Invertible matrix theorem — 12+ equivalent conditions

What we are claiming

Crucial caveat: this theorem is only for square matrices. A $3\times 5$ matrix can have independent columns or spanning columns but never both, so the IMT does not apply.

WHY are all these the same thing? (Derivation from scratch)

We don't memorize 17 facts. We build a logical loop so that following the arrows proves them all equivalent. We never need a bare formula — each link has a reason.

Link A: invertible $\Rightarrow$ trivial null space (1 ⟹ 4)

Suppose $A^{-1}$ exists and $Ax=\mathbf 0$. Multiply both sides on the left by $A^{-1}$: $A^{-1}Ax = A^{-1}\mathbf 0 \implies Ix = \mathbf 0 \implies x=\mathbf 0.$ Why this step? Left-multiplying by the inverse "undoes" $A$ and isolates $x$. So no nonzero vector can be killed by $A$.

Link B: trivial null space ⟺ independent columns (4 ⟺ 5)

Write $A=[\,a_1\ a_2\ \cdots\ a_n\,]$ with columns $a_i$. Then $Ax = x_1 a_1 + x_2 a_2 + \cdots + x_n a_n.$ Why this step? Matrix–vector multiplication is a linear combination of the columns weighted by entries of $x$. So $Ax=\mathbf 0$ means a combination of columns equals zero. "Only $x=\mathbf 0$ works" is exactly the definition of linear independence.

Link C: independent columns of a square matrix ⟹ pivot in every column ⟹ RREF $=I$ (5 ⟹ 3 ⟹ 2)

$n$ independent columns means no column is a free (non-pivot) column, so all $n$ columns are pivot columns. For an $n\times n$ matrix, $n$ pivots fill every row and column, and the reduced row echelon form must be $I_n$. Why this step? A free column would give a nonzero null-space vector, contradicting independence.

Link D: row equivalent to $I$ ⟹ invertible (2 ⟹ 1)

If $A \sim I$, then a sequence of elementary row operations turns $A$ into $I$. Each operation = multiply by an invertible elementary matrix $E_i$: $E_k\cdots E_2 E_1 A = I.$ Let $B = E_k\cdots E_1$ (a product of invertibles, hence invertible). Then $BA=I$, so $B=A^{-1}$. Why this step? Row operations are reversible — that's why each $E_i$ is invertible — so their product is a genuine inverse.

Loop closed: $1\Rightarrow4\Rightarrow5\Rightarrow3\Rightarrow2\Rightarrow1$. All five are equivalent. ✔

Hooking on the rest

• 6 (injective) $\equiv$ 4: a linear map is one-to-one iff its kernel is $\{\mathbf 0\}$ (if $Au=Av$ then $A(u-v)=\mathbf 0$).
• 7, 8, 9: "solution for every $b$" means every $b$ is a column combination $\Rightarrow$ columns span $\Rightarrow$ map is onto. For a square matrix, $n$ pivots simultaneously give independence and spanning — that's why injective and surjective coincide here.
• 13 ($\det\neq0$): elementary operations scale $\det$ by nonzero factors; $\det I=1$. So $A\sim I \iff \det A\neq 0$.
• 14: $\det A=\prod(\text{eigenvalues})$, so $\det A\neq0 \iff$ no eigenvalue is $0$. Also $0$ eigenvalue means $\exists x\neq0,\ Ax=0x=\mathbf0$ — a nonzero null vector, contradicting (4).
• 12 ($A^{\mathsf T}$): row rank $=$ column rank, and $\det A^{\mathsf T}=\det A$.
• 15, 16, 17: rank $=n$ ⟺ pivots fill columns ⟺ columns are a basis ⟺ $\dim\operatorname{Nul}A=0$.

Worked Examples

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

• Determinant — the scalar test $\det A\neq0$ (condition 13).
• Rank and Nullity — Rank–Nullity Theorem makes 15 ⟺ 17 immediate.
• Linear Independence and Span — conditions 5 and 8.
• Eigenvalues and Eigenvectors — condition 14, $0$ not an eigenvalue.
• Elementary Matrices and Row Reduction — engine behind 2 ⟹ 1.
• Linear Transformations — Injective and Surjective — conditions 6 and 9.
• Basis and Dimension — condition 16, columns as a basis.

Concept Map

Hinglish (regional understanding)