Linear-Algebra-Done-Right · Unit 17 · Video 2 · Interactive Practice

The Pseudoinverse T† and Its Three Inverse-Like Properties

IKey Formulas

Formula	Name	What it says
$T\|_{(\operatorname{null} T)^\perp} : (\operatorname{null} T)^\perp \to \operatorname{range} T$	Restriction theorem	Injective and onto $\operatorname{range} T$ , hence invertible
$T^\dagger w = (T\|_{(\operatorname{null} T)^\perp})^{-1} P_{\operatorname{range} T}\, w$	Pseudoinverse $T^\dagger \in \mathcal{L}(W, V)$	Project onto $\operatorname{range} T$ , then invert the restriction
$T$ invertible $\Rightarrow T^\dagger = T^{-1}$	Property (1)	Nothing changes when an inverse exists
$TT^\dagger = P_{\operatorname{range} T}$ , $\ T^\dagger T = P_{(\operatorname{null} T)^\perp}$	Properties (2) and (3)	Both products are orthogonal projections

Key Insight: $T^\dagger$ sends $(\operatorname{range} T)^\perp$ to $0$ , and sends each $w \in \operatorname{range} T$ to the one vector of $(\operatorname{null} T)^\perp$ that $T$ maps to $w$ .

IIBuilding $T^\dagger w$

For $T(x, y) = \big(x + y,\ \tfrac{1}{2}(x + y)\big)$ , $T^\dagger w$ projects $w$ onto $\operatorname{range} T$ , then pulls it back into $(\operatorname{null} T)^\perp$ .

💡 The solutions of $Tv = P_{\operatorname{range} T}\, w$ form the line $T^\dagger w + \operatorname{null} T$ ; the next video proves that $T^\dagger w$ is the one of smallest norm.

IIIThe Round Trip $T^\dagger T$

Sending $v$ to $W$ and back, $T^\dagger T$ keeps the $(\operatorname{null} T)^\perp$ piece of $v$ and deletes the $\operatorname{null} T$ piece.

Map

IVBoth Products Are Projections

$TT^\dagger$ and $T^\dagger T$ are orthogonal projections, and each is an identity exactly when $T$ is surjective or injective.

Map

VQuiz Questions

Problem 1 · Compute a Pseudoinverse Value

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ , $T(x, y) = (x - y,\ 0)$ — find $T^\dagger(4, 3)$ .

✅ Correct!

(2, -2)

lies in

(\operatorname{null} T)^\perp

and

T(2, -2) = (4, 0) = P_{\operatorname{range} T}(4, 3)

❌ Close, but…

T(4, 0) = (4, 0)

, so

(4, 0)

solves

Tv = P_{\operatorname{range} T} w

, but it is not in

(\operatorname{null} T)^\perp = \operatorname{span}\big((1, -1)\big)

;

T^\dagger w

must be the solution that is.

❌ Not quite. That is

P_{(\operatorname{null} T)^\perp}(4, 3)

; the projection in the definition is onto

\operatorname{range} T

W

, and it is followed by the inverse of the restriction.

❌ Not quite.

T^\dagger w = 0

only when

w \in (\operatorname{range} T)^\perp

. Here

w \notin \operatorname{range} T

, but its projection

(4, 0)

is nonzero.

❌ Not quite. Project

w

onto

\operatorname{range} T

first, then find the vector of

(\operatorname{null} T)^\perp

that

T

sends there.

Show solution

$\operatorname{range} T = \operatorname{span}\big((1, 0)\big)$ , so $P_{\operatorname{range} T}(4, 3) = (4, 0)$ .

$\operatorname{null} T = \{(x, y) : x = y\} = \operatorname{span}\big((1, 1)\big)$ , so $(\operatorname{null} T)^\perp = \operatorname{span}\big((1, -1)\big)$ .

Find $t$ with $T\big(t(1, -1)\big) = (4, 0)$ :

T(t, -t) = (2t, 0) = (4, 0) \ \Rightarrow\ t = 2.

Hence $T^\dagger(4, 3) = (2, -2)$ . The other solutions of $Tv = (4, 0)$ , such as $(4, 0)$ , differ from it by a vector of $\operatorname{null} T$ .

Problem 2 · Injective but Not Surjective

Given: $V$ is finite-dimensional and $T \in \mathcal{L}(V, W)$ is injective but not surjective — which statement is true?

✅ Correct! Injectivity gives

(\operatorname{null} T)^\perp = V

, so

T^\dagger

is a left inverse; since

\operatorname{range} T \neq W

, it is not a right inverse.

❌ Close, but…

TT^\dagger = P_{\operatorname{range} T}

, which is

I_W

only when

\operatorname{range} T = W

, that is, when

T

is surjective.

❌ Not quite.

T

is not surjective, so it is not invertible and

T^{-1}

does not exist.

❌ Not quite.

T^\dagger T

projects onto

(\operatorname{null} T)^\perp

, not onto

\operatorname{null} T

; here

\operatorname{null} T = \{0\}

, so

P_{\operatorname{null} T}

would be the zero map.

❌ Not quite. Use

TT^\dagger = P_{\operatorname{range} T}

and

T^\dagger T = P_{(\operatorname{null} T)^\perp}

Show solution

$T$ injective means $\operatorname{null} T = \{0\}$ , so $(\operatorname{null} T)^\perp = V$ and property (3) gives

T^\dagger T = P_{(\operatorname{null} T)^\perp} = P_V = I_V.

$T$ not surjective means $\operatorname{range} T \neq W$ , so property (2) gives $TT^\dagger = P_{\operatorname{range} T}$ , which sends every nonzero vector of $(\operatorname{range} T)^\perp$ to $0$ and is therefore not $I_W$ .

So $T^\dagger$ is a left inverse of $T$ but not a right inverse.

Problem 3 · A Map from F³ to F²

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $T(x, y, z) = (x + z,\ y + z)$ — find $T^\dagger(3, 0)$ and $T^\dagger T(1, 0, 0)$ .

What is $T^\dagger(3, 0)$ ?

What is $T^\dagger T(1, 0, 0)$ ?

✅ Correct!

T^\dagger(3, 0) = (2, -1, 1)

, and

T^\dagger T

removes the

\operatorname{null} T

component

(\tfrac{1}{3}, \tfrac{1}{3}, -\tfrac{1}{3})

from

(1, 0, 0)

❌ Check $T^\dagger(3, 0)$ . Every option solves

Tv = (3, 0)

; only one of them is orthogonal to

\operatorname{null} T = \operatorname{span}\big((1, 1, -1)\big)

❌ Not quite.

T^\dagger T = I

only for injective

T

, and this

T

has

\operatorname{null} T = \operatorname{span}\big((1, 1, -1)\big)

❌ Close, but… That is the projection of

(1, 0, 0)

onto

\operatorname{null} T

, the piece

T^\dagger T

deletes; property (3) keeps the other piece.

❌ Check $T^\dagger T(1, 0, 0)$ . Use

T^\dagger T = P_{(\operatorname{null} T)^\perp}

and subtract the

\operatorname{null} T

component.

Show solution

$T$ is surjective (it sends $(1, 0, 0)$ and $(0, 1, 0)$ to the standard basis), so $\operatorname{range} T = \mathbf{R}^2$ and $P_{\operatorname{range} T}(3, 0) = (3, 0)$ .

$\operatorname{null} T$ : $x + z = 0$ and $y + z = 0$ , so $\operatorname{null} T = \operatorname{span}\big((1, 1, -1)\big)$ and $(\operatorname{null} T)^\perp = \{(x, y, z) : x + y - z = 0\}$ .

Part 1. Solve $x + z = 3$ , $y + z = 0$ , $x + y - z = 0$ . From the first two, $x = 3 - z$ and $y = -z$ ; then $(3 - z) - z - z = 0$ gives $z = 1$ :

T^\dagger(3, 0) = (2, -1, 1), \qquad T(2, -1, 1) = (3, 0).

Part 2. By property (3), $T^\dagger T = P_{(\operatorname{null} T)^\perp}$ . With $n = (1, 1, -1)$ and $\|n\|^2 = 3$ :

P_{(\operatorname{null} T)^\perp}(1, 0, 0) = (1, 0, 0) - \frac{\langle (1, 0, 0), n\rangle}{3}\, n = (1, 0, 0) - \tfrac{1}{3}(1, 1, -1) = \big(\tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{1}{3}\big).

Check: $T(1, 0, 0) = (1, 0)$ and $T\big(\tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{1}{3}\big) = (1, 0)$ , and $\big(\tfrac{2}{3}, -\tfrac{1}{3}, \tfrac{1}{3}\big) \perp n$ .

Problem 4 · The Pseudoinverse of a Projection

Given: $U$ is a subspace of a finite-dimensional inner product space $V$ with $\{0\} \neq U \neq V$ , and $T = P_U \in \mathcal{L}(V)$ — find $T^\dagger$ .

✅ Correct! The restriction of

P_U

(\operatorname{null} P_U)^\perp = U

is the identity on

U

, so

T^\dagger = I_U P_U = P_U

❌ Not quite.

T^\dagger

sends

(\operatorname{range} T)^\perp = U^\perp

0

, so it cannot be

I

when

U \neq V

❌ Not quite.

P_{U^\perp}

kills

U = \operatorname{range} T

, but

T^\dagger

must send each

u \in \operatorname{range} T

to a vector that

T

maps back to

u

❌ Not quite.

T^\dagger

is defined for every

T \in \mathcal{L}(V, W)

with

V

finite-dimensional; invertibility is not needed.

❌ Not quite. Compute

\operatorname{null} P_U

and

\operatorname{range} P_U

, then apply the definition.

Show solution

For $T = P_U$ : $\operatorname{range} T = U$ and $\operatorname{null} T = U^\perp$ , so $(\operatorname{null} T)^\perp = (U^\perp)^\perp = U$ .

The restriction $T|_U$ sends each $u \in U$ to $P_U u = u$ , so it is the identity on $U$ and its inverse is the identity on $U$ too. By the definition,

T^\dagger w = (T|_U)^{-1} P_U w = P_U w \quad \text{for every } w \in V.

Hence $T^\dagger = P_U$ . Consistently with properties (2) and (3), $TT^\dagger = P_U^2 = P_U = P_{\operatorname{range} T}$ and $T^\dagger T = P_U = P_{(\operatorname{null} T)^\perp}$ .

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The Pseudoinverse T† and Its Three Inverse-Like Properties

IKey Formulas

IIBuilding T†wT^\dagger wT†w

IIIThe Round Trip T†TT^\dagger TT†T

IVBoth Products Are Projections

VQuiz Questions

IIBuilding $T^\dagger w$

IIIThe Round Trip $T^\dagger T$