Basic Properties

Necessary and sufficient condition

An $n\times n$ matrix $A$ is symmetric if and only if $A\boldsymbol{x}\cdot\boldsymbol{y}=\boldsymbol{x}\cdot A\boldsymbol{y}$ for all vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ in $\mathbb{R}^n$.

Theorem-3: An $n\times n$ matrix $A$ is symmetric if and only if $A\boldsymbol{x}\cdot\boldsymbol{y}=\boldsymbol{x}\cdot A\boldsymbol{y}$ for all vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ in $\mathbb{R}^n$ [5].

Proof:

(1) Necessity (If $A$ is symmetric): Let $\boldsymbol{x}$ and $\boldsymbol{y}$ be in $\mathbb{R}^n$. For any $n\times n$ matrix $A$:

(2) Sufficiency (If $A\boldsymbol{x}\cdot\boldsymbol{y}=\boldsymbol{x}\cdot A\boldsymbol{y}$ is valid for all vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ in $\mathbb{R}^n$): Let $\boldsymbol{a}_i$ denote $i$-th column of $A$, $e_i$ is the $i$-th base vector of $\boldsymbol{R}^n$. Then for all $1\le i,j\le n$:

i.e.,

Q.E.D.

We could simply verify its necessity using MATLAB:

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rng("default")
A = [1,3;3,3];
x = randn(2,1);
y = randn(2,1);

dot(A*x,y) % (A*x)'*y
dot(x,A*y) % x'*(A*y)

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ans =
   -7.5078
ans =
   -7.5078

Eigenvalues and eigenvectors of real symmetric matrix

All eigenvalues of real symmetric matrix are real.

Theorem-1: All eigenvalues of real symmetric matrix are real (could refer [3, 4]).

Corollary-1-1: All coefficients of the characteristic polynomial of real symmetric matrix are real.

Corollary-1-2: The trace of real symmetric matrix is real.

Corollary-1-3: The determinant of real symmetric matrix is real.

If $A$ is a real $n\times n$ symmetric matrix, the $A$ has $n$ real eigenvalues (counted by their multiplicities). And for each eigenvalue, we can find a real eigenvector associated with it.

Theorem-4: If $A$ is a real $n\times n$ symmetric matrix, the $A$ has $n$ real eigenvalues (counted by their multiplicities). And for each eigenvalue, we can find a real eigenvector associated with it [5].

Proof: Theorem-1 says that all the eigenvalues of real symmetric matrix are real (p.s., reference [5] provides another way to prove Theorem-1, and is kind of complicated than proof in blog [4]). And on another hand, fundamental theorem of algebra [6, 7] tell us “every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n $complex roots”. So, the $n$-order characteristic polynomial $\vert A-\lambda I\vert=0$ has $n$ real roots, i.e., $A$ has $n$ real eigenvalues (counted by their multiplicities) .

Since each $\lambda_i$ is real, $A-\lambda_iI$ is a real matrix and $\vert A-\lambda_iI\vert=0$ cause $\lambda_i$ is an eigenvalue. So, the real matrix equation

has nonzero real solutions $\boldsymbol{x}$ [8], i.e., there are real eigenvectors for eigenvalue $\lambda_i$.

Eigenvectors corresponding to different eigenvalues of a real symmetric matrix are orthogonal.

Theorem-2: Eigenvectors corresponding to different eigenvalues of a real symmetric matrix are orthogonal [1, 2].

Proof: Supposed that $x_1$ and $x_2$ is two eigenvectors of a real symmetric matrix, and corresponding different eigenvalues are $\lambda_1$ and $\lambda_2$, i.e., \(\begin{split} Ax_1=\lambda_1x_1\\ Ax_2=\lambda_2x_2\\ \end{split}\)

where

then we have:

due to that $x_2^Tx_1$ is a scalar, therefore $x_2^TAx_1$ is a scalar as well, so we have:

N.B., If $a$ is a scalar, $a^T=a$ always holds.

According $\eqref{eq1}$ and $\eqref{eq2}$, we have:

Cause $\lambda_1\ne\lambda_2$, so we have:

that is $x_2$ and $x_1$ is orthogonal.

Q.E.D.

References

[1] 对称矩阵 特征向量正交 - 知乎.

[2] 46 Symmetric Matrices.

[3] Some Properties of Eigenvalues and Eigenvectors - What a starry night~.

[4] Hermitian Matrix - What a starry night~.

[5] Orthogonally Diagonalizable Matrices.

[6] Fundamental theorem of algebra - Wikipedia.

[7] Fundamental Theorem of Algebra - What a starry night~.

[8] Determination of Solutions for Linear Equations - What a starry night~.