Find the singular value decomposition of a machine-precision matrix:

Format the result:

Singular value decomposition of a complex matrix:

Singular value decomposition for an exact matrix:

Singular value decomposition for an arbitrary-precision matrix:

Singular value decomposition of a symbolic matrix:

The singular value decomposition of a large numerical matrix is computed efficiently:

Singular value decomposition of a non-square matrix:

Find the singular value decomposition associated with the three largest singular values of a matrix:

Unlike the full decomposition, these matrices do not recreate any part of the matrix exactly:

Find singular value decomposition associated with the three smallest singular values:

Find the "compact" decomposition associated with the nonzero singular values:

This decomposition still has sufficient information to reconstruct the matrix:

The full singular value decomposition contains a row of zeros:

Find the thin decomposition of a rectangular matrix:

The matrix is square:

The number of singular values used equals the smaller of the dimensions of the input matrix:

This decomposition still has sufficient information to reconstruct the matrix:

The full singular value decomposition contains rows or columns of zeros in a rectangular :

Find the decomposition associated with the three largest singular values, or as many as there are if fewer:

Compute a truncated singular value decomposition for a matrix with repeated singular values:

Repeated singular values are counted separately when doing a partial decomposition:

Find the generalized singular value decomposition of a machine-precision real matrix:

Verify the result:

Find the generalized singular value decomposition of a machine-precision complex matrix:

Verify the result:

Singular value decomposition of sparse matrices:

Find the decomposition associated to the three largest singular values:

Visualize the three right-singular vectors:

Singular value decomposition of structured matrices:

Use a different structure:

The units go with the singular values:

Singular value decomposition of an identity matrix:

Verify the decomposition:

and could have been chosen to be identity matrices—the decomposition is not unique:

Singular value decomposition of HilbertMatrix :

A real rectangular matrix:

With TargetStructure->"Dense", the result of SingularValueDecomposition is a list of three dense matrices:

With TargetStructure->"Structured", the result of SingularValueDecomposition is a list containing two OrthogonalMatrix objects and a DiagonalMatrix :

A complex rectangular matrix:

With TargetStructure->"Dense", the result of SingularValueDecomposition is a list of three dense matrices:

With TargetStructure->"Structured", the result of SingularValueDecomposition is a list containing two UnitaryMatrix objects and a DiagonalMatrix :

m is a nearly singular matrix:

To machine precision, the matrix is effectively singular:

With a smaller tolerance, the nonzero singular value is detected:

The default tolerance is based on precision, so the small value is detected with precision 20:

Compute the singular value decomposition of the 2×2 matrix :

The action of TemplateBox[{v}, Transpose] is a rotation and possibly—as happens for the axis in this case—a reflection:

The action of is a scaling—either a dilation or compression—along each axis:

The action of is a rotation and possibly—though not in this case—a reflection in the target space:

Compute the singular value decomposition of the 3×2 matrix :

After the rotation in the plane by the TemplateBox[{v}, Transpose] matrix, the matrix embeds the unit circle as an ellipse in 3D:

The matrix rotates the ellipse in three dimensions:

Compute the singular value decomposition of the 2×2 matrix :

Let and denote the columns, respectively, of and :

is the direction in which TemplateBox[{{m, ., x}}, Norm] is maximized, and the maximum value is :

Similarly, is the direction in which TemplateBox[{{m, ., x}}, Norm] is minimized, and the minimum value is :

Visualize , and the unit circle along with their image under the multiplication on the left by :

is the direction in which TemplateBox[{{x, ., m}}, Norm] is maximized, and again the maximum value is :

Similarly, is the direction in which TemplateBox[{{x, ., m}}, Norm] is minimized, and again the minimum value is :

Visualize , and the unit circle along with their image under the multiplication on the right by :

Compute the singular value decomposition of the 3×2 matrix :

Let and denote the columns, respectively, of and :

is the direction in which TemplateBox[{{m, ., x}}, Norm] is maximized, and the maximum value is :

Similarly, is the direction in which TemplateBox[{{m, ., x}}, Norm] is minimized, and the minimum value is :

Visualize , and the image of the unit circle in the plane under left-multiplication by :

is the direction in which TemplateBox[{{x, ., m}}, Norm] is maximized, and again the maximum value is :

minimizes TemplateBox[{{x, ., m}}, Norm]—the minimum is zero, as the sphere is compressed into an ellipse in the plane:

maximizes TemplateBox[{{x, ., m}}, Norm] subject to the constraint , and the maximum value is :

Visualize , , and the image of the unit sphere in the plane under right-multiplication by :

Compute the singular value decomposition of the 3×3 matrix :

Let and denote the columns, respectively, of and :

is the direction in which TemplateBox[{{m, ., x}}, Norm] is maximized, and the maximum value is :

is the direction in which TemplateBox[{{m, ., x}}, Norm] is maximized if , and the maximum value is :

Similarly, is the direction in which TemplateBox[{{m, ., x}}, Norm] is minimized, and the minimum value is :

Visualize , , and the unit sphere along with their image under the multiplication on the left by :

is the direction in which TemplateBox[{{x, ., m}}, Norm] is maximized, and again the maximum value is :

is the direction in which TemplateBox[{{x, ., m}}, Norm] is maximized if , and again the maximum value is :

Similarly, is the direction in which TemplateBox[{{x, ., m}}, Norm] is minimized, and again the minimum value is :

Visualize , , and the unit sphere along with their image under the multiplication on the right by :

If the linear system has no solution, the best approximate solution is the least-squares solution. That is the solution to , where is the orthogonal projection of onto the column space of , which can be computed using the singular value decomposition. Consider the following and :

The linear system is inconsistent:

Find the matrix of the compact singular value decomposition of . Its columns are orthonormal and span :

Compute the orthogonal projection of onto the space spanned by the columns of :

Visualize , its projections onto the columns of and :

Solve :

Confirm the result using LeastSquares :

Solve the least-squares problem for the following and using only the singular value decomposition:

Compute the compact singular value decomposition where only the nonzero singular values are kept:

Let x=v.TemplateBox[{sigma}, Inverse].TemplateBox[{u}, ConjugateTranspose].b:

By definition, , so m.x=u.sigma.TemplateBox[{sigma}, Inverse].TemplateBox[{u}, ConjugateTranspose].b=u.TemplateBox[{u}, ConjugateTranspose].b, the orthogonal projection of onto :

Thus, is the solution to the least-squares problem, as confirmed by LeastSquares :

Solve the least-squares problem for the following and two different ways: by projecting onto the column space of using just the matrix of the singular value decomposition, and the direct solution using the full decomposition. Compare and explain the results:

Compute the compact singular value decomposition of:

Compute the orthogonal projection of onto :

Solve :

The direct solution can be found using x=v.TemplateBox[{sigma}, Inverse].TemplateBox[{u}, ConjugateTranspose].b=v.TemplateBox[{sigma}, Inverse].TemplateBox[{u}, Transpose].b, as both and are real-valued:

While x and xPerp are different, both solve the least-squares problem because m.x==m.xPerp:

The two solutions differ by an element of NullSpace [m]:

Note that LeastSquares [m,b] gives the result using the direct method:

For the matrices and that follow, find a matrix that minimizes TemplateBox[{{{m, ., x}, -, b}}, Norm]_F:

One solution, in this case unique, is given by x=v.TemplateBox[{sigma}, Inverse].TemplateBox[{u}, Transpose].b:

This result could also have been obtained using LeastSquares [m,b]:

Confirm the answer using Minimize :

SingularValueDecomposition can be used to find a best-fit curve to data. Consider the following data:

Extract the and coordinates from the data:

Construct a design matrix, whose columns are and , for fitting to a line :

Get the coefficients and for a linear least‐squares fit using a thin singular value decomposition:

Verify the coefficients using Fit :

Plot the best-fit curve along with the data:

Find the best-fit parabola to the following data:

Extract the and coordinates from the data:

Construct a design matrix, whose columns are , and , for fitting to a line :

Get the coefficients , and for a least‐squares fit:

Verify the coefficients using Fit :

Plot the best-fit curve along with the data: