Sparse SVD (svds) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Sparse SVD on a Simple Sparse Matrix

What is the shape of the matrix U returned by svds when decomposing a 5x5 sparse matrix with k=2?

SciPy

from scipy.sparse import csr_matrix
from scipy.sparse.linalg import svds
import numpy as np

A = csr_matrix(np.array([[1, 0, 0, 0, 2],
                         [0, 0, 3, 0, 0],
                         [0, 4, 0, 0, 0],
                         [0, 0, 0, 5, 0],
                         [6, 0, 0, 0, 0]]))

U, s, Vt = svds(A, k=2)
print(U.shape)

A (5, 2)

B (2, 5)

C (2, 2)

D (5, 5)

💡 Hint

Remember that U has the same number of rows as the original matrix and k columns.

The svds function returns U with shape (m, k) where m is the number of rows in the input matrix and k is the number of singular values requested.

❓ data_output

intermediate

Singular Values from Sparse SVD

Given a sparse matrix A and svds(A, k=3), what is the length of the singular values vector s?

SciPy

from scipy.sparse import diags
from scipy.sparse.linalg import svds

A = diags([1, 2, 3, 4, 5])
U, s, Vt = svds(A, k=3)
print(len(s))

A 1

B 5

C 3

D 0

💡 Hint

The singular values vector length equals k, the number of singular values requested.

The svds function returns exactly k singular values in the vector s.

🔧 Debug

advanced

Error Raised by svds with k Larger Than Matrix Dimensions

What error does the following code raise? from scipy.sparse import csr_matrix from scipy.sparse.linalg import svds import numpy as np A = csr_matrix(np.eye(4)) U, s, Vt = svds(A, k=5)

SciPy

from scipy.sparse import csr_matrix
from scipy.sparse.linalg import svds
import numpy as np

A = csr_matrix(np.eye(4))
U, s, Vt = svds(A, k=5)

A ValueError: k must be smaller than or equal to min(A.shape) - 1

B TypeError: svds() got an unexpected keyword argument 'k'

C RuntimeError: svds did not converge

D No error, returns U, s, Vt with shapes (4,5), (5,), (5,4)

💡 Hint

Check the allowed range for k relative to matrix dimensions.

The svds function requires k to be less than or equal to min(matrix dimensions) - 1. Here, k=5 is too large for a 4x4 matrix.

🚀 Application

advanced

Using svds to Approximate a Matrix

You have a large sparse matrix A of shape (1000, 1000). You want to approximate it using svds with k=10. Which of the following code snippets correctly reconstructs the approximate matrix?

SciPy

from scipy.sparse.linalg import svds

U, s, Vt = svds(A, k=10)
# Which code reconstructs the approximation?

A approx = U * s * Vt

B approx = U @ np.diag(s) @ Vt

C approx = np.dot(U, s, Vt)

D approx = U + s + Vt

💡 Hint

Recall how to multiply matrices and diagonal matrices in numpy.

The correct reconstruction is U times the diagonal matrix of singular values times Vt. Using @ operator and np.diag(s) is correct.

🧠 Conceptual

expert

Why Use svds Instead of Full SVD on Sparse Matrices?

Which reason best explains why svds is preferred over full SVD for large sparse matrices?

A svds can only be used on dense matrices, not sparse

B svds always produces more accurate singular values than full SVD

C svds converts sparse matrices to dense internally for faster computation

D svds computes only a few singular values and vectors, saving time and memory

💡 Hint

Think about computational cost and memory when working with large sparse data.

svds is designed to compute a small number of singular values and vectors efficiently without converting the whole matrix to dense form, saving time and memory.