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Singular Value Decomposition · Tutorial 776 of 1000

Singular Values From A Transpose A

Learn to read singular values and right singular vectors from the eigenvalues and eigenspaces of \(\mathbf{A}^{\top}\mathbf{A}\), including repeated and zero values.

Advanced 9 min read

What You'll Learn

  • Form the Gram matrix \(\mathbf{A}^{\top}\mathbf{A}\) for a real rectangular matrix.
  • Define singular values as nonnegative square roots of its eigenvalues.
  • Count singular values with multiplicity and place them in nonincreasing order.
  • Identify right singular vectors as unit eigenvectors of the Gram matrix.
  • Relate zero singular values to the null space of the original matrix.
  • Recognise why repeated singular values allow different choices of right singular vectors.

From a Matrix to Its Singular Values

Eigenvalues describe how a square matrix acts on its eigenvectors. Singular values provide a related measurement for any real matrix, including rectangular ones. To define them, begin with the square matrix \(\mathbf{A}^{\top}\mathbf{A}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\). This matrix is called the Gram matrix of \(\mathbf{A}\); it has \(n\) rows and \(n\) columns, so its eigenvalues correspond to vectors in the input space \(\mathbb{R}^n\).

The Gram matrix is symmetric, since \((\mathbf{A}^{\top}\mathbf{A})^{\top}=\mathbf{A}^{\top}\mathbf{A}\). The Spectral Theorem for real symmetric matrices says that it has real eigenvalues and an orthonormal basis of eigenvectors. We also use the standard result that the eigenvalues of \(\mathbf{A}^{\top}\mathbf{A}\) are nonnegative; its proof comes from examining why Gram matrices have nonnegative eigenvalues. These facts ensure that the square roots in the definition below are real.

Definition: Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\). The singular values of \(\mathbf{A}\) are the nonnegative square roots of the eigenvalues of \(\mathbf{A}^{\top}\mathbf{A}\), counted with algebraic multiplicity and listed in nonincreasing order: \(\sigma_1\ge\sigma_2\ge\cdots\ge\sigma_n\ge0\). A right singular vector corresponding to \(\sigma_i\) is a unit eigenvector \(\mathbf{v}_i\in\mathbb{R}^n\) of \(\mathbf{A}^{\top}\mathbf{A}\) with eigenvalue \(\sigma_i^2\). Equivalently, \(\mathbf{A}^{\top}\mathbf{A}\mathbf{v}_i=\sigma_i^2\mathbf{v}_i\) and \(\|\mathbf{v}_i\|=1\).

There are \(n\) singular values, not necessarily \(\min(m,n)\) of them: the definition counts the \(n\) eigenvalues of the \(n\times n\) matrix \(\mathbf{A}^{\top}\mathbf{A}\), including zero eigenvalues. Some conventions list only the positive singular values, but here the full list is used. Repeated eigenvalues are repeated in the list, so repeated singular values are repeated as well. The right singular vectors are chosen from the corresponding eigenspaces; they need not be uniquely determined when an eigenspace has dimension greater than one.

Computing the Values and Vectors

The calculation has a consistent order: form \(\mathbf{A}^{\top}\mathbf{A}\), find its eigenvalues with multiplicity, take their nonnegative square roots, and find unit eigenvectors for those eigenvalues. The vector associated with a singular value \(\sigma\) is an eigenvector for \(\sigma^2\), not for \(\sigma\). The transpose in \(\mathbf{A}^{\top}\mathbf{A}\) also matters: the resulting eigenvectors are in \(\mathbb{R}^n\), the space of inputs to \(\mathbf{A}\).

Worked Example: A Diagonal Matrix

Find the singular values and a corresponding set of right singular vectors for \(\mathbf{A}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\).

Here \(\mathbf{A}^{\top}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\), so

$$ \mathbf{A}^{\top}\mathbf{A} = \begin{bmatrix}3&0\\0&2\end{bmatrix} \begin{bmatrix}3&0\\0&2\end{bmatrix} = \begin{bmatrix}9&0\\0&4\end{bmatrix}. $$

The eigenvalues are \(9\) and \(4\), with respective unit eigenvectors \(\mathbf{e}_1=(1,0)^{\top}\) and \(\mathbf{e}_2=(0,1)^{\top}\). Taking nonnegative square roots and putting the values in nonincreasing order gives \(\sigma_1=3\) and \(\sigma_2=2\). The corresponding right singular vectors are \(\mathbf{v}_1=\mathbf{e}_1\) and \(\mathbf{v}_2=\mathbf{e}_2\).

Check: \(\mathbf{A}^{\top}\mathbf{A}\mathbf{e}_1=(9,0)^{\top}=9\mathbf{e}_1=\sigma_1^2\mathbf{e}_1\), and \(\mathbf{A}^{\top}\mathbf{A}\mathbf{e}_2=(0,4)^{\top}=4\mathbf{e}_2=\sigma_2^2\mathbf{e}_2\). Both vectors have norm \(1\), as required.

Worked Example: A Rectangular Matrix with a Zero Singular Value

For \(\mathbf{A}=\begin{bmatrix}1&0&0\\0&2&0\end{bmatrix}\), find all singular values and a corresponding set of right singular vectors.

The transpose is \(\mathbf{A}^{\top}=\begin{bmatrix}1&0\\0&2\\0&0\end{bmatrix}\). Multiplication gives

$$ \mathbf{A}^{\top}\mathbf{A} = \begin{bmatrix}1&0\\0&2\\0&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&2&0\end{bmatrix} = \begin{bmatrix}1&0&0\\0&4&0\\0&0&0\end{bmatrix}. $$

The eigenvalues are \(4,1,0\), with unit eigenvectors \(\mathbf{e}_2,\mathbf{e}_1,\mathbf{e}_3\), respectively. Their nonnegative square roots, in order, are \(\sigma_1=2,\sigma_2=1,\sigma_3=0\). Thus a matching list of right singular vectors is \(\mathbf{v}_1=\mathbf{e}_2,\mathbf{v}_2=\mathbf{e}_1,\mathbf{v}_3=\mathbf{e}_3\).

Check: multiplying the Gram matrix by these vectors gives \((0,4,0)^{\top}=4\mathbf{e}_2\), \((1,0,0)^{\top}=1\mathbf{e}_1\), and \(\mathbf{0}=0\mathbf{e}_3\), respectively. Each vector has norm \(1\), including the vector for the zero singular value.

Multiplicity and Choice of Right Singular Vectors

When an eigenvalue of \(\mathbf{A}^{\top}\mathbf{A}\) is repeated, its square root is repeated in the singular-value list. The corresponding right singular vectors belong to the same eigenspace. If that eigenspace has dimension greater than one, there is no unique choice of unit eigenvectors: any orthonormal basis of the eigenspace gives valid right singular vectors for that repeated singular value. The multiplicity records how many vectors are needed in an orthonormal eigenbasis, not a unique set of directions.

Worked Example: A Repeated Singular Value

Find the singular values of \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&0\end{bmatrix}\), and exhibit two different orthonormal choices for the right singular vectors associated with the repeated value.

Direct multiplication gives

$$ \mathbf{A}^{\top}\mathbf{A} = \begin{bmatrix}2&0\\0&2\\0&0\end{bmatrix} \begin{bmatrix}2&0&0\\0&2&0\end{bmatrix} = \begin{bmatrix}4&0&0\\0&4&0\\0&0&0\end{bmatrix}. $$

The eigenvalue \(4\) has multiplicity two, with eigenspace \(\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\). The eigenvalue \(0\) has eigenvector \(\mathbf{e}_3\). Therefore the singular values are \(2,2,0\). One orthonormal choice of right singular vectors is \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\).

A different orthonormal basis for the eigenspace for \(4\) is \(\mathbf{w}_1=(1/\sqrt{2},1/\sqrt{2},0)^{\top}\) and \(\mathbf{w}_2=(1/\sqrt{2},-1/\sqrt{2},0)^{\top}\). Their squared norms are \(1/2+1/2=1\), and their dot product is \(1/2-1/2=0\), so they are orthonormal. Keep \(\mathbf{e}_3\) for the zero eigenvalue.

Check: multiplying \(\mathbf{A}^{\top}\mathbf{A}\mathbf{w}_1\) gives \((4/\sqrt{2},4/\sqrt{2},0)^{\top}=4\mathbf{w}_1\); multiplying \(\mathbf{A}^{\top}\mathbf{A}\mathbf{w}_2\) gives \((4/\sqrt{2},-4/\sqrt{2},0)^{\top}=4\mathbf{w}_2\). Also \(\mathbf{A}^{\top}\mathbf{A}\mathbf{e}_3=\mathbf{0}\). The two choices differ, but both satisfy the eigenvector definition.

Theorem: For \(\mathbf{A}\in\mathbb{R}^{m\times n}\), the eigenspace of \(\mathbf{A}^{\top}\mathbf{A}\) for eigenvalue \(0\) equals \(\operatorname{Nul}\mathbf{A}\). Consequently, the number of zero singular values, counted with multiplicity, is \(\dim(\operatorname{Nul}\mathbf{A})\).

Proof. Let \(\mathbf{x}\in\mathbb{R}^n\). If \(\mathbf{x}\) is in the zero eigenspace, then \(\mathbf{A}^{\top}\mathbf{A}\mathbf{x}=\mathbf{0}\). Multiplying on the left by \(\mathbf{x}^{\top}\) gives \(\mathbf{x}^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{x}=0\). Since \(\mathbf{x}^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{x}=(\mathbf{A}\mathbf{x})^{\top}(\mathbf{A}\mathbf{x})\), this is the sum of the squares of the entries of \(\mathbf{A}\mathbf{x}\). A sum of real squares is zero only if every entry is zero, so \(\mathbf{A}\mathbf{x}=\mathbf{0}\), which means \(\mathbf{x}\in\operatorname{Nul}\mathbf{A}\). Conversely, if \(\mathbf{A}\mathbf{x}=\mathbf{0}\), then \(\mathbf{A}^{\top}\mathbf{A}\mathbf{x}=\mathbf{A}^{\top}\mathbf{0}=\mathbf{0}\), so \(\mathbf{x}\) belongs to the zero eigenspace. The two spaces are equal. Because \(\mathbf{A}^{\top}\mathbf{A}\) is real symmetric, the Spectral Theorem gives equality of the algebraic and geometric multiplicities of each eigenvalue. Thus the multiplicity of eigenvalue \(0\), and hence the number of zero singular values, is the dimension of that eigenspace, namely \(\dim(\operatorname{Nul}\mathbf{A})\). \(\square\)

This result explains why zero singular values must be retained in the full list. They correspond precisely to directions in the input space that the matrix sends to zero. In particular, a rectangular matrix can have a nontrivial null space even though its Gram matrix is square. If the Rank-Nullity Theorem is used, the number of zero singular values can also be written as \(n-\operatorname{rank}\mathbf{A}\).

Common Mistakes

Using the eigenvalues themselves as singular values. For the first example, a wrong calculation would stop at the eigenvalues \(9\) and \(4\) of \(\mathbf{A}^{\top}\mathbf{A}\) and report singular values \(9\) and \(4\). The correction is to take nonnegative square roots: \(\sqrt{9}=3\) and \(\sqrt{4}=2\). Substituting back confirms the relationship: \(3^2=9\) and \(2^2=4\).

Dropping the zero values or their vectors. In the rectangular example, the Gram matrix has three eigenvalues, including \(0\), so the full list has three singular values, \(2,1,0\). The right singular vector for \(0\) is not an exception to the definition: it is a unit eigenvector with eigenvalue \(0^2\).

Assuming a repeated value determines one unique vector. A repeated singular value specifies an eigenspace, not a single right singular vector. Check that a proposed vector is a unit eigenvector of \(\mathbf{A}^{\top}\mathbf{A}\) for the squared value; for a repeated value, different orthonormal choices may pass the same check.

For any real \(m\times n\) matrix, the defining data can be summarised as follows.

ObjectHow it is obtainedMultiplicity or size
Gram matrix\(\mathbf{A}^{\top}\mathbf{A}\)\(n\times n\)
Singular value \(\sigma\)Nonnegative square root of an eigenvalue \(\lambda\)\(\sigma=\sqrt{\lambda}\)
Right singular vectorUnit eigenvector for \(\lambda=\sigma^2\)Belongs to \(\mathbb{R}^n\)
Zero singular valuesSquare roots of zero eigenvaluesCount is \(\dim(\operatorname{Nul}\mathbf{A})\)

Check Your Understanding

For each question, connect the proposed singular values and vectors to the eigenvalues and eigenspaces of \(\mathbf{A}^{\top}\mathbf{A}\).

  1. If the eigenvalues of \(\mathbf{A}^{\top}\mathbf{A}\) are \(16,4,0\), what are the singular values in nonincreasing order?
  2. For a singular value \(\sigma=5\), what eigenvalue must its right singular vector have for \(\mathbf{A}^{\top}\mathbf{A}\)?
  3. Why does a matrix \(\mathbf{A}\in\mathbb{R}^{3\times 5}\) have five singular values under the full-list convention?
  4. If the zero eigenspace of \(\mathbf{A}^{\top}\mathbf{A}\) has dimension two, how many zero singular values does \(\mathbf{A}\) have, counted with multiplicity?
  5. When an eigenvalue of \(\mathbf{A}^{\top}\mathbf{A}\) has multiplicity greater than one, what condition must a chosen set of right singular vectors satisfy?

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