What Is the Relationship Between Eigendecomposition and SVD?

Eigendecomposition and singular value decomposition (SVD) are closely related, but they do not decompose a general matrix in the same way. The key link is that the singular vectors of a matrix are eigenvectors of two matrices built from it: $A^*A$ and $AA^*$. Here, $A^*$ means the conjugate transpose; for a real matrix, it is the ordinary transpose.

This relationship also explains a special case: for a symmetric or Hermitian matrix, an eigendecomposition can be converted directly into an SVD. For a general matrix, however, the SVD is not usually the eigendecomposition of the matrix itself.

The two decompositions

For an $m\times n$ matrix $A$, an SVD has the form

$$A=U\Sigma V^*.$$

The columns of $V$ are right singular vectors, the columns of $U$ are left singular vectors, and the diagonal entries of $\Sigma$ are the singular values $\sigma_i\geq 0$. The columns of $U$ and $V$ are orthonormal. The SVD exists for every real or complex matrix, including rectangular matrices.

An eigendecomposition, when it exists, writes a square matrix as

$$A=Q\Lambda Q^{-1},$$

where the columns of $Q$ are eigenvectors and the diagonal entries of $\Lambda$ are eigenvalues. A matrix is diagonalizable only if it has enough linearly independent eigenvectors. Its eigenvalues may be negative or complex, unlike singular values, which are always nonnegative.

The main connection: diagonalize $A^*A$ and $AA^*$

Starting from the SVD and using the orthonormality of $U$ and $V$ gives

$$\begin{aligned} A^*A &= V\Sigma^*\Sigma V^*,\\ AA^* &= U\Sigma\Sigma^* U^*. \end{aligned}$$

Both $A^*A$ and $AA^*$ are Hermitian positive semidefinite, so they have orthonormal eigenvectors and nonnegative eigenvalues. Their eigenvalues are the squared singular values of $A$ (with additional zeros if needed because of the matrix dimensions). Thus:

• The eigenvectors of $A^*A$ are right singular vectors of $A$.
• The eigenvectors of $AA^*$ are left singular vectors of $A$.
• The singular values are the square roots of the corresponding eigenvalues.

For a nonzero singular value $\sigma_i$, the matching left and right singular vectors satisfy

$$Av_i=\sigma_i u_i, \qquad A^*u_i=\sigma_i v_i.$$

Equivalently, once you find a unit eigenvector $v_i$ of $A^*A$ with eigenvalue $\sigma_i^2>0$, you can obtain its left singular vector from $u_i=Av_i/\sigma_i$. Singular vectors associated with zero singular values require choosing orthonormal bases for the relevant null spaces.

When the matrix is symmetric or Hermitian

A real symmetric or complex Hermitian matrix has an orthonormal eigendecomposition

$$A=Q\Lambda Q^*.$$

Its eigenvalues are real. If $A$ is also positive semidefinite, all eigenvalues are nonnegative, so this is already an SVD: take $U=V=Q$ and $\Sigma=\Lambda$.

If some eigenvalues are negative, the singular values are their absolute values. Let $D$ be diagonal, with entry $+1$ for each nonnegative eigenvalue and $-1$ for each negative eigenvalue. Then one valid SVD is

$$A=(QD)|\Lambda|Q^*,$$

where $|\Lambda|$ is the diagonal matrix of absolute eigenvalues. Zero eigenvalues can be assigned either sign in $D$. The signs move into the left singular vectors, while the singular values remain nonnegative.

Worked example

Consider the real matrix

$$A=\begin{pmatrix}1&1\\0&0\end{pmatrix}.$$

First compute

$$A^T A=\begin{pmatrix}1&1\\1&1\end{pmatrix}.$$

This matrix has eigenvalues $2$ and $0$, with unit eigenvectors

$$v_1=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}, \qquad v_2=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}.$$

The singular values of $A$ are therefore $\sqrt{2}$ and $0$. For the nonzero singular value,

$$u_1=\frac{Av_1}{\sqrt{2}}=\begin{pmatrix}1\\0\end{pmatrix}.$$

Complete $u_1$ to an orthonormal basis with $u_2=(0,1)^T$. Taking $U$ to have columns $u_1,u_2$, $V$ to have columns $v_1,v_2$, and $\Sigma=\operatorname{diag}(\sqrt{2},0)$ gives $A=U\Sigma V^T$.

Common confusion

The SVD of $A$ is generally not the eigendecomposition of $A$. The direct eigenvalue connection is to $A^*A$ and $AA^*$, not necessarily to $A$ itself. This matters especially for rectangular matrices, which do not have an ordinary eigendecomposition, and for square matrices that are not diagonalizable. The SVD still exists in both cases.