Symmetric matrices

A symmetric matrix is a square matrix that is equal to its own transpose. In other words, if $A$ is a symmetric matrix, then we have:

$$A = A^T.$$

This means that the entry in row $i$ and column $j$ is the same as the entry in row $j$ and column $i$, for all $i$ and $j$. That is, $A_{ij} = A_{ji}$.

For example, a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is symmetric if $b = c$. So, a symmetric $2 \times 2$ matrix looks like:

$$\begin{bmatrix} a & b \\ b & d \end{bmatrix}.$$

Similarly, a $3 \times 3$ matrix $A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is symmetric if $b=d$, $c=g$, and $f=h$. It looks like:

$$\begin{bmatrix} a & b & c \\ b & e & f \\ c & f & i \end{bmatrix}.$$

The elements along the main diagonal (from top-left to bottom-right) can have any value, but the elements mirrored across this diagonal must be identical.

Determine which of the following matrices are symmetric:

$$A = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad C = \begin{bmatrix} 5 & 1 & 7 \\ 1 & 8 & -2 \\ 7 & -2 & 0 \end{bmatrix}.$$

Solution

Let's check each matrix in turn:

$A^T = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} = A$, so $A$ is symmetric.

$B^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \neq B$, so $B$ is not symmetric.

$C^T = C$, so $C$ is symmetric.

Symmetric matrices have two fundamental properties:

1. All eigenvalues are real numbers.

If $A$ is symmetric, then every solution $\lambda$ of $\det(A - \lambda I) = 0$ is a real number. This means we never have to deal with complex eigenvalues when working with symmetric matrices.

2. The eigenvectors can be chosen to be orthonormal.

If $A$ is a symmetric matrix with two different eigenvalues $\lambda_1 \neq \lambda_2$, and corresponding eigenvectors $\vec{v}_1$ and $\vec{v}_2$, then these eigenvectors are guaranteed to be orthogonal:

$$\vec{v}_1 \cdot \vec{v}_2 = 0.$$

This means that eigenvectors associated with distinct eigenvalues always point in perpendicular directions. Even when eigenvalues are repeated, it is still possible to select eigenvectors that are mutually orthogonal.

Finally, by normalising each eigenvector to have length $1$, we can obtain an orthonormal set of eigenvectors $\{\hat{\vec{v}}_1, \hat{\vec{v}}_2, \dots\}$ satisfying

$$\hat{\vec{v}}_i \cdot \hat{\vec{v}}_j = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \neq j. \end{cases}$$

In other words, every eigenvector in the set has unit length, and any two different eigenvectors are perpendicular (i.e. their dot product is $0$). Note we use the "hat" notation $\hat{\vec{v}}$ above a vector $\vec{v}$ to denote that it's a unit vector.

Together, these properties make symmetric matrices especially well-behaved. Their eigenvectors form a clean, perpendicular coordinate system, and each eigenvector points along a direction that the matrix simply stretches or compresses.

Because the eigenvectors are orthogonal, these directions do not overlap or interfere with each other, which is why symmetric matrices are so useful in practice.

Visualisation

Have a play with the visualisation below.

Consider the symmetric matrix

$$A = \begin{bmatrix} 4 & 2 \\ 2 & 4 \end{bmatrix}.$$

The eigenvalues of $A$ are $\lambda_1 = 6$ and $\lambda_2 = 2$, with corresponding eigenvectors

$$\vec{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$

1. Check that $\vec{v}_1$ and $\vec{v}_2$ are orthogonal.
2. Normalise them to form an orthonormal set of eigenvectors $\{\hat{\vec{u}}_1, \hat{\vec{u}}_2 \}$.

Solution

1. Check orthogonality

$$\vec{v}_1 \cdot \vec{v}_2 = (1)(-1) + (1)(1) = -1 + 1 = 0.$$

So the eigenvectors are orthogonal.

2. Normalise each eigenvector

The length of $\vec{v}_1$ is:

$$|\vec{v}_1| = \sqrt{1^2 + 1^2} = \sqrt{2}.$$

So the unit vector, which we'll call $\hat{\vec{u}}_1$, is given by:

$$\hat{\vec{u}}_1 = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

The length of $\vec{v}_2$ is:

$$|\vec{v}_2| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}.$$

So the unit vector, which we'll call $\hat{\vec{u}}_2$, is:

$$\hat{\vec{u}}_2 = \dfrac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \end{bmatrix}.$$

The orthonormal set of eigenvectors, $\{\hat{\vec{u}}_1, \hat{\vec{u}}_2 \}$, of $A$ is therefore:

$$\left\{ \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \dfrac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \end{bmatrix}\right\}.$$

A key property of symmetric matrices is that they can always be written in a special form using their eigenvectors and eigenvalues. Specifically, if $A$ is a symmetric matrix, then there exists an orthogonal matrix $Q$ and a diagonal matrix $D$ such that:

$$A = Q D Q^T.$$

Here:

• $Q$ is an orthogonal matrix, meaning its columns are orthonormal eigenvectors of $A$ (each has length $1$, and they are perpendicular to each other).
• $D$ is a diagonal matrix, which just means all the non‑zero entries are on the main diagonal (from top-left to bottom-right). Those diagonal entries are the eigenvalues of $A$.

This way of writing $A$ is called orthogonal diagonalisation. It tells us that every symmetric matrix can be “simplified” into a diagonal form by choosing the right orthonormal eigenvectors.

Consider the symmetric matrix:

$$A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}.$$

This matrix has eigenvalues $\lambda_1 = 3$, $\lambda_2 = 1$, with corresponding orthonormal eigenvectors:

$$\hat{\vec{v}}_1 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 \\ 1\end{bmatrix}, \quad \hat{\vec{v}}_2 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 \\ -1\end{bmatrix}.$$

Verify that $A$ can be written in the form:

$$A = Q D Q^T.$$

where $Q$ is an orthogonal matrix and $D$ is a diagonal matrix.

Solution

We form $Q$ by putting the eigenvectors as the columns:

$$Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[6pt] \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix},$$

and $D$ by putting the corresponding eigenvalues on the diagonal:

$$D = \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix}.$$

We need to check that $A = Q D Q^T$ holds:

$$\begin{align*} Q D Q^T &= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[6pt] \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[6pt] \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \\ &= \begin{bmatrix} \frac{3}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[6pt] \frac{3}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\[6pt] \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \\ &= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \\ &= A. \end{align*}$$

And so we have verified that $A$ can indeed be written in the form $A = Q D Q^T$. This is an example of orthogonal diagonalisation.