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Symmetric Matrices and the Spectral Theorem · Tutorial 721 of 1000

Symmetric Matrices and the Transpose Criterion

Learn to test whether a real square matrix is symmetric by comparing its transpose with the original matrix, or equivalently by checking its entries across the main diagonal.

Advanced 9 min read

What You'll Learn

  • Define a real symmetric matrix using the transpose criterion
  • Explain why symmetry is a property of square matrices
  • Translate the transpose criterion into an entry-by-entry test
  • Identify symmetric and nonsymmetric matrices in worked examples
  • Check parameterized matrices for the values that make them symmetric
  • Avoid confusing a transpose with an inverse

What Symmetry Means for a Matrix

Some matrices have a balanced pattern around their main diagonal: an entry above that diagonal is matched by an equal entry in the corresponding position below it. The transpose criterion gives a concise way to describe this pattern. It is a direct equality test, not a condition about eigenvalues, orthogonality, or whether a matrix can be inverted. In this tutorial all matrices are real, and symmetry will mean precisely equality with the transpose.

If \(\mathbf{A}\) is an \(m\times n\) matrix with entries \(a_{ij}\), its transpose \(\mathbf{A}^{\top}\) is the \(n\times m\) matrix whose entry in row \(i\), column \(j\) is \(a_{ji}\). Thus transposing a matrix exchanges its rows and columns. For instance, the first row of \(\mathbf{A}\) becomes the first column of \(\mathbf{A}^{\top}\). This definition also applies to square matrices, where the transpose has the same dimensions as the original.

Definition: A real square matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is symmetric if \(\mathbf{A}^{\top}=\mathbf{A}\). If this equality fails, \(\mathbf{A}\) is nonsymmetric.

The square-matrix condition matters. If \(\mathbf{A}\) has size \(m\times n\) with \(m\ne n\), then \(\mathbf{A}^{\top}\) has size \(n\times m\), so the two matrices do not even have matching dimensions. Under this definition, a nonsquare matrix is not called symmetric. A square matrix, by contrast, is a candidate for symmetry, but it must still pass the equality test.

The Entry-by-Entry Test

For an \(n\times n\) matrix, the entry in row \(i\), column \(j\) of \(\mathbf{A}^{\top}\) is \(a_{ji}\). Equality of two matrices means equality of their corresponding entries. Therefore the transpose criterion can be checked one pair of positions at a time: compare the entry in row \(i\), column \(j\) with the entry in row \(j\), column \(i\). Entries on the main diagonal are compared with themselves, so they do not create an additional restriction.

Theorem (Entry criterion for symmetry): A real matrix \(\mathbf{A}=[a_{ij}]\in\mathbb{R}^{n\times n}\) is symmetric if and only if \(a_{ij}=a_{ji}\) for every pair \(i,j\in\{1,\dots,n\}\).

Proof. First suppose \(\mathbf{A}\) is symmetric, so \(\mathbf{A}^{\top}=\mathbf{A}\). For any \(i,j\), the entry in row \(i\), column \(j\) of \(\mathbf{A}^{\top}\) is \(a_{ji}\). Since the corresponding entry of \(\mathbf{A}\) is \(a_{ij}\), equality of the matrices gives \(a_{ji}=a_{ij}\). This holds for every pair \(i,j\). Conversely, suppose \(a_{ij}=a_{ji}\) for every pair \(i,j\). The entry in row \(i\), column \(j\) of \(\mathbf{A}^{\top}\) is then \(a_{ji}=a_{ij}\), the entry in the same position of \(\mathbf{A}\). Thus all corresponding entries agree, so \(\mathbf{A}^{\top}=\mathbf{A}\) and \(\mathbf{A}\) is symmetric. \(\square\)

This is an if-and-only-if test: symmetry implies the entry equalities, and the entry equalities imply symmetry. Practically, it is enough to inspect one side of the main diagonal. The entries on the other side must match in reflected positions, while every diagonal entry is allowed to be any real number. For a \(3\times3\) matrix, for example, the checks are \(a_{12}=a_{21}\), \(a_{13}=a_{31}\), and \(a_{23}=a_{32}\).

Worked Example: Checking a Symmetric Matrix

Determine whether the following matrix is symmetric by computing its transpose and by checking the entries across the main diagonal.

$$ \mathbf{A}= \begin{bmatrix} 2&-1&4\\ -1&0&3\\ 4&3&5 \end{bmatrix}. $$

Interchange the rows and columns to calculate the transpose. The first row \((2,-1,4)\) becomes the first column; the second row \((-1,0,3)\) becomes the second column; and the third row \((4,3,5)\) becomes the third column. Hence

$$ \mathbf{A}^{\top}= \begin{bmatrix} 2&-1&4\\ -1&0&3\\ 4&3&5 \end{bmatrix} =\mathbf{A}. $$

The entry test gives the same conclusion. The off-diagonal pairs match: \(a_{12}=-1=a_{21}\), \(a_{13}=4=a_{31}\), and \(a_{23}=3=a_{32}\). The three diagonal entries \(2,0,5\) each occupy the same position when transposed. The criterion therefore confirms that \(\mathbf{A}\) is symmetric.

Check: The displayed transpose agrees with the original matrix in all nine positions, including the three diagonal positions. Thus \(\mathbf{A}^{\top}-\mathbf{A}=\mathbf{O}\), as required by the transpose criterion.

Finding the First Mismatch

A matrix is nonsymmetric as soon as one pair of reflected entries fails to match. It is not necessary to compare every entry once such a mismatch has been found. The most reliable approach is to keep the indices explicit: for each off-diagonal position \((i,j)\), compare \(a_{ij}\) with \(a_{ji}\). Looking only at whether the matrix “looks balanced” can lead to mistakes, especially when the entries are negative or the matrix is larger than \(2\times2\).

Worked Example: A Nonsymmetric Matrix

Test the matrix below for symmetry. Show the transpose and identify a specific failure of the entry criterion.

$$ \mathbf{B}= \begin{bmatrix} 1&2&-3\\ 2&4&0\\ -1&0&6 \end{bmatrix}. $$

Transposing exchanges rows and columns, giving

$$ \mathbf{B}^{\top}= \begin{bmatrix} 1&2&-1\\ 2&4&0\\ -3&0&6 \end{bmatrix}. $$

The matrices are not equal: for instance, the entry in row 1, column 3 is \(-3\) in \(\mathbf{B}\), while it is \(-1\) in \(\mathbf{B}^{\top}\). Equivalently, the reflected entries in \(\mathbf{B}\) satisfy \(b_{13}=-3\ne -1=b_{31}\). One mismatch is enough to conclude that \(\mathbf{B}\) is nonsymmetric.

Check: The other off-diagonal pairs are \(b_{12}=2=b_{21}\) and \(b_{23}=0=b_{32}\). These agreements do not repair the unequal pair \(b_{13}\ne b_{31}\); the entry criterion requires equality for every pair.

Symmetry with an Unknown Entry

The same criterion can determine an unknown matrix entry. The unknown must be chosen so that every reflected pair agrees. This is a small system of equality conditions, not a process of row-reducing the matrix. If more than one pair contains unknowns, each pair contributes its own condition, and all of those conditions must hold at once.

Worked Example: Choosing a Parameter for Symmetry

For which real values of \(t\) is the following matrix symmetric?

$$ \mathbf{C}= \begin{bmatrix} 0&t+1&2\\ 3&5&-4\\ 2&-4&t \end{bmatrix}. $$

The pair in positions \((1,2)\) and \((2,1)\) requires \(t+1=3\), so \(t=2\). The pair in positions \((1,3)\) and \((3,1)\) already agrees, since both entries are \(2\). The pair in positions \((2,3)\) and \((3,2)\) also agrees, since both entries are \(-4\). Thus the only condition is \(t=2\).

Substituting \(t=2\) gives

$$ \mathbf{C}= \begin{bmatrix} 0&3&2\\ 3&5&-4\\ 2&-4&2 \end{bmatrix}, \qquad \mathbf{C}^{\top}= \begin{bmatrix} 0&3&2\\ 3&5&-4\\ 2&-4&2 \end{bmatrix}. $$

Check: The three off-diagonal comparisons are \(3=3\), \(2=2\), and \(-4=-4\). Therefore the substituted matrix is symmetric. If \(t\ne2\), then \(t+1\ne3\), so the pair in positions \((1,2)\) and \((2,1)\) fails; no other choice of \(t\) works.

Two Basic Consequences of the Criterion

The entry test also makes it possible to recognize simple families without repeatedly writing out a transpose. For a diagonal matrix, every off-diagonal entry is zero, so each entry is matched by the zero in its reflected position. More generally, sums and real scalar multiples of symmetric matrices remain symmetric. These consequences follow directly from the defining equality and are often useful when checking a matrix assembled from simpler pieces.

Theorem: If \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\) are symmetric and \(c\in\mathbb{R}\), then \(\mathbf{A}+\mathbf{B}\) and \(c\mathbf{A}\) are symmetric.

Proof. Since \(\mathbf{A}\) and \(\mathbf{B}\) are symmetric, \(\mathbf{A}^{\top}=\mathbf{A}\) and \(\mathbf{B}^{\top}=\mathbf{B}\). Transposition exchanges entries position by position, so the transpose of a sum is the sum of the transposes, and the transpose of a real scalar multiple is the same scalar times the transpose. Therefore

$$ (\mathbf{A}+\mathbf{B})^{\top} =\mathbf{A}^{\top}+\mathbf{B}^{\top} =\mathbf{A}+\mathbf{B}, \qquad (c\mathbf{A})^{\top} =c\mathbf{A}^{\top} =c\mathbf{A}. $$

Each matrix equals its transpose, so both are symmetric by definition. \(\square\)

This result concerns sums and scalar multiples; it does not say that every operation on symmetric matrices preserves symmetry. In particular, do not assume a product of two symmetric matrices is symmetric without checking it. As a concrete counterexample, take

$$ \mathbf{A}= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, \qquad \mathbf{B}= \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix}. $$

Both matrices are symmetric, but their products in the two orders are

$$ \mathbf{A}\mathbf{B}= \begin{bmatrix} 0&1\\ 0&0 \end{bmatrix}, \qquad (\mathbf{A}\mathbf{B})^{\top}= \begin{bmatrix} 0&0\\ 1&0 \end{bmatrix}. $$

These matrices differ, so \(\mathbf{A}\mathbf{B}\) is not symmetric. This calculation is a useful warning: the definition always asks whether a particular matrix equals its own transpose, not whether its ingredients happen to be symmetric.

Common Mistake: Confusing Transpose and Inverse

A transpose is formed by exchanging rows and columns. An inverse, when it exists, is a different matrix defined by multiplication with the original to produce the identity. Symmetry requires \(\mathbf{A}^{\top}=\mathbf{A}\), not \(\mathbf{A}^{-1}=\mathbf{A}\). For example, the matrix \(\begin{bmatrix}2&1\\1&3\end{bmatrix}\) is symmetric because its off-diagonal entries match, even though it is not its own inverse.

A second common error is to compare an entry with the wrong partner. The reflection of \(a_{ij}\) is \(a_{ji}\), not an entry chosen from the same row or the next column. For a \(3\times3\) matrix, \(a_{13}\) must be compared with \(a_{31}\). In the nonsymmetric example above, checking \(b_{13}\) against \(b_{12}\) would give \(-3\ne2\), but that is not the criterion; the relevant comparison is \(b_{13}\) against \(b_{31}\). A correct test should state the indices being compared.

A practical checklist is therefore:

  1. Confirm that the matrix is square.
  2. Write or mentally form the transpose by exchanging rows and columns.
  3. Compare reflected off-diagonal entries \(a_{ij}\) and \(a_{ji}\).
  4. Conclude symmetry only if every required pair agrees.

The test is exact and finite. No special diagonal values are required, and no condition is imposed on an off-diagonal pair beyond equality. If even one reflected pair differs, the matrix fails the definition.

Check Your Understanding

For each question, use the transpose criterion or the entry-by-entry test, and support your conclusion with the relevant comparisons.

  1. Is \(\begin{bmatrix}4&-2\\-2&7\end{bmatrix}\) symmetric? State its transpose.
  2. Identify one pair of reflected entries that proves \(\begin{bmatrix}1&0&5\\0&2&3\\5&-3&4\end{bmatrix}\) is not symmetric.
  3. Find all real \(t\) for which \(\begin{bmatrix}1&2&t\\2&0&-1\\t&-1&6\end{bmatrix}\) is symmetric.
  4. Explain why a \(2\times3\) matrix does not satisfy the definition of a symmetric matrix used here.
  5. If \(\mathbf{A}\) and \(\mathbf{B}\) are symmetric real square matrices of the same size, why is \(\mathbf{A}+\mathbf{B}\) symmetric?

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