Recipe

Find singular values and vectors from the maximum-stretch definition

Use when: $A$ is diagonal (or already diagonalized) and you want $\sigma_i$ , $\mathbf v_i$ directly from the variational definition, not from $A^{\mathsf T}A$

Goal: every singular value and its matching right singular vector, found one maximization at a time

Taught in Singular values and singular vectors

Quick recipe

Write $\lVert A\mathbf x\rVert^2$ as a weighted sum
Maximize by putting all the weight on the largest coefficient
Restrict to what's left and repeat
Verify each pair directly

1
For diagonal $A$ with entries $c_1,\dots,c_n$ on the diagonal, $\lVert A\mathbf x\rVert^2=c_1^2x_1^2+\cdots+c_n^2x_n^2$ , a weighted sum of squares.

Example

For $A=\text{diag}(3,1,2)$ : $\lVert A\mathbf x\rVert^2=9x_1^2+x_2^2+4x_3^2$ , using each diagonal entry squared as the weight on its own coordinate.

Why: this rewrites the abstract maximum-stretch definition as an ordinary constrained-optimization problem in the unit vector's own coordinates.
2
Among unit vectors, a weighted sum of squares is maximized by setting the coordinate with the largest weight to $1$ and every other coordinate to $0$ .

Example

The largest weight above is $9$ (the $x_1^2$ term), so the maximum is achieved at $\mathbf x=(1,0,0)$ , giving $\sigma_1^2=9$ , so $\sigma_1=3$ and $\mathbf v_1=(1,0,0)$ .

⚡ Shortcut

For a diagonal matrix, skip the optimization argument entirely: sort the absolute values of the diagonal entries in decreasing order, and that sorted list already is $\sigma_1,\sigma_2,\dots$

Why: this is exactly the constrained maximum $\sigma_1=\max_{\lVert\mathbf x\rVert=1}\lVert A\mathbf x\rVert$ , restated as ordinary algebra once $A$ is diagonal.
3
Set every coordinate already claimed by an earlier singular vector to zero, then maximize the remaining weighted sum among the coordinates that are left.

Example

With $x_1$ fixed at $0$ , maximize $x_2^2+4x_3^2$ subject to $x_2^2+x_3^2=1$ : the larger weight is $4$ , giving $\mathbf v_2=(0,0,1)$ , $\sigma_2=2$ . Only one coordinate remains, forcing $\mathbf v_3=(0,1,0)$ , $\sigma_3=1$ .
- every remaining coefficient is distinct — the next singular vector is forced, the standard basis vector for the next-largest remaining weight.
- two or more remaining coefficients tie for largest — any pair of orthogonal directions spanning that tied space is a valid choice.
Why: restricting to what's left after every earlier choice guarantees each new singular vector comes out perpendicular to the earlier ones automatically.
4
Compute $A\mathbf v_i$ and confirm its length equals the claimed $\sigma_i$ , then confirm every pair of singular vectors has zero dot product.

Example

$A(1,0,0)=(3,0,0)$ , length $3$ , matching $\sigma_1$ . $(1,0,0)\cdot(0,0,1)=0$ , confirming $\mathbf v_1$ and $\mathbf v_2$ are orthogonal.

Why: this catches a bookkeeping slip, a wrong coordinate flagged as the maximizer, before it propagates into an incorrect $\Sigma$ , $U$ , or $V$ .

Common mistakes

Reporting a negative diagonal entry as a negative singular value; $\sigma_i$ is always $\lvert c_i\rvert$ , and the sign shows up in $\mathbf u_i$ instead.
Forgetting to restrict to what's left before finding the next singular vector, and re-maximizing over the full space instead.
Picking an arbitrary vector in a tied direction without checking it's actually orthogonal to every earlier singular vector.
Skipping the length check in Step 4 and simply trusting whichever coordinate looked biggest.