-symmetry, each pair must add to zero, and the three diagonal entries must also be zero.

First, separate transpose from negative

Transposing changes positions. Taking the negative changes signs. These are different operations.

For example:

A = [  3   8 ]
    [ -2   5 ]

A^T = [ 3  -2 ]
      [ 8   5 ]

-A = [ -3  -8 ]
     [  2  -5 ]

The 8 moves from the first row to the first column when we transpose. It stays positive. The -2 moves too, and stays negative.

To form -A, however, every entry changes sign and stays in its original position. This distinction prevents a surprisingly common mistake: changing signs while writing the transpose and then changing them again during the comparison.

A rectangular matrix can certainly be transposed. A 2 by 3 matrix becomes a 3 by 2 matrix. But it cannot equal its transpose because their shapes differ, so it cannot be symmetric or skew-symmetric under these definitions.

Worked example 1: recognise a symmetric matrix

Consider:

S = [  5  -3   7 ]
    [ -3   2   4 ]
    [  7   4  -6 ]

The three pairs are -3 and -3, 7 and 7, and 4 and 4. Each pair matches. Writing the transpose gives exactly the same matrix, so S is symmetric.

Notice what did not matter: the diagonal contains 5, 2, and -6. They do not have to be equal, positive, or zero. On transposing, each diagonal entry returns to its own position.

A general 3 by 3 symmetric matrix therefore has this form:

[ a  b  c ]
[ b  d  e ]
[ c  e  f ]

Choose the diagonal and the entries above it. The entries below it are then fixed. This is a useful way to construct your own examples for revision.

Worked example 2: recognise a skew-symmetric matrix

Now consider:

K = [  0   6  -4 ]
    [ -6   0   9 ]
    [  4  -9   0 ]

Its transpose is:

K^T = [  0  -6   4 ]
      [  6   0  -9 ]
      [ -4   9   0 ]

That is exactly -K. All three off-diagonal pairs are negatives of one another, and the diagonal is zero. Therefore, K is skew-symmetric.

Why must the diagonal vanish? A diagonal entry stays where it is after transposing, but skew-symmetry requires it to equal its own negative:

a = -a
2a = 0
a = 0

The general form is:

[  0   p   q ]
[ -p   0   r ]
[ -q  -r   0 ]

The letters p, q, and r can themselves be negative or zero. Do not interpret the printed minus signs as an instruction that every entry below the diagonal must be negative. If p is -5, then -p is 5.

Neither type, and the one matrix that is both

Here is a matrix with a zero diagonal:

N = [ 0   4 ]
    [ 7   0 ]

It is not symmetric because 4 != 7. It is not skew-symmetric because 4 != -7. Its diagonal passes one necessary check, but its paired entries fail the deciding check.

Now consider a square matrix that is both types. Its transpose would have to satisfy A^T = A and A^T = -A. Hence A = -A, giving 2A = O, where O is the zero matrix. Every entry must be zero.

So the square zero matrix is both symmetric and skew-symmetric, and it is the only real matrix with both properties.

For a 1 by 1 matrix, the same reasoning still works. Every [a] is symmetric; only [0] is skew-symmetric.

Worked example 3: find unknown entries

When a question contains letters, turn each pair into an equation. For skew-symmetry, begin with the diagonal because it often gives the quickest answer.

Make a matrix symmetric

Find x and y if this matrix is symmetric:

M = [  4    x + 2   2y ]
    [  9      1     -3 ]
    [ y + 5  -3      6 ]

The first pair gives x + 2 = 9, so x = 7.

The second gives 2y = y + 5, so y = 5.

The remaining pair is already -3 and -3. Substituting the answers gives:

M = [  4   9  10 ]
    [  9   1  -3 ]
    [ 10  -3   6 ]

All pairs match. Checking the third pair matters: a question can contain a condition that makes the required matrix impossible, even when the first two equations have solutions.

Make a matrix skew-symmetric

Find x, y, and z if:

N = [ x - 4    2y + 1    -8 ]
    [  -7        0      z + 2 ]
    [   8       -5        0 ]

The diagonal gives x - 4 = 0, hence x = 4.

For the first pair, the opposite of -7 is 7:

2y + 1 = -(-7) = 7
2y = 6
y = 3

For the last pair, z + 2 = -(-5) = 5, hence z = 3. The -8 and 8 pair already works.

After substitution:

N = [  0   7  -8 ]
    [ -7   0   5 ]
    [  8  -5   0 ]

Decomposition: split any square matrix into two parts

For a real square matrix A, its symmetric part S and skew-symmetric part K are:

S = (A + A^T) / 2
K = (A - A^T) / 2

A = S + K

Dividing a matrix by 2 means dividing every entry by 2. These formulas work for a square matrix of any order.

There is a friendly way to understand the calculation. Suppose two opposite-position entries are 8 and 2. Their common middle value is 5. One original entry is 3 above that middle, while the other is 3 below it:

8 = 5 + 3
2 = 5 - 3

The symmetric part keeps 5 and 5. The skew-symmetric part keeps 3 and -3. Together, they recover 8 and 2.

For any pair a and b, use:

Shared value = (a + b) / 2
Signed difference = (a - b) / 2

The diagonal belongs entirely to S because a diagonal entry is paired with itself. Its difference from itself is zero.

Why the decomposition is unique

Suppose A = S + K, with S^T = S and K^T = -K. Transposing gives:

A^T = S - K

Add the two equations and K disappears: A + A^T = 2S. Subtract them and S disappears: A - A^T = 2K.

Therefore, the two parts are forced to be the formulas above. There is no second choice of S and K for the same A. Brown University’s Special Matrices tutorial also states this uniqueness result.

Worked example 4: a complete 3 by 3 decomposition

Decompose:

A = [  6   8  -2 ]
    [  2  -4   7 ]
    [ 10  -3   5 ]

Step 1: write the transpose carefully

A^T = [  6   2  10 ]
      [  8  -4  -3 ]
      [ -2   7   5 ]

Read each original row down the corresponding new column. No entry changes sign at this step.

Step 2: add and halve

A + A^T = [ 12  10   8 ]
          [ 10  -8   4 ]
          [  8   4  10 ]

S = [ 6   5  4 ]
    [ 5  -4  2 ]
    [ 4   2  5 ]

The opposite-position pairs in S are 5 and 5, 4 and 4, and 2 and 2. The diagonal is still 6, -4, and 5, exactly as in A.

Step 3: subtract and halve

A - A^T = [  0   6  -12 ]
          [ -6   0   10 ]
          [ 12 -10    0 ]

K = [  0   3  -6 ]
    [ -3   0   5 ]
    [  6  -5   0 ]

Two entries deserve a slow check:

Entry (1, 3): -2 - 10 = -12
Entry (2, 3):  7 - (-3) = 10

Every pair in K now adds to zero. Its diagonal is also zero.

Step 4: add the parts back together

Check the rows of S + K:

Row 1: [ 6 + 0,   5 + 3,   4 - 6 ]
     = [ 6,       8,      -2 ]

Row 2: [ 5 - 3,  -4 + 0,   2 + 5 ]
     = [ 2,      -4,       7 ]

Row 3: [ 4 + 6,   2 - 5,   5 + 0 ]
     = [ 10,     -3,       5 ]

These are exactly the rows of A. The answer passes all three required checks: S is symmetric, K is skew-symmetric, and their sum is the original matrix.

A useful interpretation: transfers between two shops

Imagine a fictional table of stock transfers. Rows identify the sending shop and columns identify the receiving shop. Shop P sends 14 cartons to Q, while Q sends 6 cartons to P:

A = [ 0  14 ]
    [ 6   0 ]

The decomposition is:

S = [  0  10 ]
    [ 10   0 ]

K = [  0   4 ]
    [ -4   0 ]

The common part is 10 cartons in each direction. The signed correction adds 4 to one direction and subtracts 4 from the other.

This is a mathematical breakdown of the table. The -4 is a correction relative to the shared value, not a shipment of negative cartons. Also, the difference between the original transfers is 8 cartons, while the upper-right entry of K is half that difference, or 4. Keeping those meanings separate makes the formula much easier to remember.

Properties you can prove with the transpose

Most proofs here use four rules:

(A^T)^T = A
(A + B)^T = A^T + B^T
(cA)^T = cA^T
(AB)^T = B^T A^T

The last rule reverses the order of multiplication. MIT’s lesson on transposes is useful further reading for this step.

Sums and scalar multiples keep the type

If S and T are symmetric matrices of the same order, then:

(S + T)^T = S^T + T^T = S + T

So their sum is symmetric. Replacing the plus with a minus proves the result for their difference. Multiplying by a real number c works too, since (cS)^T = cS.

For skew-symmetric K and L of the same order:

(K + L)^T = -K - L = -(K + L)

Their sum is skew-symmetric. Differences and real scalar multiples follow in the same way. Multiplication by zero gives the zero matrix, which still satisfies the relevant condition.

A plus its transpose, and A minus its transpose

For any square A, put B = A + A^T. Transposing B simply swaps the two terms, so B^T = B.

For C = A - A^T, transposing gives C^T = A^T - A = -C.

These observations also verify the types of the two decomposition parts. But remember the halves: adding B and C gives 2A, not A.

A product with its own transpose is symmetric

For any real matrix A, including a rectangular one:

(AA^T)^T = (A^T)^T A^T = AA^T
(A^T A)^T = A^T (A^T)^T = A^T A

Both products are symmetric. They need not have the same size. If A is 2 by 3, AA^T is 2 by 2, while A^T A is 3 by 3.

Take this example:

A = [ 1  2  0 ]
    [ 3  1  4 ]

AA^T = [ 5   5 ]
       [ 5  26 ]

A^T A = [ 10   5  12 ]
        [  5   5   4 ]
        [ 12   4  16 ]

Each result passes the pair test, even though the original A was rectangular. If the dimensions are confusing, revise matrix multiplication and the order of AB and BA first.

The product trap: two symmetric matrices need not give a symmetric product

Consider two symmetric matrices:

S = [ 2  0 ]
    [ 0  5 ]

T = [ 1  3 ]
    [ 3  4 ]

Their products are:

ST = [  2   6 ]
     [ 15  20 ]

TS = [ 2  15 ]
     [ 6  20 ]

ST is not symmetric because 6 and 15 do not match. The exact condition follows from:

(ST)^T = T^T S^T = TS

Therefore, ST is symmetric if and only if ST = TS. When multiplication gives the same answer in either order, the matrices are said to commute.

We can also construct two useful expressions:

ST + TS = [  4  21 ]
          [ 21  40 ]

ST - TS = [ 0  -9 ]
          [ 9   0 ]

For any two symmetric matrices of the same order, the first expression is symmetric and the second is skew-symmetric. To prove this generally, transpose each expression and reverse the order within each product.

What happens when a skew-symmetric matrix is squared?

If K^T = -K, then:

(K^2)^T = K^T K^T
        = (-K)(-K)
        = K^2

So K squared is symmetric. For example:

K = [  0   3 ]
    [ -3   0 ]

K^2 = [ -9   0 ]
      [  0  -9 ]

A symmetric result can have negative entries. Symmetry describes matching positions, not the signs or sizes of the numbers.

A determinant result, with its condition attached

If a real skew-symmetric matrix has odd order, its determinant is zero. The proof uses det(K^T) = det(K) and det(-K) = (-1)^n det(K) for an n by n matrix:

K^T = -K
det(K) = det(K^T) = det(-K)
det(K) = (-1)^n det(K)

When n is odd, this says det(K) = -det(K), hence det(K) = 0. Brown University’s Special Matrices tutorial gives this determinant argument.

Every 3 by 3 real skew-symmetric matrix is therefore singular. But do not drop the words “odd order”. The 2 by 2 K in the previous example has determinant 0 - 3(-3) = 9, so it is invertible.

Even order does not guarantee invertibility either. A 2 by 2 zero matrix is an immediate counterexample. To connect the result with inverses, see what singular and non-singular matrices mean.

A short checking routine for written answers

Before moving to the next question, check these points:

1. Shape: Is the matrix square when the definition requires it?
2. Positions: Did each original row become a column in the transpose?
3. Pairs: Have you checked every pair across the main diagonal?
4. Diagonal: For skew-symmetry, is every diagonal entry zero?
5. Decomposition: Did you divide both parts by 2 and verify their sum?
6. Products: Did you reverse multiplication order when transposing?

For a classification question, finish with the relevant equality, such as A^T = -A. For decomposition, state the two parts and show that they reconstruct A. These closing lines explain why your answer works.

Practice questions with answers

Try each question before reading its answer. A short calculation that you do yourself is more useful than a long solution that only looks familiar.

1. Classify this matrix

A = [  0  -5 ]
    [ -5   0 ]

Answer: Symmetric only. The entries across the diagonal match, but they are not negatives of one another. A zero diagonal does not make this matrix skew-symmetric.

2. Find the missing entries

B = [  0   p    2 ]
    [ -4   q   -7 ]
    [  r   7    0 ]

B is skew-symmetric. Find p, q, and r.

Answer: p = 4, q = 0, and r = -2. Substitute them and check that every paired sum is zero.

3. Can any value of x make this symmetric?

C = [ 1   x   2 ]
    [ 5   3   4 ]
    [ 9   4   6 ]

Answer: No. Choosing x = 5 fixes one pair, but 2 and 9 still do not match. All required conditions must hold together.

4. Decompose a matrix with fractional parts

D = [ 3   6 ]
    [ 1  -2 ]

Answer:

S = [  3   7/2 ]
    [ 7/2  -2  ]

K = [  0    5/2 ]
    [ -5/2   0  ]

The upper-right entries add to 6, and the lower-left entries add to 1. The diagonal comes entirely from S.

5. Decompose this 3 by 3 matrix

E = [  4  -1   8 ]
    [  5   0   3 ]
    [ -2   7  -6 ]

Answer:

S = [ 4  2   3 ]
    [ 2  0   5 ]
    [ 3  5  -6 ]

K = [  0  -3   5 ]
    [  3   0  -2 ]
    [ -5   2   0 ]

For a quick reconstruction check, the lower-left entry is 3 + (-5) = -2, and the middle-right entry is 5 + (-2) = 3.

6. Decide whether a product is symmetric

S and T are symmetric, and ST != TS. Is ST symmetric?

Answer: No. Its transpose is TS, which the question says differs from ST. The condition for symmetry fails.

7. Classify an expression without multiplying it out

A is any real square matrix. Classify A^2 - (A^T)^2.

Answer: Skew-symmetric. Put B = A^2. Since B^T = (A^T)^2, the expression is B - B^T, whose transpose is its negative.

8. Check a determinant claim

A student says every skew-symmetric matrix has determinant zero. What is missing?

Answer: The statement is guaranteed for odd order. A 2 by 2 matrix with rows [0, 3] and [-3, 0] is skew-symmetric and has determinant 9, disproving the unrestricted claim.

Sources and further reading

• NCERT, Mathematics Part I, Matrices, reprint 2026-27: sections 3.5 and 3.6 for transpose rules, definitions, and decomposition.
• MIT OpenCourseWare, Gilbert Strang, Linear Algebra, Fall 2011: Transposes, Permutations, Vector Spaces for transpose operations and symmetric products.
• Brown University, Special Matrices tutorial: symmetric and skew-symmetric matrices for the odd-order determinant result and uniqueness of decomposition.

Frequently asked questions

What is the difference between symmetric and skew-symmetric matrices?

A symmetric matrix satisfies A^T = A, so entries across the main diagonal match. A skew-symmetric matrix satisfies A^T = -A, so those entries are negatives of each other and every diagonal entry is zero.

Can a rectangular matrix be symmetric or skew-symmetric?

No. Both definitions require a square matrix. A non-square matrix and its transpose have different orders and cannot be equal.

Why is the diagonal of a skew-symmetric matrix zero?

Each diagonal entry stays in the same position after transposing. Skew-symmetry requires that entry to equal its own negative. For real numbers, a = -a gives a = 0.

Does a zero diagonal prove that a matrix is skew-symmetric?

No. You must also check the off-diagonal pairs. Rows [0, 4] and [7, 0] give a zero diagonal, but 4 and 7 are not negatives of each other.

Can one matrix be both symmetric and skew-symmetric?

Yes, the square zero matrix. It is the only real matrix that satisfies both transpose conditions, because those conditions together give A = -A.

Are all diagonal matrices symmetric?

Yes. Transposing a diagonal matrix leaves it unchanged. A diagonal matrix is also skew-symmetric only when every diagonal entry is zero.

How do I express a matrix as a symmetric part plus a skew-symmetric part?

For a square matrix A, calculate S = (A + A^T)/2 and K = (A - A^T)/2. Then check S^T = S, K^T = -K, and S + K = A.

Is the decomposition unique?

Yes. From A = S + K, transposing gives A^T = S - K. Adding and subtracting these equations fixes the two parts, so there is no alternative pair.

What happens if A is already symmetric or already skew-symmetric?

If A is symmetric, its symmetric part is A and its skew-symmetric part is zero. If A is skew-symmetric, its symmetric part is zero and its skew-symmetric part is A.

Is the product of two symmetric matrices always symmetric?

No. For symmetric matrices S and T of the same order, ST is symmetric exactly when ST = TS. Always reverse the order when transposing a product.

Is the square of a skew-symmetric matrix also skew-symmetric?

Its square is always symmetric, because (K^2)^T = (-K)(-K) = K^2. For a real nonzero skew-symmetric K, its square is not skew-symmetric: the diagonal entry (K^2)ii equals the negative sum of the squares of row i, so at least one diagonal entry is nonzero.

Does every skew-symmetric matrix have determinant zero?

Every real skew-symmetric matrix of odd order does. In even order, the determinant may be zero or nonzero. The matrix with rows [0, 3] and [-3, 0] has determinant 9.