Pseudoinverse: Worked examples — Maths deep dive

This page is the drill hall for the Pseudoinverse. The parent note built the three formulas; here we throw every kind of matrix at them — tall, fat, square-invertible, square-singular, rank-deficient (both tall and fat flavours), and a real-world data-fit — so no scenario can surprise you in an exam.

Before we start, one promise: every symbol below was earned in the parent note. As a lightning refresher of the only three tools — plus the four rules that define what a pseudoinverse even is:

Recall The three formulas (from the parent)

Tall & full column rank ( $n$ independent columns, $m \geq n$ ): $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . This is a left inverse: $A^{+} A = I_{n}$ .
Fat & full row rank ( $m$ independent rows, $n \geq m$ ): $A^{+} = A^{⊤} (A A^{⊤})^{- 1}$ . This is a right inverse: $A A^{+} = I_{m}$ .
Any shape, any rank: $A^{+} = V Σ^{+} U^{⊤}$ from the Singular Value Decomposition, inverting each nonzero $σ_{i}$ , leaving zeros dead.

Here $A^{⊤}$ means "flip $A$ across its diagonal" (rows become columns), and $∥ x ∥$ means the ordinary length $x_{1}^{2} + x_{2}^{2} + \dots$ .

Recall The four Penrose conditions (what "

$A^{+}$ " means) — and the two parent mistakes we lean on $A^{+}$ is the unique matrix satisfying all four:

$A A^{+} A = A$ 2. $A^{+} A A^{+} = A^{+}$ 3. $(A A^{+})^{⊤} = A A^{+}$ 4. $(A^{+} A)^{⊤} = A^{+} A$ . Condition 1 is the one we use again and again below as a "plug-back" check.

Two traps from the parent note we will trip on purpose:

Mistake 1: " $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ always." It needs full column rank; otherwise $A^{⊤} A$ is singular (Examples 5, 8).
Mistake 2: " $A^{+} A = I$ always." Only for full column rank. In general $A^{+} A$ is a projection (it can have eigenvalue 0), seen in Examples 5, 8, 9.

The scenario matrix

Every matrix you can feed a pseudoinverse falls into exactly one of these boxes. The whole point of $A^{+}$ is that it never crashes — but which formula is safe changes box to box. Below, "rank" = number of genuinely independent rows/columns (directions the matrix does not squash to zero).

Cell	Shape & rank	What goes wrong with $A^{- 1}$ ?	Which formula is safe	Example
C-tall	tall, full column rank	not square — no $A^{- 1}$ ; $A x = b$ has no exact solution	$(A^{⊤} A)^{- 1} A^{⊤}$ (least squares)	Ex 1, Ex 2
C-fat	fat, full row rank	not square; infinitely many solutions	$A^{⊤} (A A^{⊤})^{- 1}$ (min-norm)	Ex 3
C-sqinv	square, invertible	nothing! $A^{- 1}$ exists	any — all agree, $A^{+} = A^{- 1}$	Ex 4
C-sqsing	square, singular (rank deficient)	$det = 0$ , no $A^{- 1}$ ; both $A^{⊤} A$ and $A A^{⊤}$ singular	must use SVD	Ex 5
C-zero	the zero matrix (degenerate limit)	everything is nullspace	SVD gives $A^{+} = 0$ (see Ex 6 for why)	Ex 6
C-word	real data fit (tall in disguise)	overdetermined, noisy	least squares	Ex 7
C-twist-t	exam trap: rank-deficient tall	looks like C-tall but $A^{⊤} A$ is singular!	SVD, not the tall formula	Ex 8
C-twist-f	exam trap: rank-deficient fat	looks like C-fat but $A A^{⊤}$ is singular!	SVD, not the fat formula	Ex 9

The two figures below give the geometry behind the two "no exact inverse" cells: the tall case projects $b$ down (Ex 1), the fat case picks the shortest arrow (Ex 3).

Cell C-tall — least squares

Worked example Example 1 (cell

C-tall) — the projection picture $A = (11)$ , $b = (31)$ . Solve $A x = b$ as best possible ( $x$ is a single number).

Forecast: guess the single best number $x$ before reading. (The two equations $x = 3$ and $x = 1$ disagree — what's the fair compromise?)

Step 1 — spot the cell. $A$ is $2 \times 1$ : two rows, one column. One column, and it's nonzero, so full column rank → cell C-tall → use $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . Why this step? We must classify before we pick a formula; the tall formula is only legal when the column(s) are independent.

Step 2 — build $A^{⊤} A$ . $A^{⊤} A = (11) (11) = 2$ . Why this step? $A^{⊤} A$ is the tiny square matrix (here just the number 2) we must invert; it is invertible exactly because the column is independent.

Step 3 — assemble $A^{+}$ . $A^{+} = (2)^{- 1} (11) = (\frac{1}{2} \frac{1}{2})$ . Why this step? This row vector is the refund machine: multiply any $b$ by it and out comes the best $x$ .

Step 4 — apply to $b$ . $x = A^{+} b = \frac{1}{2} \cdot 3 + \frac{1}{2} \cdot 1 = 2$ . Why this step? $x$ is literally the average of 3 and 1 — the fairest compromise, exactly as intuition demands.

What it looks like: in figure s01, $b = (3, 1)$ is dropped straight down onto the line spanned by $(1, 1)$ ; the foot of that perpendicular is $A x = (2, 2)$ , and the leftover arrow $b - A x = (1, - 1)$ is the residual, perpendicular to the line.

Verify: the residual $b - A x = (3, 1) - (2, 2) = (1, - 1)$ must be orthogonal to the column $(1, 1)$ : dot product $1 \cdot 1 + (- 1) \cdot 1 = 0$ ✓. Orthogonality of the residual is the least-squares condition (see Orthogonal Projection).

Worked example Example 2 (cell

C-tall) — the parent's line-fit, re-verified Same as the parent's Example 1: $A = 111012$ , $b = 600$ , fitting $y = c + d t$ .

Forecast: three points $(0, 6), (1, 0), (2, 0)$ — no straight line hits all three. Guess the slope's sign.

Step 1 — cell. $3 \times 2$ , columns $(1, 1, 1)$ and $(0, 1, 2)$ are independent → C-tall. Why this step? Same test as before; here we confirm rank $2 =$ number of columns.

Step 2 — $A^{⊤} A$ and $A^{⊤} b$ . $A^{⊤} A = (3335)$ , $A^{⊤} b = (60)$ . Why this step? These are the normal equations $A^{⊤} A x = A^{⊤} b$ .

Step 3 — invert the $2 \times 2$ . $det = 3 \cdot 5 - 3 \cdot 3 = 6$ , so $(A^{⊤} A)^{- 1} = \frac{1}{6} (5 - 3 - 3 3)$ . Why this step? A $2 \times 2$ inverse is $\frac{1}{d e t} (d - c - b a)$ .

Step 4 — solve. $x = \frac{1}{6} (5 - 3 - 3 3) (60) = (5 - 3)$ . Best line $y = 5 - 3 t$ . Slope is negative. Why this step? Solving the normal equations yields the least-squares coefficients $(c, d) = (5, - 3)$ .

Verify: predictions $(5, 2, - 1)$ at $t = 0, 1, 2$ ; residual $r = b - A x = (6, 0, 0) - (5, 2, - 1) = (1, - 2, 1)$ . Check $A^{⊤} r = 0$ : column $(1, 1, 1) \cdot (1, - 2, 1) = 0$ ✓, column $(0, 1, 2) \cdot (1, - 2, 1) = 0 - 2 + 2 = 0$ ✓.

Cell C-fat — minimum norm

Worked example Example 3 (cell

C-fat) — the shortest arrow $A = (12)$ , $b = (4)$ . Solve $x_{1} + 2 x_{2} = 4$ with the smallest $∥ x ∥$ .

Forecast: the solutions form a whole line in the plane. Which point on that line is closest to the origin?

Step 1 — cell. $1 \times 2$ : one row, and it's nonzero → full row rank → cell C-fat → use $A^{+} = A^{⊤} (A A^{⊤})^{- 1}$ . Why this step? One equation, two unknowns: infinitely many solutions. We need the shortest one, which is the fat formula's job.

Step 2 — force $x$ into the row space. The parent's derivation writes $x = A^{⊤} w$ , where $w$ is a small helper vector (here a single unknown number) chosen so that the true equation $A x = b$ is satisfied. Substituting $x = A^{⊤} w$ turns $A x = b$ into $A A^{⊤} w = b$ . Why this step? Every solution splits into a row-space part plus nullspace junk; writing $x = A^{⊤} w$ builds $x$ purely out of the row direction $(1, 2)$ , guaranteeing zero junk — which is exactly what "smallest norm" demands. The letter $w$ is just the coefficient telling us how much of the row direction to take.

Step 3 — solve for $w$ , then $x$ . $A A^{⊤} = (12) (12) = 1 + 4 = 5$ , so $w = (A A^{⊤})^{- 1} b = \frac{1}{5} \cdot 4 = \frac{4}{5}$ . Then $x = A^{⊤} w = (12) \cdot \frac{4}{5} = (4/5 8/5)$ . Why this step? Once $w$ is known, $x = A^{⊤} w$ hands us the answer directly. Bundling the two steps gives the closed form $A^{+} = A^{⊤} (A A^{⊤})^{- 1} = (1/5 2/5)$ , and indeed $A^{+} b = (4/5 8/5)$ .

What it looks like: in figure s02 the dashed line is all solutions of $x_{1} + 2 x_{2} = 4$ ; the plum arrow to $(4/5, 8/5)$ is perpendicular to that line — the closest point to the origin.

Verify (two ways): (i) it solves the equation: $\frac{4}{5} + 2 \cdot \frac{8}{5} = \frac{4}{5} + \frac{16}{5} = \frac{20}{5} = 4$ ✓. (ii) $x$ is perpendicular to the nullspace direction $(2, - 1)$ : $\frac{4}{5} \cdot 2 + \frac{8}{5} \cdot (- 1) = \frac{8}{5} - \frac{8}{5} = 0$ ✓ (see Four Fundamental Subspaces).

Cell C-sqinv — the honest inverse

Worked example Example 4 (cell

C-sqinv) — $A^{+}$ collapses to $A^{- 1}$ $A = (2111)$ . Since $det A = 2 - 1 = 1 \neq = 0$ , $A$ is square and invertible.

Forecast: the pseudoinverse should give exactly the ordinary inverse. Predict $A^{+}$ .

Step 1 — cell. Square, $det \neq = 0$ → C-sqinv → all formulas agree and equal $A^{- 1}$ . Why this step? When $A^{- 1}$ exists it uniquely satisfies all four Penrose conditions, so $A^{+} = A^{- 1}$ .

Step 2 — ordinary inverse. $A^{- 1} = \frac{1}{1} (1 - 1 - 1 2) = (1 - 1 - 1 2)$ . Why this step? The $2 \times 2$ inverse formula with $det = 1$ .

Step 3 — sanity via the tall formula. With full column rank the tall formula must give the same thing: $A^{+} = (A^{⊤} A)^{- 1} A^{⊤}$ . Since $A^{⊤} A$ is invertible and $A$ square, $(A^{⊤} A)^{- 1} A^{⊤} = A^{- 1} (A^{⊤})^{- 1} A^{⊤} = A^{- 1}$ . Why this step? This algebraic collapse is the whole reason $A^{+}$ "generalizes" the inverse (see Matrix Inverse).

Verify: $A A^{- 1} = (2111) (1 - 1 - 1 2) = (1001) = I$ ✓.

Cell C-sqsing — square but singular

Worked example Example 5 (cell

C-sqsing) — when $det = 0$ , only SVD survives $A = (1111)$ . Here $det = 1 - 1 = 0$ : not invertible, and both $A^{⊤} A$ and $A A^{⊤}$ are singular, so neither the tall nor the fat formula is legal.

Forecast: $A$ crushes everything onto the line $(1, 1)$ . The refund can only work along that one surviving direction. Guess how "spread out" $A^{+}$ looks.

Step 1 — cell. Square, $det = 0$ → C-sqsing → must use $A^{+} = V Σ^{+} U^{⊤}$ . Why this step? The parent's Mistake 1 warned us: $A^{⊤} A$ singular ⇒ tall formula explodes.

Step 2 — find the surviving direction. $A$ has rank 1: it scales the direction $u = \frac{1}{2} (1, 1)$ by 2 and kills the perpendicular direction $u_{⊥} = \frac{1}{2} (1, - 1)$ . So the single nonzero singular value is $σ_{1} = 2$ . Why this step? $A u = \frac{1}{2} (2, 2) = 2 u$ and $A u_{⊥} = \frac{1}{2} (0, 0) = 0$ : these are the Eigenvalues and Eigenvectors $2$ and $0$ , which here coincide with the singular directions because $A$ is symmetric.

Step 3 — write out the full $U$ , $Σ$ , $V$ explicitly. Because $A$ is symmetric and its eigenvalues $2, 0$ are non-negative, we may take $U = V$ built column-by-column from $u$ (for $σ_{1} = 2$ ) then $u_{⊥}$ (for $σ_{2} = 0$ ): $U = V = (\frac{1}{2} \frac{1}{2} \frac{1}{2} - \frac{1}{2}), Σ = (2000) .$ Why this step? $U$ and $V$ must be genuine $2 \times 2$ orthonormal matrices; their columns are the unit singular vectors in singular-value order. You can check $U^{⊤} U = I$ (columns are perpendicular unit vectors) and $U Σ V^{⊤} = A$ .

Step 4 — invert nonzeros only. $Σ^{+} = (1/2 0 00)$ (flip 2 to 1/2, leave the 0 dead). Why this step? Dividing by zero is impossible; the dead direction carries no information to restore.

Step 5 — reassemble. $A^{+} = V Σ^{+} U^{⊤} = \frac{1}{2} u u^{⊤} = \frac{1}{2} \cdot \frac{1}{2} (1111) = (1/4 1/4 1/4 1/4)$ . Why this step? Only the surviving direction $u$ is rebuilt, scaled by $1/ σ_{1} = 1/2$ .

Verify (Penrose condition 1): $A A^{+} A = A$ . Compute $A^{+} A = (1/2 1/2 1/2 1/2)$ (a projection onto the row space, not $I$ — parent's Mistake 2!). Then $A \cdot (A^{+} A) = (1111) = A$ ✓. Also confirm $U Σ V^{⊤} = A$ : $U Σ = (2200)$ , times $V^{⊤}$ gives $(1111)$ ✓.

Cell C-zero — the degenerate limit

Worked example Example 6 (cell

C-zero) — the all-zero matrix $A = (0000)$ . Every direction is nullspace. What is $A^{+}$ ?

Forecast: if $A$ throws everything away, the fairest refund gives... what?

Step 1 — cell. Zero matrix → C-zero, the extreme of C-sqsing (rank 0) → SVD. Why this step? No formula involving $(A^{⊤} A)^{- 1}$ can work; there is nothing to invert.

Step 2 — SVD. All singular values are $0$ : $Σ = 0$ , hence $Σ^{+} = 0$ (invert nothing, since no $σ_{i}$ is nonzero). Why this step? The rule "invert nonzeros, leave zeros" leaves everything zero.

Step 3 — assemble. $A^{+} = V \cdot 0 \cdot U^{⊤} = (0000)$ . Why this step? With no invertible directions, the best refund is "give nothing back" — which is exactly $x = 0$ , the minimum-norm least-squares answer.

Verify (why $A^{+} = 0$ is the right answer, not a cop-out): the pseudoinverse must give the least-squares, minimum-norm solution of $A x = b$ . For $A = 0$ , every $x$ gives error $∥0 \cdot x - b ∥ = ∥ b ∥$ , so the error can't be reduced; among all $x$ tied for that error, the smallest-norm one is $x = 0$ . Since $x = A^{+} b = 0$ for every $b$ , we need $A^{+} = 0$ — the unique matrix delivering the fairest refund. All four Penrose conditions then hold trivially ( $A A^{+} A = 0 = A$ ✓).

Cell C-word — a real-world fit

Worked example Example 7 (cell

C-word) — resistance from noisy readings A physicist measures voltage $V$ across a resistor at currents $I = 1, 2, 3$ amps, getting $V = 2.1, 3.9, 6.1$ volts. Ohm's law says $V = R I$ (a line through the origin). Find the best resistance $R$ in ohms.

Forecast: the readings are near a slope-2 line. Guess $R \approx 2 Ω$ .

Step 1 — model as $A x = b$ . $A = 123$ (currents), $x = (R)$ , $b = 2.1 3.9 6.1$ (voltages). Why this step? One unknown $R$ , three noisy equations → overdetermined → cell C-word = C-tall in disguise.

Step 2 — cell & formula. $3 \times 1$ , nonzero column → full column rank → $R = (A^{⊤} A)^{- 1} A^{⊤} b$ . Why this step? Least squares over the noise, exactly like Example 1.

Step 3 — compute $A^{⊤} A$ and $A^{⊤} b$ . $A^{⊤} A = 1^{2} + 2^{2} + 3^{2} = 1 + 4 + 9 = 14$ . $A^{⊤} b = 1 (2.1) + 2 (3.9) + 3 (6.1) = 2.1 + 7.8 + 18.3 = 28.2$ . Why this step? These two numbers are all the normal equation $A^{⊤} A R = A^{⊤} b$ needs.

Step 4 — solve for $R$ . $R = \frac{A ^{⊤} b}{A ^{⊤} A} = \frac{28.2}{14} = 2.0142857 \dots \approx 2.01 Ω$ . Why this step? The formula weights higher-current readings more (larger $I^{2}$ ), which is correct: they carry more signal.

Verify (units + sanity): $[A^{⊤} b] =$ amp·volt, $[A^{⊤} A] =$ amp $^{2}$ , ratio has units volt/amp $= Ω$ ✓. Value $\approx 2.01 Ω$ matches the forecast of $\approx 2 Ω$ ✓ (application of Least Squares Regression).

Cell C-twist-t — the tall exam trap

Worked example Example 8 (cell

C-twist-t) — a tall matrix where the tall formula still fails $A = 123246$ . It's $3 \times 2$ (tall), so a hasty student reaches for $(A^{⊤} A)^{- 1} A^{⊤}$ . Trap: column 2 is exactly $2 \times$ column 1, so rank $= 1$ , and $A^{⊤} A$ is singular. Find $A^{+}$ correctly.

Forecast: will the tall formula work? (No — and this is the exam's whole point.)

Step 1 — detect the trap. Columns $(1, 2, 3)$ and $(2, 4, 6)$ are proportional → rank $1 < 2$ → not full column rank → cell C-twist-t → SVD required. Why this step? Tall shape is necessary but not sufficient (parent's Mistake 1); we need independent columns.

Step 2 — write $A$ as one outer product (rank-1). $A = 123 (12)$ , i.e. $A = c r^{⊤}$ with column-direction $c = (1, 2, 3)$ and row-direction $r = (1, 2)$ . Why this step? Every rank-1 matrix is a single column times a single row; this makes the SVD trivial to read off.

Step 3 — get the unit singular vectors and $σ_{1}$ . Normalise: the left singular vector is $u = \frac{c}{∥ c ∥} = \frac{1}{14} (1, 2, 3)$ , and the right singular vector is $v = \frac{r}{∥ r ∥} = \frac{1}{5} (1, 2)$ . The lone singular value is $σ_{1} = ∥ c ∥ ∥ r ∥ = 14 \cdot 5 = 70$ . Why this step? Since $A = c r^{⊤} = (∥ c ∥ u) (∥ r ∥ v)^{⊤} = (∥ c ∥∥ r ∥) u v^{⊤}$ , comparing to $A = σ_{1} u v^{⊤}$ forces $σ_{1} = ∥ c ∥∥ r ∥$ . Here $u$ is now explicitly defined — the normalised column direction.

Step 4 — pseudoinverse of a rank-1 matrix. For $A = σ_{1} u v^{⊤}$ , the SVD rule gives $A^{+} = \frac{1}{σ _{1}} v u^{⊤}$ . Substituting: $A^{+} = \frac{1}{70} \cdot \frac{1}{5} (12) \cdot \frac{1}{14} (123) = \frac{1}{70} (122436) .$ Why this step? $70 \cdot 5 \cdot 14 = 70 \cdot 70 = 70$ ; only the surviving direction is inverted.

Verify (Penrose condition 1): $A A^{+} A = A$ . Since $A^{+} A = v v^{⊤} = \frac{1}{5} (1224)$ (row-space projection, not $I$ — Mistake 2 again), $A (A^{+} A)$ : first column $\frac{1}{5} [(1, 2, 3) \cdot 1 + (2, 4, 6) \cdot 2] = \frac{1}{5} (5, 10, 15) = (1, 2, 3)$ ✓, recovering column 1 of $A$ ; column 2 similarly gives $(2, 4, 6)$ ✓.

Cell C-twist-f — the fat exam trap

Worked example Example 9 (cell

C-twist-f) — a fat matrix where the fat formula still fails $A = (122436)$ . It's $2 \times 3$ (fat), so a hasty student reaches for $A^{⊤} (A A^{⊤})^{- 1}$ . Trap: row 2 is exactly $2 \times$ row 1, so rank $= 1$ , and $A A^{⊤}$ is singular — the direct analogue of Example 8's trap, one dimension over.

Forecast: will the fat formula work? (No — and it's the mirror image of Ex 8.)

Step 1 — detect the trap. Rows $(1, 2, 3)$ and $(2, 4, 6)$ are proportional → rank $1 < 2$ → not full row rank → cell C-twist-f → SVD required. Note $A = A_{8}^{⊤}$ (the transpose of Example 8's matrix), so we can reuse everything. Why this step? Fat shape is necessary but not sufficient; we need independent rows.

Step 2 — write $A$ as one outer product. $A = (12) (123)$ , i.e. $A = c r^{⊤}$ with column-direction $c = (1, 2)$ and row-direction $r = (1, 2, 3)$ . Why this step? Same rank-1 reading as before, now with the roles of the two directions swapped.

Step 3 — unit singular vectors and $σ_{1}$ . $u = \frac{c}{∥ c ∥} = \frac{1}{5} (1, 2)$ (left, size 2), $v = \frac{r}{∥ r ∥} = \frac{1}{14} (1, 2, 3)$ (right, size 3), and $σ_{1} = ∥ c ∥∥ r ∥ = 5 \cdot 14 = 70$ . Why this step? Identical logic to Ex 8; here $u$ lives in $R^{2}$ and $v$ in $R^{3}$ .

Step 4 — assemble $A^{+}$ . $A^{+} = \frac{1}{σ _{1}} v u^{⊤} = \frac{1}{70} \cdot \frac{1}{14} 123 \cdot \frac{1}{5} (12) = \frac{1}{70} 123246$ . Why this step? $A^{+}$ is $3 \times 2$ as required, and it equals $(A_{8}^{+})^{⊤}$ — a good sanity check, since $(A^{⊤})^{+} = (A^{+})^{⊤}$ always.

Verify (Penrose condition 1): $A A^{+} A = A$ . Here $A A^{+} = u u^{⊤} = \frac{1}{5} (1224)$ (column-space projection onto the line $(1, 2)$ ). Then $(A A^{+}) A$ : for row 1, $\frac{1}{5} [1 \cdot (1, 2, 3) + 2 \cdot (2, 4, 6)] = \frac{1}{5} (5, 10, 15) = (1, 2, 3)$ ✓; row 2 gives $(2, 4, 6)$ ✓.

Flashcards

Best single number fitting

x = 3

and

x = 1

(Ex 1)?

x = 2

, the average; residual

(1, - 1)

⟂ column

(1, 1)

Min-norm solution of

x_{1} + 2 x_{2} = 4

(Ex 3)?

x = (4/5, 8/5)

, perpendicular to nullspace direction

(2, - 1)

Why does a tall matrix sometimes still need SVD (Ex 8)?

If its columns are dependent (rank deficient),

A^{⊤} A

is singular so the tall formula fails.

Why does a fat matrix sometimes still need SVD (Ex 9)?

If its rows are dependent (rank deficient),

A A^{⊤}

is singular so the fat formula fails.

Pseudoinverse of

(1111)

(Ex 5)?

\frac{1}{4} (1111)

; note

A^{+} A

is a projection, not

I

Pseudoinverse of the zero matrix (Ex 6)?

The zero matrix itself.

Best-fit resistance from

V = 2.1, 3.9, 6.1

I = 1, 2, 3

(Ex 7)?

R = 28.2/14 \approx 2.01 Ω

Handy identity linking transpose and pseudoinverse (Ex 9)?

(A^{⊤})^{+} = (A^{+})^{⊤}

Connections

Pseudoinverse — parent; this page is its scenario drill.
Singular Value Decomposition — the only safe tool for cells C-sqsing, C-zero, C-twist-t, C-twist-f.
Least Squares Regression — cells C-tall and C-word.
Orthogonal Projection — the residual-⟂-column fact in Ex 1.
Four Fundamental Subspaces — min-norm ⟂ nullspace in Ex 3, row/column-space projections in Ex 8–9.
Matrix Inverse — the collapse $A^{+} = A^{- 1}$ in Ex 4.
Eigenvalues and Eigenvectors — the singular values $2, 0$ in Ex 5.

Worked examples — Pseudoinverse