Symmetric Matrices

A matrix $A$ is symmetric if $A^T = A$ holds.

  • thus, $A$ must be a square matrix


What's special about $A \mathbf x = \lambda \mathbf x$? when $A$ is symmetric?

  • $A$ has real eigenvalues, and orthonormal eigenvectors
  • therefore Eigendecomposition of $A$ is $A = Q \Lambda Q^T$ instead of $A = S \Lambda S^{-1}$
  • this fact is sometimes referred as the Spectral Theorem

Orthogonal Eigenvectors

  • diagonalize $A = S \Lambda S^{-1}$
  • $A^T = (S \Lambda S^{-1})^T = (S^{-1})^T \Lambda^T S^T = (S^{-1})^T \Lambda \, S^T$
  • since $A = A^T$, we have:
  • $S \Lambda S^{-1} = (S^{-1})^T \Lambda \, S^T$
  • so maybe $S^{-1} = S^T$? Then $S$ is an Orthogonal Matrix
  • thus, the eigenvectors $\mathbf v_1 , \ ... \ , \mathbf v_n$ are orthonormal
  • So we write $A = Q \Lambda Q^T$

Let's show that

  • Suppose $A$ has two eigenvalues $\lambda_1 \ne \lambda_2$ and their eigenvalues are $\mathbf v_1$ and $\mathbf v_2$
  • so $A \mathbf v_1 = \lambda_1 \mathbf v_1$ and $A \mathbf v_2 = \lambda_2 \mathbf v_2$
  • $A \mathbf v_1 = \lambda_1 \mathbf v_1$ multiply by $\mathbf v_2$ on the left:
    • $(A \mathbf v_1)^T \mathbf v_2 = \lambda_1 \mathbf v_1^T \mathbf v_2$
    • $\mathbf v_1^T A^T \mathbf v_2 = \lambda_1 \mathbf v_1^T \mathbf v_2$
  • $A$ is symmetric, so
    • $\mathbf v_1^T A^T \mathbf v_2 = \mathbf v_1^T A \mathbf v_2 = \lambda_1 \mathbf v_1^T \mathbf v_2$
  • we have $\mathbf v_1^T A \mathbf v_2 = \lambda_1 \mathbf v_1^T \mathbf v_2$
    • since $A \mathbf v_2 = \lambda_2 \mathbf v_2$,
    • $\lambda_2 \mathbf v_1^T \mathbf v_2 = \lambda_1 \mathbf v_1^T \mathbf v_2$
  • but we assumed that $\lambda_1 \ne \lambda_2$!
    • so the only way it can be true is when $\mathbf v_1^T \mathbf v_2 = 0$
  • thus, $\mathbf v_1 \; \bot \; \mathbf v_2$


Real Eigenvalues

  • $A \mathbf x = \lambda \mathbf x$
  • let's take a Complex Conjugate: for $c = a + ib$ a conjugate is $\overline {c} = \overline{a + ib} = a - ib$
  • $A$ is real, so $\overline A = A$
  • thus, we have $A \overline {\mathbf x} = \overline {\lambda \mathbf x}$
  • so if $A$ has eigenvalue $\lambda$ and eigenvector $\mathbf x$, then $\overline \lambda$ and $\overline {\mathbf x}$ are also eigenvalue and eigenvector - for real matrices $A$
  • we want to show that $\overline \lambda = \lambda$ and $\overline {\mathbf x} = \mathbf x$, i.e. they are not complex

Let's show that:

  • Transpose $A \overline {\mathbf x} = \overline {\lambda \mathbf x}$:
    • $\overline {\mathbf x}^T A^T = \overline {\lambda \mathbf x}^T$
  • Since $A$ is symmetric, $\overline {\mathbf x}^T A = \overline {\lambda \mathbf x}^T$
    • Multiply both sides by $\mathbf x$ on the left:
    • $\overline {\mathbf x}^T A \, \mathbf x = \overline {\lambda \mathbf x}^T \mathbf x$
  • now take $A \mathbf x = \lambda \mathbf x$ and multiply both sides by $\overline {\mathbf x}^T$ on the right:
    • $\overline {\mathbf x}^T A \mathbf x = \overline {\mathbf x}^T \lambda \, \mathbf x$
  • so $\overline {\mathbf x}^T A \mathbf x = \overline {\lambda \mathbf x}^T \mathbf x$ and $\overline {\mathbf x}^T A \mathbf x = \overline {\mathbf x}^T \lambda \mathbf x$
    • they have the same right hand side
    • thus $\overline \lambda \overline {\mathbf x}^T \mathbf x = \lambda \overline {\mathbf x}^T \mathbf x$
    • given that $\overline {\mathbf x}^T \mathbf x \ne 0$, we divide by it and have:
  • $\overline \lambda = \lambda$, or, $\lambda \in \mathbb R$

Let's have a look at $\overline {\mathbf x}^T \mathbf x$ (when $\ne 0$):

  • $\Big[ \overline x_1 \ \overline x_2 \ \cdots \ \overline x_n \Big] \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \sum \overline x_i x_i$
  • for a complex number $c$, $c \cdot \overline c = (a - ib) \cdot (a + ib) = a^2 + b^2 \in \mathbb R$
  • so it's a sum of real numbers! sum of squared lengths of each component of $\mathbf x$
  • which means that the entire dot product is in $\mathbb R$


Eigenvalues are Non-Negative

  • Eigenvalues are non-negative
  • If $A$ is positive-definite, then all eigenvalues are positive


A symmetric matrix is positive-definite when

  • all eigenvalues of $A$ are greater than zero
  • so pivots are greater then zero as well


If $A$ is symmetric,

  • then its row space is the same as column space
  • i.e. $C(A) = C(A^T)$

Other properties

E.g. the identity matrix $I$: all eigenvalues $\lambda_i = 1$ and every vector is eigenvectors

Spectral Theorem

We can apply Eigendecomposition to $A$ and get

  • $A = Q \Lambda Q^T = \sum \lambda_i \mathbf q_i \mathbf q_i^T$ - sum of Outer Products
  • each of these outer products can be seen as a Projection Matrix
  • so symmetric matrix can be represented as a combination of mutually orthogonal projection matrices


Gram Matrices

$A A^T$ and $A^T A$ Symmetric

  • moreover, every symmetric matrix $B$ can be represented as $A A^T$ or $A^T A$:
  • Eigendecomposition of $B = Q \Lambda Q^T = Q \sqrt{\Lambda} \sqrt{\Lambda^T} Q^T = (Q \sqrt{\Lambda}) (Q \sqrt{\Lambda})^T = A A^T$ where $A = Q \sqrt{\Lambda}$
  • also Cholesky Decomposition would show the same
  • if $B$ is positive-definite, then such $A$ is non-singular


Identity and square diagonal matrices are symmetric



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