Example · Undergraduate
Terms
| Term | This example | Why |
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| eigenvalue |
None, since the matrix is not square
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| singular value |
$\sqrt{14}$ and $1$
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The singular values of \(A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) are $\sqrt{14}$ and $1$, the square roots of the eigenvalues of \(A A^{\top} = \begin{bmatrix} 5 & 6 \\ 6 & 10 \end{bmatrix}\). That matrix has trace $15$ and determinant $50 - 36 = 14$, so its eigenvalues are the roots of $\lambda^2 - 15 \lambda + 14 = (\lambda - 14)(\lambda - 1)$. The $3 \times 3$ matrix $A^{\top} A$ has the same nonzero eigenvalues, plus $0$.
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| full rank |
✓ |
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Distinctions using this example
- Eigenvalue vs. singular value: The $2 \times 3$ matrix has no eigenvalues, because only square matrices have them. Its singular values are $\sqrt{14}$ and $1$, the square roots of the eigenvalues $14$ and $1$ of $A A^{\top}$ (which are the nonzero eigenvalues of $A^{\top} A$).
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