A Complete Two-By-Two SVD Calculation

From a Two-by-Two Matrix to Its SVD

For a real matrix \(\mathbf{A}\), the singular value decomposition separates its action into an orthogonal change of input coordinates, a scaling along perpendicular directions, and an orthogonal change of output coordinates. In a two-by-two calculation, the main work is usually finding eigenvalues and eigenvectors of the Gram matrix \(\mathbf{A}^{\top}\mathbf{A}\), then tracking signs and normalization carefully.

For a nonsingular \(\mathbf{A}\), the eigenvalues of \(\mathbf{A}^{\top}\mathbf{A}\) are positive, and their square roots are the singular values. If \(\mathbf{v}_i\) is a unit eigenvector for eigenvalue \(\lambda_i\), the corresponding left singular vector is \(\mathbf{u}_i=\mathbf{A}\mathbf{v}_i/\sqrt{\lambda_i}\). The eigenvalues and singular values must be paired in the same order: the largest eigenvalue gives the first (largest) singular value.

The following formula makes the eigenvalue step explicit. It applies to any real symmetric two-by-two matrix. Its eigenvalues are the roots of its characteristic equation, obtained by setting the determinant of the matrix minus \(\lambda\mathbf{I}_2\) equal to zero.

Proof. Expanding the characteristic determinant gives \(\det(\mathbf{G}-\lambda\mathbf{I}_2)=(a-\lambda)(d-\lambda)-b^2=\lambda^2-(a+d)\lambda+(ad-b^2)\). The quadratic formula gives roots \(\bigl((a+d)\pm\sqrt{(a+d)^2-4(ad-b^2)}\bigr)/2\). The expression under the square root simplifies to \(a^2-2ad+d^2+4b^2=(a-d)^2+4b^2\), which proves the formula. \(\square\)

A Calculation with a Negative Eigenvalue of the Original Matrix

Why the Construction Gives a Factorization

The calculations above use a procedure that works for any nonsingular real two-by-two matrix once an orthonormal eigenbasis of its Gram matrix has been found. Earlier material on orthonormal eigenbases establishes that a real symmetric matrix has such a basis; \(\mathbf{A}^{\top}\mathbf{A}\) is symmetric. Here is a direct verification of why the procedure gives the required factors.

Proof. First, each eigenvalue is positive. For either eigenvector \(\mathbf{v}_i\ne\mathbf{0}\), \(\lambda_i\mathbf{v}_i^{\top}\mathbf{v}_i=\mathbf{v}_i^{\top}\mathbf{A}^{\top}\mathbf{A}\mathbf{v}_i=(\mathbf{A}\mathbf{v}_i)^{\top}(\mathbf{A}\mathbf{v}_i)>0\), because \(\mathbf{A}\) is nonsingular and hence \(\mathbf{A}\mathbf{v}_i\ne\mathbf{0}\). Thus \(\sigma_i>0\), so the definition of \(\mathbf{u}_i\) is valid. For \(i,j\in\{1,2\}\),

Therefore \(\mathbf{U}\) is orthogonal; \(\mathbf{V}\) is orthogonal because its columns were assumed orthonormal. Also, \(\mathbf{A}\mathbf{v}_i=\sigma_i\mathbf{u}_i\) for each \(i\), so the two columns of \(\mathbf{A}\mathbf{V}\) are exactly the columns of \(\mathbf{U}\boldsymbol{\Sigma}\). Hence \(\mathbf{A}\mathbf{V}=\mathbf{U}\boldsymbol{\Sigma}\). Multiplying on the right by \(\mathbf{V}^{\top}\), and using \(\mathbf{V}\mathbf{V}^{\top}=\mathbf{I}_2\), yields \(\mathbf{A}=\mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^{\top}\). \(\square\)

A Nonsymmetric Example with Different Left and Right Vectors

A Short Example with Coordinate Directions

Common Errors and a Reliable Check

Do not confuse eigenvalues with singular values. For the first example, the eigenvalues of \(\mathbf{A}\) itself are \(3\) and \(-1\), but its singular values are \(3\) and \(1\). The singular values come from the square roots of the eigenvalues of \(\mathbf{A}^{\top}\mathbf{A}\), not from the eigenvalues of \(\mathbf{A}\). Using \(-1\) as a diagonal entry of \(\boldsymbol{\Sigma}\) would violate the definition that singular values are nonnegative.

A wrong calculation: in that example, taking \(\mathbf{u}_2=\mathbf{v}_2=(1,-1)^{\top}/\sqrt{2}\) because the matrix is symmetric would give \(\mathbf{A}\mathbf{v}_2=(1,-1)^{\top}/\sqrt{2}\). Direct multiplication instead gives \(\mathbf{A}\mathbf{v}_2=(-1,1)^{\top}/\sqrt{2}\). The correction is to calculate \(\mathbf{u}_2=\mathbf{A}\mathbf{v}_2/\sigma_2=(-1,1)^{\top}/\sqrt{2}\). This sign change is necessary for \(\mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^{\top}\) to reproduce \(\mathbf{A}\).

For any proposed full SVD, check three things: the diagonal entries of \(\boldsymbol{\Sigma}\) are nonnegative and decreasing; the columns of \(\mathbf{U}\) and \(\mathbf{V}\) are orthonormal; and direct multiplication gives \(\mathbf{A}\). When constructing the factors from the Gram matrix, also check \(\mathbf{A}^{\top}\mathbf{A}\mathbf{v}_i=\lambda_i\mathbf{v}_i\) and \(\mathbf{A}\mathbf{v}_i=\sigma_i\mathbf{u}_i\). These checks catch incorrect eigenvectors, mismatched ordering, and lost signs.