LINEAR-ALGEBRA

Why Only Perfect Squares Can Be Truly Inverted

IKey Formulas

For an $m \times n$ matrix $A$ with rank $r$ :

Condition	Matrix Shape	Type of Inverse	The Invertible Piece
$r = m = n$	Square, full rank	Two-sided inverse $A^{-1}$	$A$ itself
$r = n < m$	Tall, full column rank	Left-inverse $(A^TA)^{-1}A^T$	$A^TA$ ( $n\times n$ )
$r = m < n$	Wide, full row rank	Right-inverse $A^T(AA^T)^{-1}$	$AA^T$ ( $m\times m$ )
$r < m,\,n$	General	Pseudo-inverse $A^{+}$	via SVD

Dimensions of the four subspaces:

Row space $= r$ , Null space $= n - r$ , Column space $= r$ , Left null space $= m - r$

IIVisualization 1 — Rank and the Four Subspaces

How the rank $r$ splits $\mathbb{R}^n$ and $\mathbb{R}^m$ into four subspaces — and which inverse that permits.

Rows m 3 Cols n 2 Rank r 2

💡 A true two-sided inverse needs both red regions ( $n-r$ and $m-r$ ) to vanish — that happens only when $r = m = n$ .

IIIVisualization 2 — What the Map Crushes and Misses

$A$ sends the row space one-to-one onto the column space; the null space collapses to $0$ .

Rows m 3 Cols n 2 Rank r 2

IVVisualization 3 — Which Gram Matrix Is Invertible?

For a tall $A$ , $A^TA$ is invertible while $AA^T$ is singular — the smaller Gram matrix wins.

Matrix A

💡 The invertible Gram matrix is always the smaller $r \times r$ one — it builds the one-sided inverse, never $A$ itself.

VQuiz Questions

Question 1

A matrix $A$ is $4 \times 4$ with rank $r = 4$ . Which statement is true?

✅ Correct! Square ( $m=n$ ) and full rank ( $r=m=n$ ) is exactly the condition for a true inverse.

❌ Not quite. When $r = m = n$ , both null spaces are trivial, so a full two-sided inverse exists.

Show solution

Solution:

Here $m = n = 4$ and $r = 4$ , so $r = m = n$ . This is Case 1: the matrix is square and full rank. Both the null space ( $\dim = n - r = 0$ ) and the left null space ( $\dim = m - r = 0$ ) collapse to $\{0\}$ .

With nothing crushed and nothing missed, the map is a perfect bijection, and a genuine two-sided inverse $A^{-1}$ exists.

Question 2

Let $A$ be a tall $5 \times 3$ matrix with full column rank $r = 3$ . Which matrix is the small invertible one used to build the left-inverse?

✅ Correct! For a tall matrix, $A^TA$ is the small $n\times n$ invertible matrix.

❌ Not quite. For a tall matrix the invertible piece is the smaller $A^TA$ ( $n\times n$ ), not $AA^T$ .

Show solution

Solution:

A tall matrix with $r = n$ has a trivial null space (injective) but a nontrivial left null space, so only a left-inverse exists: $A^{+}_{\text{left}} = (A^TA)^{-1}A^T, \qquad (A^TA)^{-1}A^T \cdot A = I.$

The actually invertible piece is $A^TA$ , which is $n \times n = 3 \times 3$ . By contrast $AA^T$ is $5 \times 5$ but has rank only $3$ , so it is singular. The invertible Gram matrix is always the smaller $r \times r$ one.

Question 3

True or False: Every square matrix has a two-sided inverse.

✅ Correct! A square matrix needs full rank ( $r = m = n$ ); dependent rows leave a nontrivial null space.

❌ Not quite. A square matrix with dependent rows has a null space and therefore no inverse.

Show solution

Solution:

This is False. Being square ( $m = n$ ) is necessary but not sufficient. The matrix also needs full rank: $r = m = n$ .

A square matrix with dependent rows has $r < n$ , which means its null space is nontrivial ( $\dim = n - r > 0$ ). Some inputs are crushed to zero, information is lost, and no inverse can recover them. Square alone is not enough — you need $r = m = n$ .

Question 4

A matrix $A$ has rank $r$ strictly less than both $m$ and $n$ . What is the correct statement about inverting it?

✅ Correct! With both null spaces nontrivial, only the SVD-based pseudo-inverse $A^+$ remains.

❌ Not quite. If $r < m,n$ both null spaces are nontrivial, so no clean inverse exists — only $A^+$ .

Show solution

Solution:

When $r < m$ and $r < n$ , both null spaces are nontrivial:

Null space has dimension $n - r > 0$ : information is destroyed on the input side.
Left null space has dimension $m - r > 0$ : some outputs are never reached.

No one-sided inverse can fix both problems. The best available repair is the pseudo-inverse $A^{+}$ , built from the singular value decomposition (SVD). Remember: a one-sided inverse is not a real inverse — it works on only one side and relies on a smaller invertible matrix, never $A$ itself.

Solved: 0 / 4