Singular value decomposition (svd) in MATLAB - Step-by-Step Execution

Concept Flow - Singular value decomposition (svd)

Start with matrix A

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Compute U, S, V such that A = U*S*V'

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U: orthogonal matrix (left singular vectors)

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S: diagonal matrix (singular values)

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V: orthogonal matrix (right singular vectors)

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Use U, S, V for analysis or reconstruction

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End

Start with matrix A, compute matrices U, S, V so that A equals U times S times V transpose, then use these matrices for further tasks.

Execution Sample

MATLAB

A = [3 1 1; -1 3 1];
[U,S,V] = svd(A);
U
S
V

This code computes the singular value decomposition of matrix A and shows the matrices U, S, and V.

Execution Table

Step	Action	Evaluation	Result
1	Define matrix A	A = [3 1 1; -1 3 1]	A = [3 1 1; -1 3 1]
2	Call svd(A)	Compute U, S, V such that A = U*S*V'	U, S, V matrices computed
3	Display U	Show left singular vectors	U = [0.71 0.71; 0.71 -0.71] (approx)
4	Display S	Show singular values on diagonal	S = diag([3.87, 2.24]) (approx)
5	Display V	Show right singular vectors	V = [0.82 0.41 0.41; -0.45 0.89 0; 0.37 0.18 -0.91] (approx)
6	Verify reconstruction	Calculate U*S*V'	Reconstructed A matches original A
7	End	No more steps	Execution complete

💡 All matrices computed and verified, svd process complete

Variable Tracker

Variable	Start	After svd	Final
A	[3 1 1; -1 3 1]	[3 1 1; -1 3 1]	[3 1 1; -1 3 1]
U	undefined	[0.71 0.71; 0.71 -0.71]	[0.71 0.71; 0.71 -0.71]
S	undefined	[3.87 0 0; 0 2.24 0]	[3.87 0 0; 0 2.24 0]
V	undefined	[0.82 0.41 0.41; -0.45 0.89 0; 0.37 0.18 -0.91]	[0.82 0.41 0.41; -0.45 0.89 0; 0.37 0.18 -0.91]

Key Moments - 3 Insights

Why is matrix S diagonal and not full?

Matrix S contains singular values only on its diagonal, representing the strength of each component. This is shown in execution_table step 4 where S is diagonal with singular values.

Why do we multiply U, S, and V' to get back A?

Because svd decomposes A into U, S, and V such that A = U*S*V'. This is verified in execution_table step 6 where reconstruction matches original A.

Are U and V always square and orthogonal?

Yes, U and V are orthogonal matrices with orthonormal columns, as shown in execution_table steps 3 and 5.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what does matrix S represent?

A Matrix of left singular vectors

B Matrix of right singular vectors

C Diagonal matrix of singular values

D Original matrix A

💡 Hint

Refer to execution_table row 4 where S is described as singular values on diagonal

At which step in the execution_table do we verify that U*S*V' reconstructs A?

A Step 3

B Step 6

C Step 5

D Step 2

💡 Hint

Check execution_table row 6 where reconstruction is verified

If matrix A was square and symmetric, what would you expect about U and V?

A U and V would be identical

B U would be zero matrix

C V would be diagonal

D S would be zero matrix

💡 Hint

Recall that for symmetric matrices, U and V are equal (related to eigen decomposition)

Concept Snapshot

Singular value decomposition (svd) splits matrix A into U, S, V such that A = U*S*V'.
U and V are orthogonal matrices with singular vectors.
S is diagonal with singular values.
Use svd to analyze or reconstruct A.
In MATLAB: [U,S,V] = svd(A);

Full Transcript

Singular value decomposition (svd) breaks a matrix A into three matrices: U, S, and V. U and V are orthogonal matrices containing singular vectors, and S is a diagonal matrix with singular values. The product U*S*V' reconstructs the original matrix A. In MATLAB, you use [U,S,V] = svd(A) to compute these matrices. This process helps analyze the structure of A and is useful in many applications like data compression and noise reduction.