Eigenvalue vs. singular value
An eigenvalue of a square matrix $A$ is a factor $\lambda$ by which $A$ scales a particular nonzero vector $x$, called a eigenvalue: $Ax = \lambda x$. A singular value of any matrix $A$ is a factor $\sigma$ by which $A$ stretches vectors along some direction, allowing it them turn. Singular values are real and nonnegative, but eigenvalues can be negative or complex.
Definitions
Consider a square matrix $A$, which stretches and turns vectors when it multiplies them. An eigenvalue of $A$ is a number $\lambda$ such that $A$ stretches some nonzero vector $v$ by the factor $\lambda$ without turning it off its line: $A v = \lambda v$.
Consider a square matrix $A$. An eigenvalue of $A$ is a number $\lambda$ such that $A v = \lambda v$ for some nonzero vector $v$, called an eigenvector.
Examples
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The matrix \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)
: $i$ and $-i$
The eigenvalues of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) are $i$ and $-i$, because \(A \begin{bmatrix} 1 \\ -i \end{bmatrix} = \begin{bmatrix} i \\ 1 \end{bmatrix} = i \begin{bmatrix} 1 \\ -i \end{bmatrix}\) and \(A \begin{bmatrix} 1 \\ i \end{bmatrix} = \begin{bmatrix} -i \\ 1 \end{bmatrix} = -i \begin{bmatrix} 1 \\ i \end{bmatrix}\). $A$ rotates every nonzero real vector by $90°$, so it has no real eigenvalues.
-
The matrix \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\)
: None, since the matrix is not square
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 $A^{\top} A$).
-
The matrix \(\begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}\)
: $2$ and $-3$
The eigenvalues are $2$ and $-3$, because \(\begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ -3 \end{bmatrix} = -3 \begin{bmatrix} 0 \\ 1 \end{bmatrix}\).
-
The matrix \(\begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\)
: $2$ (twice)
The eigenvalue of \(A = \begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix}\) is $2$, because \(A \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix} = 2 \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).
-
The matrix \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)
: $1$ (twice)
The shear \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has the eigenvalue $1$ twice, and it fixes the vector $(1, 0)$. Its singular values are $\varphi \approx 1.618$ and $1 / \varphi \approx 0.618$, the square roots of the eigenvalues $\frac{3 \pm \sqrt{5}}{2}$ of $A^{\top} A$, because the shear stretches some directions and shrinks others.
Properties
- The eigenvalues of $A$ are the roots of its characteristic polynomial $\det(\lambda I - A)$.
- The origin of $\dot{x} = A x$ is asymptotically stable if and only if every eigenvalue of $A$ has negative real part ($A$ is Hurwitz).
Consider an $m \times n$ real matrix $A$. The singular values of $A$ are the square roots of the eigenvalues of the $n \times n$ matrix $A^{\top} A$, which are nonnegative.
Examples
-
The matrix \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\)
: $1$ and $1$
The singular values of \(A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) are $1$ and $1$, because $A^{\top} A = I$, whose eigenvalues are $1$ and $1$. A rotation does not change the length of any vector.
-
The matrix \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\)
: $\sqrt{14}$ and $1$
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 $0$.
-
The matrix \(\begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}\)
: $3$ and $2$
The singular values of \(A = \begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}\) are $3$ and $2$, the square roots of the eigenvalues $9$ and $4$ of \(A^{\top} A = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix}\). $A$ is symmetric, so its singular values are the absolute values of its eigenvalues.
-
The matrix \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)
: $\varphi$ and $1 / \varphi$, where $\varphi = \frac{1 + \sqrt{5}}{2}$
The shear \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has the eigenvalue $1$ twice, and it fixes the vector $(1, 0)$. Its singular values are $\varphi \approx 1.618$ and $1 / \varphi \approx 0.618$, the square roots of the eigenvalues $\frac{3 \pm \sqrt{5}}{2}$ of $A^{\top} A$, because the shear stretches some directions and shrinks others.
Properties
- Geometrically, $A$ maps the unit sphere to an ellipsoid, and the lengths of the ellipsoid’s semi-axes are the singular values of $A$.
- The largest singular value of $A$ is its operator norm $\lVert A \rVert_2$, the most that $A$ stretches a vector.
- For a symmetric matrix, the singular values are the absolute values of the eigenvalues.
Examples
Each cell lists the eigenvalues or the singular values of the matrix in that row.
| Example | eigenvalue | singular value | Why |
|---|---|---|---|
| The matrix \(\begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}\) | $2$ and $-3$ | $3$ and $2$ | The diagonal matrix with entries $2$ and $-3$ has eigenvalues $2$ and $-3$, its diagonal entries. Its singular values are their absolute values, $3$ and $2$, as for every symmetric matrix. |
| The matrix \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\) | $i$ and $-i$ | $1$ and $1$ | The rotation by $90°$ has the eigenvalues $\pm i$, because it turns every nonzero real vector off its line. Its singular values are $1$ and $1$, because a rotation does not change lengths. |
| The matrix \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) | $1$ (twice) | $\varphi$ and $1 / \varphi$, where $\varphi = \frac{1 + \sqrt{5}}{2}$ | The shear \(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) has the eigenvalue $1$ twice, and it fixes the vector $(1, 0)$. Its singular values are $\varphi \approx 1.618$ and $1 / \varphi \approx 0.618$, the square roots of the eigenvalues $\frac{3 \pm \sqrt{5}}{2}$ of $A^{\top} A$, because the shear stretches some directions and shrinks others. |
| The matrix \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}\) | None, since the matrix is not square | $\sqrt{14}$ and $1$ | 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 $A^{\top} A$). |
Notes
- For a normal matrix ($A^{\top} A = A A^{\top}$, such as a symmetric or orthogonal matrix), the singular values are the absolute values of the eigenvalues.
- For a square matrix, the product of the singular values is $\lvert \det A \rvert$, the absolute value of the product of the eigenvalues.