All eigenvalues of a hermitian matrix are real

Dependencies:

  1. Matrix
  2. Eigenvalues and Eigenvectors
  3. Transpose of product
  4. Matrix multiplication is associative

Let $A$ be an $n$ by $n$ hermitian matrix. Then all its eigenvalues are real.

Proof

Let $(\lambda, v)$ be an eigenvalue-eigenvector pair of $A$. \[ v^*v = \sum_{i=1}^n |v_i|^2 \ge 0 \] Since $v$ is an eigenvector, $v_i$ cannot be 0 for all $i$. Therefore, $v^*v > 0$.

\[ v^*Av = v^*(\lambda v) = \lambda v^*v \] \[ v^*Av = v^*A^*v = (Av)^*v = (\lambda v)^*v = \overline{\lambda}v^*v \] \[ \lambda v^*v = \overline{\lambda}v^*v \implies (\lambda - \overline{\lambda})v^*v = 0 \implies (\lambda - \overline{\lambda}) = 0 \implies \lambda = \overline{\lambda} \]

Since $\lambda = \overline{\lambda}$, $\lambda$ is real.

Dependency for:

  1. Bound on eigenvalues of sum of matrices
  2. Bounding matrix quadratic form using eigenvalues
  3. Positive definite iff eigenvalues are positive

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  1. /linear-algebra/vector-spaces/condition-for-subspace
  2. /linear-algebra/matrices/gauss-jordan-algo
  3. /sets-and-relations/composition-of-bijections-is-a-bijection
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