A matrix equal to its own transpose is the best behaved of all square matrices. Its eigenvalues are real, and eigenvectors for different eigenvalues are at right angles, so it stretches space along perpendicular directions and never turns or shears it.

Pulling a square sheet of rubber from its edges stretches it along two directions at right angles, and a thread drawn along either of them only lengthens. A symmetric matrix is a pull of that kind, in any number of dimensions.

The symmetric matrix $A=\begin{pmatrix}2&2\\2&5\end{pmatrix}$ has eigenvalues $6$ and $1$, with eigenvectors $\mathbf{u}=(1,2)$ and $\mathbf{v}=(2,-1)$. The two lines of eigenvectors cross at a right angle, $\mathbf{u}\cdot\mathbf{v}=2-2=0$: the map stretches by $6$ along $\mathbf{u}$ and leaves the line of $\mathbf{v}$ as it is.

A square matrix is symmetric when $A^{T}=A$. Then the matrix can be moved across a dot product, since $(A\mathbf{x})\cdot\mathbf{y}=\mathbf{x}^{T}A^{T}\mathbf{y}=\mathbf{x}^{T}A\mathbf{y}=\mathbf{x}\cdot(A\mathbf{y})$. Two facts follow. Eigenvectors for different eigenvalues are orthogonal, the theorem below. And every eigenvalue is real: that is proved below for $2\times2$ matrices, while for larger ones the proof needs complex numbers, which this course does not use, so it is stated here and used.

• For $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ the characteristic polynomial $\lambda^2-(a+d)\lambda+(ad-b^2)$ has discriminant $(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2\ge0$, so both eigenvalues are real.
• An $n\times n$ symmetric matrix has $n$ real eigenvalues, counted with multiplicity. This is the fact taken on trust.
• $A^{T}A$ is symmetric for every $A$, and its eigenvalues are never negative: $A^{T}A\mathbf{v}=\lambda\mathbf{v}$ gives .
• Without symmetry both facts fail: the quarter turn has no real eigenvalue, and $\begin{pmatrix}2&1\\0&3\end{pmatrix}$ has the eigenvectors $(1,0)$ and , which are not orthogonal.

Move $A$ across the dot product: $(A\mathbf{u})\cdot\mathbf{v}=\mathbf{u}\cdot(A\mathbf{v})$. The left side is $\lambda\,\mathbf{u}\cdot\mathbf{v}$ and the right side $\mu\,\mathbf{u}\cdot\mathbf{v}$. Their difference $(\lambda-\mu)\,\mathbf{u}\cdot\mathbf{v}$ is zero, and $\lambda-\mu$ is not, so $\mathbf{u}\cdot\mathbf{v}=0$.

A symmetric matrix can be moved across the dot product.

Replace $A\mathbf{u}$ by $\lambda\mathbf{u}$ and $A\mathbf{v}$ by $\mu\mathbf{v}$.

Subtract one side from the other.

$\lambda-\mu\neq0$, so the dot product must vanish.

For a symmetric $2\times2$ matrix the discriminant of the characteristic polynomial is a sum of squares, so it is never negative and both eigenvalues are real.

For a symmetric 2 by 2 matrix with two different eigenvalues, the eigenvectors of one are at right angles to those of the other. One eigenvector is enough to find the other line.

• Symmetric: $A^{T}=A$, so $(A\mathbf{x})\cdot\mathbf{y}=\mathbf{x}\cdot(A\mathbf{y})$.
• Its eigenvalues are real: proved for $2\times2$, taken on trust beyond.
• Eigenvectors for different eigenvalues are orthogonal.
• $A^{T}A$ is symmetric with eigenvalues never negative.

Great work! You now know how to:

• recognise a symmetric matrix and move it across a dot product;
• show that a symmetric 2 by 2 matrix has real eigenvalues;
• prove that eigenvectors for different eigenvalues are orthogonal;
• use one eigenvector to find the other in the plane.