The Spectral Theorem has animated the past few sections. In particular, we applied the fact that symmetric matrices can be orthogonally diagonalized to simplify quadratic forms, which enabled us to use principal component analysis to reduce the dimension of a dataset. But what can we do with matrices that are not symmetric or even square? For instance, the following matrices are not diagonalizable, much less orthogonally so:
\begin \begin 2 & 1 \\ 0 & 2 \end, \hspace \begin 1 & 1 & 0 \\ -1 & 0 & 1 \end. \endIn this section, we will develop a description of matrices called the singular value decomposition that is, in many ways, analogous to an orthogonal diagonalization. For example, we have seen that any symmetric matrix can be written in the form \(QDQ^T\) where \(Q\) is an orthogonal matrix and \(D\) is diagonal. A singular value decomposition will have the form \(U\Sigma V^T\) where \(U\) and \(V\) are orthogonal and \(\Sigma\) is diagonal. Most notably, we will see that every matrix has a singular value decomposition whether it's symmetric or not.
We will begin by explaining what a singular value decomposition is and how we can find one for a given matrix \(A\text<.>\)
Recall how the orthogonal diagonalization of a symmetric matrix is formed: if \(A\) is symmetric, we write \(A = QDQ^T\) where the diagonal entries of \(D\) are the eigenvalues of \(A\) and the columns of \(Q\) are the associated eigenvectors. Moreover, the eigenvalues are related to the maximum and minimum values of the associated quadratic form \(q_A(\mathbf u)\) among all unit vectors.
A general matrix, particularly a matrix that is not square, may not have eigenvalues and eigenvectors, but we can discover analogous features, called singular values and singular vectors, by studying a function somewhat similar to a quadratic form. More specifically, any matrix \(A\) defines a function
\begin l_A(\mathbf x) = |A\mathbf x|, \endwhich measures the length of \(A\mathbf x\text<.>\) For example, the diagonal matrix \(D=\begin 3 & 0 \\ 0 & -2 \\ \end\) gives the function \(l_D(\mathbf x) = \sqrt\text<.>\) The presence of the square root means that this function is not a quadratic form. We can, however, define the singular values and vectors by looking for the maximum and minimum of this function \(l_A(\mathbf u)\) among all unit vectors \(\mathbf u\text<.>\)
While \(l_A(\mathbf x)\) is not itself a quadratic form, it becomes one if we square it:
\begin \left(l_A(\mathbf x)\right)^2 = |A\mathbf x|^2 = (A\mathbf x)\cdot(A\mathbf x) = \mathbf x\cdot(A^TA\mathbf x)=q_(\mathbf x)\text <.>\end
We call \(G=A^TA\text\) the Gram matrix associated to \(A\) and note that
\begin l_A(\mathbf x) = \sqrtThis is important in the next activity, which introduces singular values and singular vectors.
The following interactive figure will help us explore singular values and vectors geometrically before we begin a more algebraic approach. This figure is also available at gvsu.edu/s/0YE.
The four sliders at the top of this figure enable us to choose a \(2\times2\) matrix \(A\text<.>\) Below on the left, we see the unit circle and the red unit vector \(\mathbf x\text\) which may be varied by clicking in the head of the vector and dragging it to a new unit vector.
Figure 7.4.1. Singular values, right singular vectors and left singular vectorsSelect the matrix \(A=\begin 1 & 2 \\ -2 & - 1 \\ \end\text<.>\) As we vary the vector \(\mathbf x\text\) we see the vector \(A\mathbf x\) in gray while the height of the blue bar to the right tells us \(l_A(\mathbf x) = |A\mathbf x|\text<.>\)
\begin U = \begin\mathbf u_1 & \mathbf u_2 \end, \hspace \Sigma = \begin \sigma_1 & 0 \\ 0 & \sigma_2 \\ \end, \hspace V = \begin\mathbf v_1 & \mathbf v_2 \end \end
As this activity shows, the singular values of \(A\) are the maximum and minimum values of \(l_A(\mathbf x)=|A\mathbf x|\) among all unit vectors and the right singular vectors \(\mathbf v_1\) and \(\mathbf v_2\) are the directions in which they occur. The key to finding the singular values and vectors is to utilize the Gram matrix \(G\) and its associated quadratic form \(q_G(\mathbf x)\text<.>\) We will illustrate with some more examples.
We will find a singular value decomposition of the matrix \(A=\begin 1 & 2 \\ -1 & 2 \end \text<.>\) Notice that this matrix is not symmetric so it cannot be orthogonally diagonalized.
We begin by constructing the Gram matrix \(G = A^TA = \begin 2 & 0 \\ 0 & 8 \\ \end\text<.>\) Since \(G\) is symmetric, it can be orthogonally diagonalized with
\begin D = \begin 8 & 0 \\ 0 & 2 \\ \end,\hspace Q = \begin 0 & 1 \\ 1 & 0 \\ \end\text \endWe now know that the maximum value of the quadratic form \(q_G(\mathbf x)\) is 8, which occurs in the direction \(\twovec01\text<.>\) Since \(l_A(\mathbf x) = \sqrt
In the same way, we also know that the second singular value \(\sigma_2=\sqrt\) with associated right singular vector \(\mathbf v_2=\twovec10\text<.>\)
The first left singular vector \(\mathbf u_1\) is defined by \(A\mathbf v_1 = \twovec22 = \sigma_1\mathbf u_1\text<.>\) Because \(\sigma_1 = \sqrt\text\) we have \(\mathbf u_1 = \twovec>>\text<.>\) Notice that \(\mathbf u_1\) is a unit vector because \(\sigma_1 = |A\mathbf v_1|\text<.>\)
In the same way, the second left singular vector is defined by \(A\mathbf v_2 = \twovec1 = \sigma_2\mathbf u_2\text\) which gives us \(\mathbf u_2 = \twovec>>\text<.>\)
We then construct
\begin U & <>=<> \begin \mathbf u_1 & \mathbf u_2 \end = \begin 1/\sqrt & 1/\sqrt \\ 1/\sqrt & -1/\sqrt \\ \end\\ \Sigma & <>=<> \begin \sigma_1 & 0 \\ 0 & \sigma_2 \\ \end = \begin \sqrt & 0 \\ 0 & \sqrt \\ \end\\ V & <>=<> \begin \mathbf v_1 & \mathbf v_2 \end = \begin 0 & 1 \\ 1 & 0 \\ \end \end
We now have \(AV=U\Sigma\) because
\begin AV = \begin A\mathbf v_1 & A\mathbf v_2 \end = \begin \sigma_1\mathbf u_1 & \sigma_2\mathbf u_2 \end = \Sigma U\text <.>\end
Because the right singular vectors, the columns of \(V\text\) are eigenvectors of the symmetric matrix \(G\text\) they form an orthonormal basis, which means that \(V\) is orthogonal. Therefore, we have \((AV)V^T = A = U\Sigma V^T\text<.>\) This gives the singular value decomposition
\begin A = \begin 1 & 2 \\ -1 & 2 \\ \end = \begin 1/\sqrt & 1/\sqrt \\ 1/\sqrt & -1/\sqrt \\ \end \begin \sqrt & 0 \\ 0 & \sqrt \\ \end \begin 0 & 1 \\ 1 & 0 \\ \end^T = U\Sigma V^T\text <.>\end
To summarize, we find a singular value decomposition of a matrix \(A\) in the following way:
\begin \sigma_i\sigma_j(\mathbf u_i\cdot\mathbf u_j) & <>=<> (\sigma_i\mathbf u_i)\cdot(\sigma_j\mathbf u_j) = (A\mathbf v_i)\cdot(A\mathbf v_j)\\ & <>=<> \mathbf v_i\cdot(A^TA\mathbf v_j) \\ & <>=<> \mathbf v_i\cdot(G\mathbf v_j) = \lambda_j\mathbf v_i\cdot\mathbf v_j = 0 \end
Let's find a singular value decomposition for the symmetric matrix \(A=\begin 1 & 2 \\ 2 & 1 \end\text<.>\) The associated Gram matrix is
\begin G = A^TA = \begin 5 & 4 \\ 4 & 5 \\ \end\text \endwhich has an orthogonal diagonalization with
\begin D = \begin 9 & 0 \\ 0 & 1 \\ \end,\hspace Q = \begin 1/\sqrt & 1/\sqrt \\ 1/\sqrt & -1/\sqrt \\ \end\text <.>\end
This gives singular values and vectors
and the singular value decomposition \(A=U\Sigma V^T\) where
\begin U = \begin 1/\sqrt & -1/\sqrt \\ 1/\sqrt & 1/\sqrt \end,\hspace \Sigma = \begin 3 & 0 \\ 0 & 1 \end,\hspace V = \begin 1/\sqrt & 1/\sqrt \\ 1/\sqrt & -1/\sqrt \end. \end
This example is special because \(A\) is symmetric. With a little thought, it's possible to relate this singular value decomposition to an orthogonal diagonalization of \(A\) using the fact that \(G=A^TA = A^2\text<.>\)
In this activity, we will construct the singular value decomposition of \(A=\begin 1 & 0 & -1 \\ 1 & 1 & 1 \end\text<.>\) Notice that this matrix is not square so there are no eigenvalues and eigenvectors associated to it.
We will find a singular value decomposition of the matrix \(A=\begin 2 & -2 & 1 \\ -4 & -8 & -8 \\ \end\text<.>\)
Finding an orthogonal diagonalization of \(G=A^TA\) gives
\begin D=\begin 144 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 0 \\ \end,\hspace Q=\begin 1/3 & 2/3 & 2/3 \\ 2/3 & -2/3 & 1/3 \\ 2/3 & 1/3 & -2/3 \\ \end\text \end
which gives singular values \(\sigma_1=\sqrt=12\text\) \(\sigma_2 = \sqrt= 3\text\) and \(\sigma_3 = 0\text<.>\) The right singular vectors \(\mathbf v_i\) appear as the columns of \(Q\) so that \(V = Q\text<.>\)
\begin A\mathbf v_1 = \twovec = 12\mathbf u_1, \hspace & \mathbf u_1 = \twovec\\ A\mathbf v_2 = \twovec = 3\mathbf u_1, \hspace & \mathbf u_1 = \twovec10\\ A\mathbf v_3 = \twovec \end
Notice that it's not possible to find a third left singular vector since \(A\mathbf v_3=\zerovec\text<.>\) We thereform form the matrices
\begin U = \begin 0 & 1 \\ -1 & 0 \\ \end,\hspace \Sigma = \begin 12 & 0 & 0 \\ 0 & 3 & 0 \\ \end,\hspace V=\begin 1/3 & 2/3 & 2/3 \\ 2/3 & -2/3 & 1/3 \\ 2/3 & 1/3 & -2/3 \\ \end\text \end
which gives the singular value decomposition \(A=U\Sigma V^T\text<.>\)
Notice that \(U\) is a \(2\times2\) orthogonal matrix because \(A\) has two rows, and \(V\) is a \(3\times3\) orthogonal matrix because \(A\) has three columns.
As we'll see in the next section, some additional work may be needed to construct the left singular vectors \(\mathbf u_j\) if more of the singular values are zero, but we won't worry about that now. For the time being, let's record our work in the following theorem.
An \(m\times n\) matrix \(A\) may be written as \(A=U\Sigma V^T\) where \(U\) is an orthogonal \(m\times m\) matrix, \(V\) is an orthogonal \(n\times n\) matrix, and \(\Sigma\) is an \(m\times n\) matrix whose entries are zero except for the singular values of \(A\) which appear in decreasing order on the diagonal.
Notice that a singular value decomposition of \(A\) gives us a singular value decomposition of \(A^T\text<.>\) More specifically, if \(A=U\Sigma V^T\text\) then
\begin A^T = (U\Sigma V^T)^T = V\Sigma^T U^T. \endIf \(A=U\Sigma V^T\text\) then \(A^T = V\Sigma^T U^T\text<.>\) In other words, \(A\) and \(A^T\) share the same singular values, and the left singular vectors of \(A\) are the right singular vectors of \(A^T\) and vice-versa.
As we said earlier, the singular value decomposition should be thought of a generalization of an orthogonal diagonalization. For instance, the Spectral Theorem tells us that a symmetric matrix can be written as \(QDQ^T\text<.>\) Many matrices, however, are not symmetric and so they are not orthogonally diagonalizable. However, every matrix has a singular value decomposition \(U\Sigma V^T\text<.>\) The price of this generalization is that we usually have two sets of singular vectors that form the orthogonal matrices \(U\) and \(V\) whereas a symmetric matrix has a single set of eignevectors that form the orthogonal matrix \(Q\text<.>\)
Now that we have an understanding of what a singular value decomposition is and how to construct it, let's explore the ways in which a singular value decomposition reveals the underlying structure of the matrix. As we'll see, the matrices \(U\) and \(V\) in a singular value decomposition provide convenient bases for some important subspaces, such as the column and null spaces of the matrix. This observation will provide the key to some of our uses of these decompositions in the next section.
Let's suppose that a matrix \(A\) has a singular value decomposition \(A=U\Sigma V^T\) where
\begin U=\begin \mathbf u_1 & \mathbf u_2 & \mathbf u_3 & \mathbf u_4 \end,\hspace \Sigma = \begin 20 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end,\hspace V=\begin \mathbf v_1 & \mathbf v_2 & \mathbf v_3 \end. \end
This activity shows how a singular value decomposition of a matrix encodes important information about its null and column spaces. This is, in fact, the key observation that makes singular value decompositions so useful: the left and right singular vectors provide orthonormal bases for \(Nul(A)\) and \(Col(A)\text<.>\)
Suppose we have a singular value decomposition \(A=U\Sigma V^T\) where \(\Sigma = \begin \sigma_1 & 0 & 0 & 0 & 0 \\ 0 & \sigma_2 & 0 & 0 & 0 \\ 0 & 0 & \sigma_3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end \text<.>\) This means that \(A\) has four rows and five columns just as \(\Sigma\) does.
As in the activity, if \(\mathbf x = c_1 \mathbf v_1 + c_2\mathbf v_2 + \ldots + c_5\mathbf v_5\text\) we have
\begin A\mathbf x = \sigma_1c_1\mathbf u_1 + \sigma_2c_2\mathbf u_2 + \sigma_3c_3\mathbf u_3\text <.>\end
If \(\mathbf b\) is in the \(Col(A)\text\) then \(\mathbf b\) must have the form
\begin \mathbf b = \sigma_1c_1\mathbf u_1 + \sigma_2c_2\mathbf u_2 + \sigma_3c_3\mathbf u_3\text \end
which says that \(\mathbf b\) is a linear combination of \(\mathbf u_1\text\) \(\mathbf u_2\text\) and \(\mathbf u_3\text<.>\) These three vectors therefore form a basis for \(Col(A)\text<.>\) In fact, since they are columns in the orthogonal matrix \(U\text\) they form an orthonormal basis for \(Col(A)\text<.>\)
Remembering that \(Rank(A)=dimCol(A)\text\) we see that \(Rank(A) = 3\text\) which results from the three nonzero singular values. In general, the rank \(r\) of a matrix \(A\) equals the number of nonzero singular values, and \(\mathbf u_1, \mathbf u_2, \ldots,\mathbf u_r\) form an orthonormal basis for \(Col(A)\text<.>\)
Moreover, if \(\mathbf x = c_1 \mathbf v_1 + c_2\mathbf v_2 + \ldots + c_5\mathbf v_5\) satisfies \(A\mathbf x = \zerovec\text\) then
\begin A\mathbf x = \sigma_1c_1\mathbf u_1 + \sigma_2c_2\mathbf u_2 + \sigma_3c_3\mathbf u_3=\zerovec\text \end
which implies that \(c_1=0\text\) \(c_2=0\text\) and \(c_3=0\text<.>\) Therefore, \(\mathbf x = c_4\mathbf v_4+c_5\mathbf v_5\) so \(\mathbf v_4\) and \(\mathbf v_5\) form an orthonormal basis for \(Nul(A)\text<.>\)
More generally, if \(A\) is an \(m\times n\) matrix and if \(Rank(A) = r\text\) the last \(n-r\) right singular vectors form an orthonormal basis for \(Nul(A)\text<.>\)
Generally speaking, if the rank of an \(m\times n\) matrix \(A\) is \(r\text\) then there are \(r\) nonzero singular values and \(\Sigma\) has the form
\begin \begin \sigma_1 & \ldots & 0 & \ldots & 0 \\ 0 & \ldots & 0 & \ldots & 0 \\ 0 & \ldots & \sigma_r & \ldots & 0 \\ 0 & \ldots & 0 & \ldots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \ldots & 0 & \ldots & 0 \\ \end, \end
The first \(r\) columns of \(U\) form an orthonormal basis for \(Col(A)\text\)
\begin U = \left[ \underbrace<\mathbf u_1 ~ \ldots ~ \mathbf u_r>_ \hspace \mathbf u_ ~ \ldots ~ \mathbf u_m \right] \endand the last \(n-r\) columns of \(V\) form an orthonormal basis for \(Nul(A)\text\)
\begin V = \left[ \mathbf v_1 ~ \ldots ~ \mathbf v_r\hspace \underbrace <\mathbf v_
In fact, we can say more. Remember that Proposition 7.4.6 says that \(A\) and its transpose \(A^T\) share the same singular values. Since the rank of a matrix equals its number of nonzero singular values, this means that \(Rank(A)=Rank(A^T)\text\) a fact that we cited back in Section 6.2.
For any matrix \(A\text\)
\begin Rank(A) = Rank(A^T)\text \endIf we have a singular value decomposition of an \(m\times n\) matrix \(A=U\Sigma V^T\text\) Proposition 7.4.6 also tells us that the left singular vectors of \(A\) are the right singular vectors of \(A^T\text<.>\) Therefore, \(U\) is the \(m\times m\) matrix whose columns are the right singular vectors of \(A^T\text<.>\) This means that the last \(m-r\) vectors form an orthonormal basis for \(Nul(A^T)\text<.>\) Therefore, the columns of \(U\) provide orthonormal bases for \(Col(A)\) and \(Nul(A^T)\text\)
\begin U = \left[ \underbrace<\mathbf u_1 ~ \ldots ~ \mathbf u_r>_ \hspace \underbrace <\mathbf u_This reflects the familiar fact that \(Nul(A^T)\) is the orthogonal complement of \(Col(A)\text<.>\)
In the same way, \(V\) is the \(n\times n\) matrix whose columns are the left singular vectors of \(A^T\text\) which means that the first \(r\) vectors form an orthonormal basis for \(Col(A^T)\text<.>\) Because the columns of \(A^T\) are the rows of \(A\text\) this subspace is sometimes called the row space of \(A\) and denoted \(Row(A)\text<.>\) While we have yet to have an occasion to use \(Row(A)\text\) there are times when it is important to have an orthonormal basis for it. Fortunately, a singular value decomposition provides just that. To summarize, the columns of \(V\) provide orthonormal bases for \(Col(A^T)\) and \(Nul(A)\text\)
\begin V = \left[ \underbrace<\mathbf v_1 ~ \ldots ~ \mathbf v_r>_\hspace \underbrace <\mathbf v_Considered altogether, the subspaces \(Col(A)\text\) \(Nul(A)\text\) \(Col(A^T)\text\) and \(Nul(A^T)\) are called the four fundamental subspaces associated to \(A\text<.>\) In addition to telling us the rank of a matrix, a singular value decomposition gives us orthonormal bases for all four fundamental subspaces.
Suppose \(A\) is an \(m\times n\) matrix having a singular value decomposition \(A=U\Sigma V^T\text<.>\) Then
When we previously outlined a procedure for finding a singular decomposition of an \(m\times n\) matrix \(A\text\) we found the left singular vectors \(\mathbf u_j\) using the expression \(A\mathbf v_j = \sigma_j\mathbf u_j\text<.>\) This produces left singular vectors \(\mathbf u_1, \mathbf u_2,\ldots,\mathbf u_r\text\) where \(r=Rank(A)\text<.>\) If \(r\lt m\text\) however, we still need to find the left singular vectors \(\mathbf u_,\ldots,\mathbf u_m\text<.>\) Theorem 7.4.9 tells us how to do that: because those vectors form an orthonormal basis for \(Nul(A^T)\text\) we can find them by solving \(A^T\mathbf x = \zerovec\) to find a basis for \(Nul(A^T)\) and applying the Gram-Schmidt algorithm.
We won't worry about this issue too much, however, as we will frequently use software to find singular value decompositions for us.
As we'll see in the next section, there are times when it is helpful to express a singular value decomposition in a slightly different form.
Suppose we have a singular value decomposition \(A = U\Sigma V^T\) where
\begin U = \begin \mathbf u_1 & \mathbf u_2 & \mathbf u_3 & \mathbf u_4 \end,\hspace \Sigma = \begin 18 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end,\hspace V = \begin \mathbf v_1 & \mathbf v_2 & \mathbf v_3 \end\text <.>\end
\begin U\Sigma = \begin \mathbf u_1 & \mathbf u_2 \end \begin 18 & 0 & 0 \\ 0 & 4 & 0 \\ \end\text <.>\end
\begin \begin 18 & 0 & 0 \\ 0 & 4 & 0 \\ \endV^T = \begin 18 & 0 \\ 0 & 4 \\ \end \begin \mathbf v_1 & \mathbf v_2 \end^T\text <.>\end
We call this a reduced singular value decomposition.
If \(A\) is an \(m\times n\) matrix having rank \(r\text\) then \(A=U_r \Sigma_r V_r^T\) where
In Example 7.4.4, we found the singular value decomposition
\begin A=\begin 2 & -2 & 1 \\ -4 & -8 & -8 \\ \end = \begin 0 & 1 \\ -1 & 0 \\ \end \begin 12 & 0 & 0 \\ 0 & 3 & 0 \\ \end \begin 1/3 & 2/3 & 2/3 \\ 2/3 & -2/3 & 1/3 \\ 2/3 & 1/3 & -2/3 \\ \end^T\text <.>\end
Since there are two nonzero singular values, \(Rank(A) =2\) so that the reduced singular value decomposition is
\begin A=\begin 2 & -2 & 1 \\ -4 & -8 & -8 \\ \end = \begin 0 & 1 \\ -1 & 0 \\ \end \begin 12 & 0 \\ 0 & 3 \\ \end \begin 1/3 & 2/3 \\ 2/3 & -2/3 \\ 2/3 & 1/3 \\ \end^T\text <.>\end
This section has explored singular value decompositions, how to find them, and how they organize important information about a matrix.
Consider the matrix \(A = \begin 1 & 2 & 1 \\ 0 & -1 & 2 \\ \end \text<.>\)
Find singular value decompositions for the following matrices:
Consider the matrix \(A = \begin 2 & 1 \\ 1 & 2 \end \text<.>\)
Suppose that the matrix \(A\) has the singular value decomposition
\begin \begin -0.46 & 0.52 & 0.46 & 0.55 \\ -0.82 & 0.00 & -0.14 & -0.55 \\ -0.04 & 0.44 & -0.85 & 0.28 \\ -0.34 & -0.73 & -0.18 & 0.55 \end \begin 6.2 & 0.0 & 0.0 \\ 0.0 & 4.1 & 0.0 \\ 0.0 & 0.0 & 0.0 \\ 0.0 & 0.0 & 0.0 \end \begin -0.74 & 0.62 & -0.24 \\ 0.28 & 0.62 & 0.73 \\ -0.61 & -0.48 & 0.64 \end. \end
Consider the matrix \(A = \begin 1 & 0 & -1 \\ 2 & 2 & 0 \\ -1 & 1 & 2\\ \end \text<.>\)
Consider the matrix \(B=\begin 1 & 0 \\ 2 & -1 \\ 1 & 2 \end\) and notice that \(B=A^T\) where \(A\) is the matrix in Exercise 7.4.5.1.
Suppose that \(A\) is a square \(m\times m\) matrix with singular value decomposition \(A=U\Sigma V^T\text<.>\)
If \(A\) is a matrix and \(G=A^TA\) its Gram matrix, remember that
\begin \mathbf x\cdot(G\mathbf x) = \mathbf x\cdot(A^TA\mathbf x) = (A\mathbf x)\cdot(A\mathbf x) = \len
Determine whether the following statements are true or false and explain your reasoning.
Suppose that \(A=U\Sigma V^T\) is a singular value decomposition of the \(m\times n\) matrix \(A\text<.>\) If \(\sigma_1,\ldots,\sigma_r\) are the nonzero singular values, the general form of the matrix \(\Sigma\) is
\begin \Sigma = \begin \sigma_1 & \ldots & 0 & \ldots & 0 \\ 0 & \ldots & 0 & \ldots & 0 \\ 0 & \ldots & \sigma_r & \ldots & 0 \\ 0 & \ldots & 0 & \ldots & 0 \\ 0 & \vdots & 0 & \vdots & 0 \\ 0 & \ldots & 0 & \ldots & 0 \\ \end. \end
This page titled 7.4: Singular Value Decompositions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by David Austin via source content that was edited to the style and standards of the LibreTexts platform.
