Proof: Eigen value of real symmetric matrix

Question

If $A$ is the real symmetric matrix then all of it's eigen values are real. prove this ?

Answer 1

proof for:

If $A$ is the real symmetric matrix then all of it's eigen values are real.

$A$ is symmetric matrix:

$A=A^{T}$      ....................(1)

Let $v$ be the eigen vector corresponding to eigen value $\lambda$.

$Av = \lambda v$      ....................(2)

taking conjugate transpose of the both sides of equation (2),

$(\overline{Av})^{T} = (\overline{\lambda v})^{T}$  or  $\overline{(Av)^{T}} = \overline{(\lambda v)^{T}}$  (both are same)

$(\overline{A} \cdot \overline{v})^{T} = (\overline{\lambda} \cdot \overline{v})^{T}$

$(A \cdot \overline{v})^{T} = (\overline{\lambda} \cdot \overline{v})^{T}$  (for any matrix with only real entries, $\overline{A} = A$)

$(\overline{v})^{T} \cdot A^{T} = \overline{\lambda} \cdot (\overline{v})^{T}$

$(\overline{v})^{T} \cdot A =\overline{\lambda} \cdot (\overline{v})^{T}$  (from equation (1))

let's represent $(\overline{v})^{T}$ from $v^{*}$,

$v^{*} A = \overline{\lambda} v^{*}$      ....................(3)

post-multiply $v$ in both sides of equation (3),

$v^{*} A v = \overline{\lambda} v^{*} v$      ....................(4)

pre-multiply $v^{*}$ in both sides of equation (2),

$v^{*}Av = v^{*} \lambda v$

$v^{*}Av = \lambda v^{*} v$      ....................(5)

from equation (4) and (5),

$\overline{\lambda} v^{*} v =  \lambda v^{*} v$

$\overline{\lambda} v^{*} v - \lambda v^{*} v = 0$

$\overline{\lambda} v^{*} v - \lambda v^{*} v = 0$

$( \overline{\lambda} - \lambda ) v^{*} v = 0$      ....................(6)

now, a slight detour because $v^{*} v = ?$

let's take an example, $v = \begin{bmatrix} x+iy \\ p+iq \\a+ib \end{bmatrix}$ then,  $v^{*} = \begin{bmatrix} x-iy & p-iq & a-ib \end{bmatrix}$

so, $v^{*} v = \begin{bmatrix} x-iy & p-iq & a-ib \end{bmatrix} \cdot \begin{bmatrix} x+iy \\ p+iq \\a+ib \end{bmatrix}$

$v^{*} v = x^{2} + y^{2} + p^{2} + q^{2} + a^{2} + b^{2}$

here, $v^{*}v$ can't be zero because $v^{*}v$ can be zero in only one case where all $x, y, p, q, a$ and $b$ are all zeros and in that case $v$ itself will become zero but that is contradicting with the definition of an eigen vector.

hence, $v^{*} v > 0$      ....................(7)

So, from equation (6) and (7),

$\overline{\lambda} - \lambda = 0$

$\overline{\lambda} = \lambda$

that means $\lambda$ is a real number.

Reference: Conjugate Transpose of a Matrix

Answer 2

Every real symmetric matrix A has an associated quadratic form Q(x) =x Transpose A x . Where x is the eigen vector corresponding to an eigen value lambda . We substitute Ax = lambda x then we will get Q(x) = x transpose lamda x . Now lambda is a scalar so Q(x) = lambda x transpose x . Eigen vectors of a real symmetric matrix are LI and orthonormal . This means x transpose x is 1 . Now Q(x) = lambda . For a real symmetric matrix its quadratric form will always be >=  0 for all values of x . Therefore Q(x)>=0 . This means lambda has to be >=0.  Therefore eigen values have to be real .