The Invertible Matrix Theorem
Before The theorem joins invertibility with the independence of the columns. Builds on: Matrix Operations and Invertibility · Linear Independence
Next Determinants give 1 more test for the list, and eigenvalues use the theorem to find where A - λI is not invertible. Leads to: Determinants · Eigenvalues and Eigenvectors
Before You Start: Prerequisite Check
Can you do these? (Click to reveal self-test)
Test yourself on these prerequisite skills:
Invertibility: What is the definition of an invertible matrix?
Check
A square matrix $A$ is invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Not every square matrix is invertible.
Linear independence: What does it mean for the columns of $A$ to be linearly independent?
Check
The columns are linearly independent when the only solution to $A\mathbf{x} = \mathbf{0}$ is $\mathbf{x} = \mathbf{0}$ (the trivial solution). Equivalently, no column is a linear combination of the others.
Pivot count: How many pivots does a $3 \times 3$ invertible matrix have?
Check
Exactly 3: one in each row and one in each column. The RREF of an invertible $n \times n$ matrix is $I_n$.
If you struggled: Review Matrix Operations and Invertibility and Linear Independence.
Try This First
For $A = \begin{pmatrix}2&4\\1&2\end{pmatrix}$, check all three of the following:
(a) Are the columns of $A$ linearly independent?
(b) Does $A\mathbf{x} = \mathbf{0}$ have only the trivial solution?
(c) Is $A$ invertible?
Then repeat for $B = \begin{pmatrix}2&1\\4&3\end{pmatrix}$.
Show Solution
For $A = \begin{pmatrix}2&4\\1&2\end{pmatrix}$:
Row-reduce: $R_2 \to R_2 - \frac{1}{2}R_1$ gives $[0, 0]$. One pivot, one free column.
(a) Columns $\begin{pmatrix}2\\1\end{pmatrix}$ and $\begin{pmatrix}4\\2\end{pmatrix} = 2\begin{pmatrix}2\\1\end{pmatrix}$: linearly dependent.
(b) Free variable means nontrivial solutions exist: $A\mathbf{x} = \mathbf{0}$ does NOT have only the trivial solution.
(c) $A$ is not invertible (the left block cannot be reduced to $I_2$).
For $B = \begin{pmatrix}2&1\\4&3\end{pmatrix}$:
Row-reduce: $R_2 - 2R_1$ gives $[0, 1]$. Two pivots.
(a) Columns $\begin{pmatrix}2\\4\end{pmatrix}$ and $\begin{pmatrix}1\\3\end{pmatrix}$: neither is a multiple of the other. Linearly independent.
(b) No free variable: only the trivial solution.
(c) $B$ is invertible.
The pattern: all three conditions fail together for $A$, and all three hold together for $B$. The Invertible Matrix Theorem states that this is not a coincidence: these conditions are equivalent.
The Equivalence Idea
For numbers, knowing “$a \neq 0$” gives you many facts at once: $a$ has a multiplicative inverse, $ax = b$ has a unique solution, and so on. For matrices, the Invertible Matrix Theorem plays the same role: a single test (say, row-reducing to $I_n$) simultaneously certifies a chain of equivalent conditions.
Each condition can be used to prove the others. If you know the columns are linearly independent, you immediately know the system $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, that $A$ row-reduces to $I_n$, that $A^T$ is also invertible, and so on.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Course | MATH 231 |
| Difficulty | Intermediate |
| Time | ~30 minutes |
| Source | Lay, Linear Algebra and Its Applications, Section 2.3 |
Key Concepts
The Invertible Matrix Theorem
Theorem (Lay, Section 2.3): Let $A$ be an $n \times n$ matrix. The following statements are all equivalent: each one is true precisely when all the others are, and each one is false precisely when all the others are.
| # | Condition |
|---|---|
| 1 | $A$ is invertible. |
| 2 | $A$ is row equivalent to $I_n$. |
| 3 | $A$ has $n$ pivot positions. |
| 4 | $A\mathbf{x} = \mathbf{0}$ has only the trivial solution. |
| 5 | The columns of $A$ are linearly independent. |
| 6 | The linear transformation $\mathbf{x} \mapsto A\mathbf{x}$ is one-to-one. |
| 7 | $A\mathbf{x} = \mathbf{b}$ has at least one solution for every $\mathbf{b} \in \mathbb{R}^n$. |
| 8 | The columns of $A$ span $\mathbb{R}^n$. |
| 9 | The linear transformation $\mathbf{x} \mapsto A\mathbf{x}$ is onto. |
| 10 | There is a matrix $C$ with $CA = I_n$ (left inverse exists). |
| 11 | There is a matrix $D$ with $AD = I_n$ (right inverse exists). |
| 12 | $A^T$ is invertible. |
The critical word is equivalent: if any single condition holds, all twelve hold. If any single condition fails, all twelve fail.
Source: Lay, Linear Algebra and Its Applications, Section 2.3: Characterizations of Invertible Matrices.
Why Equivalence Is Useful
You choose whichever condition is easiest to check in context:
- Fast check: row-reduce $A$ and count pivots (condition 3). If you get $n$ pivots, $A$ is invertible and all other conditions hold automatically.
- Geometric: columns span $\mathbb{R}^n$ (condition 8), or the transformation is onto (condition 9).
- Null space: if $\text{Nul}(A) = \{\mathbf{0}\}$ (condition 4), then $A$ is invertible.
Conversely, to show $A$ is NOT invertible, you only need to find one condition that fails (e.g., a row without a pivot, or a free variable in $A\mathbf{x} = \mathbf{0}$).
A Key Remark: Square Matrices Only
The IMT applies exclusively to square ($n \times n$) matrices. For a non-square matrix, the twelve conditions above are not equivalent to each other, and the concept of “invertible” does not apply in this form.
For a $2 \times 3$ matrix, for example, the columns can be linearly independent (condition 5 holds) even though the transformation cannot be onto $\mathbb{R}^2$ with only 3-dimensional inputs... wait, actually a $2\times 3$ matrix can be onto. The point is that one-to-one and onto are no longer equivalent for non-square matrices. Only for square matrices do all twelve conditions collapse into one.
Worked Example: Using the IMT
Problem: Is $A = \begin{pmatrix}1&3&5\\2&4&6\\1&1&0\end{pmatrix}$ invertible?
Row-reduce $A$:
$R_2 - 2R_1$: $[0, -2, -4]$; $R_3 - R_1$: $[0, -2, -5]$.
$R_3 - R_2$: $[0, 0, -1]$.
\[ \begin{pmatrix}1&3&5\\0&-2&-4\\0&0&-1\end{pmatrix} \]
Three pivots. By the IMT (condition 3), $A$ is invertible. Simultaneously: the columns span $\mathbb{R}^3$, the system $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b}$, the null space contains only $\mathbf{0}$, $A^T$ is invertible, and so on.
Using the IMT
The IMT is most useful as a one-step conclusion: verify any single condition, then immediately state all the rest.
Example chain:
“The columns of $A$ are linearly independent (condition 5). By the IMT, $A$ is invertible (condition 1), and the equation $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b} \in \mathbb{R}^n$ (condition 7), and the linear transformation $T(\mathbf{x}) = A\mathbf{x}$ is both one-to-one (condition 6) and onto (condition 9).”
Contrapositive:
“The system $A\mathbf{x} = \mathbf{0}$ has a nontrivial solution (negation of condition 4). By the IMT, $A$ is not invertible (negation of condition 1), so $A$ does not row-reduce to $I_n$, the columns are linearly dependent, the transformation is neither one-to-one nor onto, and $A^T$ is also not invertible.”
Common Pitfalls
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| Applying the IMT to a non-square matrix | The IMT requires $n \times n$; the conditions are not all equivalent for rectangular matrices | For non-square matrices, check each property (onto, one-to-one, etc.) separately |
| Thinking some IMT conditions are “stronger” than others | All twelve conditions are logically equivalent; none implies the others in a one-directional sense | If one holds, all hold; if one fails, all fail |
| Concluding that “not invertible” means the system $A\mathbf{x} = \mathbf{b}$ has no solution | If $A$ is not invertible, the system may have no solution OR infinitely many solutions, depending on $\mathbf{b}$ | The IMT says: for some $\mathbf{b}$, the system has no solution. Row-reduce $[A \mid \mathbf{b}]$ to decide for a specific $\mathbf{b}$ |
| Checking only one direction of the equivalence | “Condition X implies $A$ is invertible” must also work in reverse | Check that the condition is one of the twelve in the IMT, not just an implication of one of them |
Practice Problems
Without computing the inverse, determine whether each matrix is invertible. State which IMT condition you used.
(a) $A = \begin{pmatrix}1&0\\0&0\end{pmatrix}$
(b) $B = \begin{pmatrix}5&2\\3&1\end{pmatrix}$
(c) $C = \begin{pmatrix}1&2&3\\0&1&4\\0&0&1\end{pmatrix}$
Suppose row-reducing $A$ (a $4 \times 4$ matrix) reveals exactly 4 pivot positions.
Without any further computation, state whether each of the following is true or false. Cite the IMT condition.
(a) $A\mathbf{x} = \mathbf{b}$ has a unique solution for every $\mathbf{b} \in \mathbb{R}^4$.
(b) The columns of $A$ are linearly dependent.
(c) The transformation $T(\mathbf{x}) = A\mathbf{x}$ is onto $\mathbb{R}^4$.
(d) $A^T$ is not invertible.
Let $A = \begin{pmatrix}2&-4&6\\-1&2&-3\\3&-6&9\end{pmatrix}$.
(a) Row-reduce $A$ and state the pivot count.
(b) Which IMT conditions fail? List at least four.
(c) For $\mathbf{b} = \begin{pmatrix}4\\-2\\6\end{pmatrix}$, determine whether $A\mathbf{x} = \mathbf{b}$ is consistent or inconsistent.
Suppose $A$ and $B$ are $n \times n$ matrices and $AB = I_n$.
(a) Explain why $A$ must be invertible. (Hint: use IMT condition 9 or 7.)
(b) Does it follow that $BA = I_n$? Justify using the IMT, not by computation.
Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear transformation with standard matrix $A$.
Suppose you know that $T$ is one-to-one.
(a) What does “one-to-one” mean for $T$? (Give the formal definition and an equivalent matrix condition.)
(b) List three additional properties that must also hold (using the IMT), with brief justification.
(c) A classmate claims: “If $T$ is one-to-one, then $T$ must map distinct inputs to distinct outputs, but it might not hit all of $\mathbb{R}^3$.” Evaluate this claim.
Common Misconceptions
the Invertible Matrix Theorem applies to any matrix. The IMT is a theorem about square $n \times n$ matrices. For a $2 \times 3$ or $3 \times 2$ matrix, the conditions in the theorem are not all equivalent, and “invertible” in the square-matrix sense does not apply. A $3 \times 2$ matrix can have linearly independent columns (condition 5 of the IMT analog) but cannot have a pivot in every row (condition 3 fails), and the transformation cannot be onto $\mathbb{R}^3$.
some IMT conditions are stronger than others. All twelve conditions are logically equivalent: each one implies all the others, and the negation of each one implies the negation of all the others. There is no hierarchy. Conditions that “look weaker” (like “there exists a left inverse”) are just as strong as “there exist $n$ pivots.” Any single condition is a complete test for invertibility.
if $A$ is not invertible, then $A\mathbf{x} = \mathbf{b}$ has no solution. The IMT says: if $A$ is not invertible, then condition 7 fails, meaning there EXIST some $\mathbf{b}$ for which the system has no solution. It says nothing about specific right-hand sides. For a particular $\mathbf{b}$ that happens to be in the column space of $A$, the system may still be consistent (but it will then have infinitely many solutions). Whether a specific $A\mathbf{x} = \mathbf{b}$ is consistent requires checking the augmented matrix.
Mastery Checklist
Novice (Level 1-2):
Competent (Level 3-4):
Proficient (Level 5):
Mental Model
The Master Key
The Invertible Matrix Theorem is a master key: it contains twelve different locks (conditions), and they all either open together or stay shut together. If any one lock opens, the key fits all twelve. If any one lock refuses, none of them open.
This means you never need to check more than one condition to know everything about an $n \times n$ matrix: one pivot check, or one null space computation, or one column independence test, and you know all twelve facts at once.
The theorem is remarkable because these conditions look very different in flavor: some are geometric (span $\mathbb{R}^n$), some are algebraic (invertible), some are computational (row equivalent to $I_n$), some are transformation-theoretic (one-to-one, onto). Yet for square matrices they are completely equivalent.
Connections
Looking ahead in MATH 231:
- Determinants: the determinant adds a 13th equivalent condition: $A$ is invertible iff $\det(A) \neq 0$
- Eigenvalues and Eigenvectors: $A - \lambda I$ is invertible iff $\lambda$ is not an eigenvalue; the IMT governs which $\lambda$ values are eigenvalues
Real-world connections:
- In numerical computing, the condition number of an invertible matrix measures how close it is to being singular, guiding algorithmic choices in solving linear systems.
- In statistics, a data matrix’s Gram matrix $X^TX$ is invertible exactly when the data columns are linearly independent (no multicollinearity), which determines whether ordinary least squares has a unique solution.
Source: Lay, Linear Algebra and Its Applications, Section 2.3: Characterizations of Invertible Matrices.
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|---|---|---|
| Matrix Operations and Invertibility | MATH231 Skills Index | Determinants |
Last updated: 2026-06-16
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