SVD Calculator

Singular value decomposition (SVD) is a way of factorizing a matrix: any real matrix AA of size m×nm×n decomposes as

where UU and VV are orthogonal matrices of sizes m×mm×m and n×nn×n respectively, and ΣΣ is a rectangular matrix of the same size as AA (so m×nm×n) which has non-negative numbers on its diagonal and zeroes everywhere else. The diagonal elements of ΣΣ are in fact the singular values of AA.

🙋 If AA is complex, replace the transposition (we also have matrix transpose calculator to help you with that) VTVT with the complex conjugation V∗V∗. UU and VV then become unitary matrices, but ΣΣ still features real non-negative numbers on its diagonal.

Here's a visualization of singular value decomposition of a 4×34×3 matrix MM.

Once we know what the singular value decomposition of a matrix is, it'd be beneficial to see some examples. Calculating SVD by hand is a time-consuming procedure, as we will see in the section on How to calculate SVD of a matrix. We bet the quickest way to generate examples of SVD is to use singular value decomposition calculator!

Working with this SVD calculator is simple!

1. Pick the matrix size: the number of rows and the number of columns in AA.
2. Enter the matrix entries in their dedicated fields.
3. The components of singular value decomposition UU, ΣΣ and VTVT will appear at the bottom of the calculator.

Do you want to verify the results? Just perform the matrix multiplication of the result's three matrices and compare that outcome with your initial matrix. Remember that numerical computations and rounding may cause tiny discrepancies!

Do you want to understand how the SVD calculator got its results? In the next section, we will discuss all the theory that stands behind the singular value decomposition and explain step-by-step how to find the SVD of a matrix. Ready?

Here's how to calculate the singular value decomposition of a m×nm×n matrix AA by hand. We will see that SVD is closely related to the eigenvalues and eigenvectors of AA.

As we remember, we can easily find the eigenvalues and eigenvectors for square matrices, yet AA can be rectangular in SVD. What can we do? Let's consider two square matrices that are closely related to AA: these matrices are ATAATA and AATAAT:

• The columns of VV are eigenvectors of ATAATA.
• The non-zero elements of ΣΣ are the non-zero singular values of AA, i.e., they are the square roots of the non-zero eigenvalues of ATAATA.
• Once we know VV and ΣΣ, we can recover UU from the SVD formula (A=UΣVTA=UΣVT).

In more details, to find SVD by hand:

1. Compute ATAATA.

2. Compute the eigenvalues and eigenvectors of ATAATA.

3. Draw a matrix of the same size as AA and fill in its diagonal entries with the square roots of the eigenvalues you found in Step 2. This is ΣΣ.

4. Write down the matrix whose columns are the eigenvectors you found in Step 2. This is VV.

5. The SVD equation A=UΣVTA=UΣVT transforms to AV=UΣAV=UΣ. We can rewrite this in terms of columns as Avi​=σi​ui​Avi​=σi​ui​. This tells us how to compute UU, as the columns of UU set ui​=σi​1​Avi​ui​=σi​1​Avi​ for every ii such that σi​=0σi​=0.

6. If UU needs more columns to fill its size, you can pick arbitrary vectors, but you have to make sure that UU is an orthogonal matrix. Therefore, you must pick vectors that have unit length and are orthogonal to all the columns in UU (and the ones you're adding).

No, the SVD is not unique. Even if we agree to have the diagonal elements of Σ in descending order (which makes Σ unique), the matrices U and V are still non-unique.