Recipe

Compute a matrix's pseudoinverse

Use when: you need $A^{+}$ for a matrix $A$ that may be singular, rectangular, or otherwise not invertible

Goal: the pseudoinverse $A^{+}=V\Sigma^{+}U^{\mathsf T}$ , matching $A^{-1}$ exactly whenever $A$ genuinely is invertible

Taught in The pseudoinverse

Quick recipe

Find the SVD of $A$
Identify which singular values are nonzero
Build $\Sigma^{+}$ by inverting the nonzero entries
Multiply back: $A^{+}=V\Sigma^{+}U^{\mathsf T}$
Check the result against what you'd expect

1
Write $A=U\Sigma V^{\mathsf T}$ , with $\Sigma$ 's diagonal holding the singular values in decreasing order.

Example

$A=\begin{bmatrix}2&2\\2&2\end{bmatrix}$ is symmetric, so its SVD coincides with its eigendecomposition. Its eigenvalues are $4$ and $0$ (trace $4$ , determinant $0$ ), with normalized eigenvectors $(1,1)/\sqrt2$ and $(1,-1)/\sqrt2$ . $\Sigma=\begin{bmatrix}4&0\\0&0\end{bmatrix}, \quad U=V=\frac{1}{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$

⚡ Shortcut

If $A$ is already known to be square and invertible, skip the SVD entirely: $A^{+}=A^{-1}$ , the ordinary inverse, exactly.

Why: the pseudoinverse is built directly from the SVD's three pieces, so the whole construction has to start by finding them.
2
Read off $\Sigma$ 's diagonal and note which entries are strictly positive; call that count the rank $r$ .

Example

Here $\Sigma=\text{diag}(4,0)$ : the first singular value is nonzero, the second is exactly $0$ , so $r=1$ .
- every singular value is nonzero — $A$ is invertible (if square), and $A^{+}$ will equal $A^{-1}$ exactly.
- at least one singular value is zero — $A$ is singular or rank-deficient, and $A^{+}$ genuinely differs from any two-sided inverse.
Why: $\Sigma^{+}$ inverts only the nonzero singular values, so this check decides which entries actually get inverted in the next step.
3
Replace each nonzero singular value with its reciprocal, keep every zero exactly as zero, and match $\Sigma^{+}$ 's shape to $\Sigma^{\mathsf T}$ 's own shape.

Example

$\Sigma=\text{diag}(4,0)$ is $2\times2$ , so $\Sigma^{+}$ is also $2\times2$ : invert the $4$ to $1/4$ , leave the $0$ as $0$ . $\Sigma^{+}=\begin{bmatrix}1/4&0\\0&0\end{bmatrix}$

Why: this is the one place the pseudoinverse departs from an ordinary inverse; a zero singular value would make an ordinary inverse blow up, but $\Sigma^{+}$ passes it through as zero.
4
Carry out the matrix product in exactly that order: $V$ first, then $\Sigma^{+}$ , then $U$ transposed.

Example

$A^{+}=\frac1{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}1/4&0\\0&0\end{bmatrix}\frac1{\sqrt2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}=\begin{bmatrix}1/8&1/8\\1/8&1/8\end{bmatrix}$

Why: reassembling in this exact order is what turns the three separate pieces back into one genuine matrix acting on the original space.
5
If $A$ was invertible, confirm $A^{+}$ equals $A^{-1}$ exactly; if $A$ was singular, confirm $AA^{+}A=A$ instead.

Example

$AA^{+}=\begin{bmatrix}2&2\\2&2\end{bmatrix}\begin{bmatrix}1/8&1/8\\1/8&1/8\end{bmatrix}=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}$ , and multiplying that by $A$ again gives back $\begin{bmatrix}2&2\\2&2\end{bmatrix}=A$ .

Why: this confirms the construction satisfies the one property every genuine pseudoinverse must have, even though $AA^{+}$ itself is visibly not the identity.

Common mistakes

Inverting a zero singular value instead of leaving it as zero; this produces a bogus, undefined entry.
Forgetting to match $\Sigma^{+}$ 's shape to $\Sigma^{\mathsf T}$ when $A$ , and therefore $\Sigma$ , is not square.
Assuming $AA^{+}$ or $A^{+}A$ equals the identity for a singular or rectangular $A$ ; neither one generally does.
Multiplying $U$ , $\Sigma^{+}$ , $V$ in the wrong order; it must be $V\Sigma^{+}U^{\mathsf T}$ , never $U\Sigma^{+}V^{\mathsf T}$ .