In the article The Spectral Decomposition of Symmetric Matrices, it has been shown that every symmetric matrix A can be expressed as A = EΛE’ where E is a matrix whose columns are the eigenvectors of A with the norm 1 and Λ is a diagonal matrix whose elements are the eigenvalues of A. Now, what if the matrix A is not a symmetric matrix, but any matrix? The singular value decomposition does the same “work” as the spectral decomposition, so it can be viewed as a more general decomposition or an extension of the spectral decomposition.

**The Singular Value Decomposition**

If A is any m x n matrix all elements of which are real numbers then there is an orthogonal m x m matrix U and an orthogonal n x n matrix V such that A = UΣV’ where Σ = [σ_{ij}] is an m x n matrix with σ_{ii} ≥ 0 for i = 1, 2, …, k, σ_{11} ≥ σ_{22} ≥ … ≥ σ_{kk}, and σ_{ij} = 0 if i ≠ j; k = min(m,n). The positive constants σ_{ii} are called the singular values of A.

**How to determine U and V?**

Let A = UΣV’. As a result, AA’ = (UΣV’)(VΣ’U’) = U(ΣΣ’)U’, which is a symmetric matrix. Noting that AA’ = U(ΣΣ’)U’, it can be concluded that: 1) U in A = UΣV’ can be obtained by applying the spectral decomposition to AA’ and 2) the diagonal elements of ΣΣ’ are the squares of the diagonal elements of Σ. The matrix V is obtained similarly, namely by applying the spectral decomposition to A’A. (Note that A’A = (VΣ’U’)(UΣV’) = VΣ’ΣV’.)

**Example 1**

Let . Determine the singular value decomposition of A.

*Answer*

U can be obtained from the spectral decomposition of symmetric matrix AA’. Note that:

By solving |λI – AA’| = 0, it turns out that the the eigenvalues of AA’ are as follows: λ_{1} = 3, λ_{2} = 1, and λ_{3} = 0. The eigenvalues λ_{1}, λ_{2}, and λ_{3} (respectively) result in the eigenspaces E_{1}, E_{2}, and E_{3} as shown below.

Consequently, we have the orthogonal matrix U as follows.

Note that if A = UΣV’ then A’ = VΣ’U’ and A’U = VΣ’ = VΣ. Since A’U = VΣ, it follows that where , = i-th column vector of U and = i-th column vector of V. Therefore, applies. As a consequence,

So, and the singular value decomposition of A is:

**Example 2**

Find the singular value decomposition of .

*Answer*

V can be determined from the spectral decomposition of the symmetric matrix A’A. Note that:

By finding the roots of |λI – A’A| = 0, we have the eigenvalues of AA’, i.e. λ_{1} = 25, λ_{2} = 9, and λ_{3} = 0. The eigenvalues λ_{1}, λ_{2}, and λ_{3} , respectively, result in the eigenspaces E_{1}, E_{2}, and E_{3} as follows.

As a consequence, we have the orthogonal matrix V as follows:

Note that if A = UΣV’ then A’ = VΣ’U’ and AV = UΣ. From the relationship AV = UΣ we have where , = i-th column vector of V and = i-th column vector of U. So, applies. As a consequence, it follows that:

So, and the singular value decomposition of A is: