LINEAR-ALGEBRA · Interactive Practice | Unit 28 · Video 2

Killing U to Find V: The AᵀA Shortcut for SVD

IKey Formulas

Formula	Meaning
$A = U \Sigma V^T$	The SVD decomposition
$A^T A = V \Sigma^2 V^T$	Symmetric eigendecomposition; columns of $V$ are eigenvectors
$A A^T = U \Sigma^2 U^T$	Symmetric eigendecomposition; columns of $U$ are eigenvectors
$u_i = \frac{1}{\sigma_i} A v_i$	Recover $U$ from $V$ to lock signs

The singular values satisfy $\sigma_i = \sqrt{\lambda_i}$ , where $\lambda_i$ are the eigenvalues of $A^T A$ .

IIVisualization 1 — Watching $U$ Vanish

Forming $A^T A$ from $A = U\Sigma V^T$ collapses it to $V\Sigma^2 V^T$ — every trace of $U$ disappears.

σ₁ 2.00 σ₂ 1.00 V angle 30° U angle 60°

IIIVisualization 2 — Shared Nonzero Eigenvalues

$A^T A$ and $A A^T$ have different sizes, yet share every nonzero eigenvalue — each one equals a $\sigma_i^2$ .

a₁₁ 2 a₁₂ 1 a₁₃ 0 a₂₁ 0 a₂₂ 1.5 a₂₃ 2

IVVisualization 3 — The Sign Trap

Eigenvectors carry a sign ambiguity; deriving $u_i = \tfrac{1}{\sigma_i} A v_i$ locks $U$ 's signs to $V$ .

U method

VQuiz Questions

Test your understanding of the $A^T A$ shortcut.

Question 1

When we form $A^T A$ starting from $A = U \Sigma V^T$ , which property of $U$ causes it to "disappear" from the expression?

✅ Correct! Orthogonality of $U$ collapses the inner product to the identity.

❌ Not quite. The cancellation requires $U^T U = I$ , which holds because $U$ is orthogonal.

Show solution

Solution:

$A^T A = (V \Sigma^T U^T)(U \Sigma V^T)$

The middle factor is $U^T U$ . Because $U$ is orthogonal, $U^T U = I$ , and the expression collapses to:

$A^T A = V \Sigma^T \Sigma V^T = V \Sigma^2 V^T$

$U$ vanishes entirely. The fact that $U$ is square or that $\Sigma$ has a particular shape is not what makes the cancellation work — orthogonality is.

Question 2

Suppose $A^T A$ has eigenvalues $\{25, 9, 0\}$ . What are the singular values of $A$ ?

✅ Correct! $\sigma_i = \sqrt{\lambda_i}$ .

❌ Not quite. Singular values are the square roots of the eigenvalues of $A^T A$ , and zero is included.

Show solution

Solution:

The singular values are the square roots of the eigenvalues of $A^T A$ :

$\sigma_i = \sqrt{\lambda_i}$

So $\sigma_1 = \sqrt{25} = 5$ , $\sigma_2 = \sqrt{9} = 3$ , and $\sigma_3 = \sqrt{0} = 0$ .

Zero is a valid singular value — it indicates that $A$ has a non-trivial null space. A common mistake is to forget the square root and report the eigenvalues themselves.

Question 3

Let $A$ be a $3 \times 5$ matrix. Which of the following is true about the eigenvalues of $A^T A$ and $A A^T$ ?

✅ Correct! The larger matrix carries the extra zeros.

❌ Not quite. Nonzero eigenvalues are shared. The bigger matrix ( $5 \times 5$ here) carries the extra zeros.

Show solution

Solution:

$A^T A$ is $5 \times 5$ and $A A^T$ is $3 \times 3$ . The general fact $AB$ and $BA$ share their nonzero eigenvalues tells us both matrices have the same nonzero eigenvalues — these are $\sigma_1^2, \sigma_2^2, \sigma_3^2$ (at most $\min(m, n) = 3$ of them).

The larger matrix, $A^T A$ ( $5 \times 5$ ), simply carries extra zeros. It has $5 - 3 = 2$ additional zero eigenvalues compared to $A A^T$ .

Question 4

True or False: After computing $V$ from the eigendecomposition of $A^T A$ , the safest way to obtain $U$ is to independently compute the eigendecomposition of $A A^T$ .

✅ Correct! Independent computation risks sign disagreements; instead use $u_i = (1/\sigma_i) A v_i$ .

❌ Not quite. Independent computation can break the sign convention. The safe recovery is $u_i = (1/\sigma_i) A v_i$ .

Show solution

Solution: False.

Eigenvectors are only defined up to a sign. Computing $V$ and $U$ independently can produce sign mismatches: $v_i$ and $u_i$ might end up "pointing the wrong way" relative to each other, breaking $A = U \Sigma V^T$ .

The fix is to compute $V$ first, then derive $U$ from $V$ using:

$u_i = \frac{1}{\sigma_i} A v_i$

This locks the sign of $u_i$ to the choice made for $v_i$ , guaranteeing the reconstruction $U \Sigma V^T$ equals $A$ exactly.

Solved: 0 / 4

Killing U to Find V: The AᵀA Shortcut for SVD

IKey Formulas

IIVisualization 1 — Watching UUU Vanish

IIIVisualization 2 — Shared Nonzero Eigenvalues

IVVisualization 3 — The Sign Trap

VQuiz Questions

IIVisualization 1 — Watching $U$ Vanish