Linear Algebra Quiz: Singular Value Decomposition Svd

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Singular Value Decomposition SvdQuestion 1 of 8

For a matrix $A$ with singular value decomposition $A = U\Sigma V^T$ , suppose the singular values are $\sigma_1 = 8$ , $\sigma_2 = 3$ , and $\sigma_3 = 0$ . If $B = A^T A$ , what is the trace of $B$ ?

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Linear Algebra Quiz

Linear Algebra Quiz: Singular Value Decomposition Svd

Practice Singular Value Decomposition Svd in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Singular Value Decomposition Svd, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

73 (correct answer)
64
11
24

Explanation: Since

A = U\Sigma V^T

, we have

A^T = V\Sigma^T U^T

. Therefore,

B = A^T A = V\Sigma^T U^T U\Sigma V^T = V\Sigma^T\Sigma V^T

. The eigenvalues of

B

are the diagonal entries of

\Sigma^T\Sigma

, which are

\sigma_1^2 = 64

\sigma_2^2 = 9

, and

\sigma_3^2 = 0

. The trace equals the sum of eigenvalues:

64 + 9 + 0 = 73

. Choice B uses only

\sigma_1^2

. Choice C uses

\sigma_1 + \sigma_2

. Choice D uses

3 \times \sigma_1

Question 2

Consider a matrix $A$ with SVD $A = U\Sigma V^T$ where $\Sigma = \text{diag}(6, 4, 2, 0)$ . If we form the rank-2 approximation $A_2$ using the first two singular values, what percentage of the Frobenius norm of $A$ is preserved in $A_2$ ?

Approximately 76.9%
Approximately 83.3%
Exactly 50%
Approximately 92.3% (correct answer)

Explanation: When you encounter SVD approximation problems, you're dealing with how well lower-rank matrices can represent the original data. The key insight is that the Frobenius norm has a direct relationship with singular values. For any matrix with SVD

A = U\Sigma V^T

, the Frobenius norm squared equals the sum of squared singular values:

||A||_F^2 = \sigma_1^2 + \sigma_2^2 + \sigma_3^2 + \sigma_4^2

. With

\Sigma = \text{diag}(6, 4, 2, 0)

, we get

||A||_F^2 = 36 + 16 + 4 + 0 = 56

. The rank-2 approximation

A_2

uses only the first two singular values, so

||A_2||_F^2 = 36 + 16 = 52

. The percentage preserved is

\frac{52}{56} = \frac{13}{14} \approx 0.923 = 92.3\%

. Looking at the wrong answers: Choice A (76.9%) might come from incorrectly using

\frac{6+4}{6+4+2+0} = \frac{10}{12}

, which adds singular values instead of squaring them. Choice B (83.3%) could result from the calculation

\frac{10}{12} = 0.833

, also from the linear addition error. Choice C (50%) might stem from thinking half the singular values means half the norm is preserved, ignoring that larger singular values contribute disproportionately. Remember: In SVD approximations, always square the singular values when working with Frobenius norms. The largest singular values capture most of the matrix's "energy," so low-rank approximations often preserve surprisingly high percentages of the original norm.

Question 3

Consider the matrix $A = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$ . If the largest singular value of $A$ is approximately 5.46, what is the smallest singular value approximately?

0.73
1.46
2.00
0.37 (correct answer)

Explanation: For a

2 \times 2

matrix, the product of singular values equals

|\det(A)|

. Here,

\det(A) = 3 \cdot 2 - 1 \cdot 4 = 2

. If

\sigma_1 \approx 5.46

, then

\sigma_1 \sigma_2 = 2

, so

\sigma_2 = \frac{2}{5.46} \approx 0.37

. Choice A uses

\frac{4}{5.46}

. Choice B subtracts:

5.46 - 4

. Choice C assumes the singular values are eigenvalues and uses symmetry incorrectly.

Question 4

A data matrix $X$ has SVD $X = U\Sigma V^T$ with singular values $\{10, 6, 3, 1\}$ . For principal component analysis, if we want to retain at least 90% of the variance, how many principal components should we keep?

3
2 (correct answer)
4
1

Explanation: In PCA, the variance explained by each component is proportional to

\sigma_i^2

. Total variance:

10^2 + 6^2 + 3^2 + 1^2 = 100 + 36 + 9 + 1 = 146

. We need 90% of 146 = 131.4. First component: 100 (68.5%). First two: 100 + 36 = 136 (93.2%). Since 136 > 131.4, we need 2 components. Choice A uses 3 components (unnecessary). Choice C uses all components. Choice D uses only 1 component (insufficient at 68.5%).

Question 5

If $A$ is a $3 \times 4$ matrix with singular values $\sigma_1 = 7$ , $\sigma_2 = 5$ , $\sigma_3 = 0$ , what is the dimension of the null space of $A^T$ ?

3
2
1 (correct answer)
0

Explanation: When you encounter singular values in a linear algebra problem, you're dealing with the Singular Value Decomposition (SVD), which reveals crucial information about a matrix's fundamental subspaces. The key insight is that zero singular values directly tell you about the dimensions of the null spaces. For any

m \times n

matrix, the SVD gives you

r

non-zero singular values, where

r

is the rank of the matrix. Since

A

has singular values

\sigma_1 = 7

\sigma_2 = 5

, and

\sigma_3 = 0

, only two are non-zero, so

\text{rank}(A) = 2

. The dimension of the null space of

A^T

equals

m - \text{rank}(A)

, where

m

is the number of rows. Since

A

3 \times 4

, we have

\dim(\text{null}(A^T)) = 3 - 2 = 1

. This makes C correct. Looking at the wrong answers: A) 3 incorrectly assumes the null space dimension equals the number of rows. B) 2 mistakenly uses the rank itself rather than subtracting it from the row count. D) 0 would only be true if

A

had full row rank (rank 3), which would require all three singular values to be non-zero. Remember this pattern: for an

m \times n

matrix with

r

non-zero singular values,

\dim(\text{null}(A^T)) = m - r

and

\dim(\text{null}(A)) = n - r

. The number of zero singular values always equals the total dimension of both null spaces combined.

Question 6

Let $A$ be a $4 \times 5$ matrix with rank 3. In the singular value decomposition $A = U\Sigma V^T$ , what are the dimensions of the matrices $U$ , $\Sigma$ , and $V$ ?

$U$ is $4 \times 4$ , $\Sigma$ is $4 \times 5$ , $V$ is $5 \times 5$ (correct answer)
$U$ is $4 \times 3$ , $\Sigma$ is $3 \times 3$ , $V$ is $5 \times 3$
$U$ is $4 \times 5$ , $\Sigma$ is $5 \times 5$ , $V$ is $5 \times 4$
$U$ is $3 \times 4$ , $\Sigma$ is $4 \times 5$ , $V$ is $3 \times 5$

Explanation: In the full SVD,

U

is always

m \times m

(here

4 \times 4

\Sigma

m \times n

(here

4 \times 5

), and

V

n \times n

(here

5 \times 5

). The rank only affects how many singular values are nonzero, not the matrix dimensions. Choice B gives the reduced SVD dimensions. Choice C reverses some dimensions. Choice D incorrectly uses the rank as a matrix dimension.

Question 7

In the reduced SVD of an $m \times n$ matrix $A$ with rank $r$ , which statement about the orthogonal matrices is correct?

$U$ has $m$ orthonormal columns and $V$ has $n$ orthonormal columns
$U$ has $r$ orthonormal columns and $V$ has $r$ orthonormal columns (correct answer)
$U$ has $r$ orthonormal rows and $V$ has $r$ orthonormal rows
$U$ has $\min(m,r)$ orthonormal columns and $V$ has $\min(n,r)$ orthonormal columns

Explanation: When you encounter SVD questions, focus on understanding what "reduced" means and how rank affects matrix dimensions. The reduced SVD is a compact form that eliminates unnecessary zero singular values and their corresponding vectors. In the reduced SVD

A = U\Sigma V^T

, the matrix

A

has rank

r

, meaning it has exactly

r

non-zero singular values. The reduced form constructs matrices to match this rank:

U

becomes

m \times r

\Sigma

becomes

r \times r

, and

V^T

becomes

r \times n

. Since

U

has dimensions

m \times r

, it contains

r

orthonormal columns. Similarly,

V

has dimensions

n \times r

, so it also contains

r

orthonormal columns. Choice A describes the full SVD, not the reduced version. In the full SVD,

U

would indeed have

m

orthonormal columns and

V

would have

n

orthonormal columns, but this includes columns corresponding to zero singular values that are eliminated in the reduced form. Choice C incorrectly focuses on rows rather than columns. While

U

and

V

do have

r

rows each in their transpose forms, the question asks about the orthogonal matrices themselves, where we count orthonormal columns. Choice D suggests the dimensions depend on

\min(m,r)

and

\min(n,r)

, but since

r

cannot exceed either

m

n

(rank is bounded by both dimensions), these expressions simply equal

r

. Study tip: Remember that "reduced" SVD means "rank-sized" - everything scales to match the rank

r

Question 8

For a matrix $A$ with SVD $A = U\Sigma V^T$ , which expression correctly represents the pseudoinverse $A^+$ ?

$V^T\Sigma^+ U$ where $\Sigma^+$ has $1/\sigma_i$ for nonzero $\sigma_i$ and 0 elsewhere
$U\Sigma^{-1} V^T$ where $\Sigma^{-1}$ has $1/\sigma_i$ for all $\sigma_i$
$V\Sigma^+ U^T$ where $\Sigma^+$ has $1/\sigma_i$ for nonzero $\sigma_i$ and 0 elsewhere (correct answer)
$(V\Sigma U^T)^{-1}$ computed using standard matrix inversion

Explanation: When you encounter pseudoinverse problems, remember that the pseudoinverse extends the concept of matrix inversion to non-square or singular matrices using the Singular Value Decomposition (SVD). The pseudoinverse formula derives directly from the SVD structure. If

A = U\Sigma V^T

, then the pseudoinverse is

A^+ = V\Sigma^+ U^T

. Notice how this "reverses" the SVD: the

V

and

U

matrices swap positions and lose their transposes, while

\Sigma^+

is the pseudoinverse of the diagonal matrix

\Sigma

. For

\Sigma^+

, you take the reciprocal of each nonzero singular value and leave zeros as zeros (since

1/0

is undefined). This construction ensures

A^+

exists even when

A

isn't invertible. Answer C correctly gives

V\Sigma^+ U^T

with the proper handling of

\Sigma^+

. Answer A incorrectly uses

V^T\Sigma^+ U

, which has the wrong transposes—this would give you something like

(A^T)^+

instead of

A^+

. Answer B attempts

U\Sigma^{-1} V^T

, but

\Sigma^{-1}

doesn't exist when

A

has zero singular values, and the matrix order is wrong anyway. Answer D suggests using standard matrix inversion on

V\Sigma U^T

, but this expression isn't even equal to

A

, and standard inversion fails for non-invertible matrices. Study tip: Remember the pseudoinverse formula as "flip and transpose": from

U\Sigma V^T

V\Sigma^+ U^T

. The matrices swap positions and drop their transposes, while

\Sigma^+

uses reciprocals of nonzero values only.