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When we calculate singular values in Singular value decomposition we use the common eigenvalues (positive square roots) of $A^TA$ or $AA^T$, where $A$ is an $m\times n$ real matrix. We know that singular values of A may be repeated. Now I am trying to understand in which situation we will get repeated singular values of $A$.

I am thinking like that as we know $A^TA$ and $AA^T$ are both symmetric matrix, the singular values of $A$ will repeat when eigenvalues of $A^TA$ and $AA^T$ will repeat, i.e. symmetric matrix has repeated eigenvalues. But I did not find out when. Can any one help me to tell when eigenvalues of a symmetric matrix will repeat and also some example?

Thanks a lot.

dantopa
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P Pradhan
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  • All singular values of the identity matrix are the same... For a more general case, consider a diagonal matrix that is not a multiple of the identity. – lhf Aug 26 '15 at 10:06
  • The symmetric matrix $\left(\begin{smallmatrix}1&-1\-1&1\end{smallmatrix}\right)$ doesn't have repeated eigenvalues. – Arthur Aug 26 '15 at 10:08
  • The title does not reflect the question asked at the end. – lhf Aug 26 '15 at 10:12
  • Any real symmetric matrix can have repeated eigenvalues. However, if you are computing the eigenvalues of a symmetric matrix (without any special structure or properties), do not expect repeated eigenvalues. Due to floating-point errors in computation, there won't be any repeated eigenvalues. However, there could be clusters of nearly equal eigenvalues. – Vini Mar 12 '17 at 10:33

1 Answers1

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"Can any one help me to tell when eigenvalues of a symmetric matrix will repeat and also some example?"

The characteristic equation will have a form like $$ p ( \lambda ) = \prod_{k=1}^{n} \left( \lambda - \xi_{k} \right)^{m_{k}} $$ where $n$ is the number of distinct eigenvalues $\xi$, and $m_{k}$ is the algebraic multiplicity of each eigenvalue.

For the $2 \times 2$ matrix, the eigenvalue $\xi$ will be repeated twice: $$ p(\lambda) = \left( \lambda - \xi \right)^{2} = \lambda^{2} - 2 \lambda \xi + \xi^{2} $$ To construct matrices $\mathbf{A}$ with repeated eigenvalues, we can exploit the expression for the characteristic in terms of the trace and determinant: $$ p(\lambda) = \lambda^{2} - \lambda \text{ tr}\left(\mathbf{A}\right) + \det \left( \mathbf{A} \right) $$ Given $$ \mathbf{A} = \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right], $$ control the eigenvalues via $$ \begin{align} \text{ tr} \left( \mathbf{A} \right) &= a+d = 2\xi ,\\ \det \left( \mathbf{A} \right) &= ad-bc = \xi^{2}. \end{align} $$ Simplest examples are $$ \left[ \begin{array}{cc} \xi & 0 \\ \alpha & \xi \\ \end{array} \right], \quad \left[ \begin{array}{cc} 0 & -\xi \\ \xi & 2\xi \\ \end{array} \right] $$ with $\alpha$ an arbitrary complex constant.

For the symmetric matrix $$ \left[ \begin{array}{cc} a & b \\ b & c \\ \end{array} \right] $$ the constraints are $$ \begin{align} \text{ tr} \left( \mathbf{A} \right) &= a+c = \xi ,\\ \det \left( \mathbf{A} \right) &= ac-b^{2} = \xi^{2}. \end{align} $$ Examples of include $$ \left[ \begin{array}{cc} 0 & 1 \\ 1 & 2 i \\ \end{array} \right] $$ which repeats the eigenvalue $i=\sqrt{-1}$.

dantopa
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