3.108 min read

Injectivity and Linear Independence

There is a beautiful connection between injectivity of $A$ and the columns of $A$ . $A$ is injective if and only if its columns are linearly independent.

Why? $A\mathbf{x} = x_1\mathbf{a}_1 + \cdots + x_n\mathbf{a}_n = \mathbf{0}$ has only the trivial solution iff the columns $\mathbf{a}_j$ are linearly independent (by definition of linear independence).

This means: "the map is injective" = "the columns are independent" = "the null space is trivial." Three equivalent ways to say the same thing.

Formal View

Theorem 3.10

A \in \mathbb{R}^{m \times n}

is injective

\Longleftrightarrow

its columns

\mathbf{a}_1, \ldots, \mathbf{a}_n

are linearly independent.

Why This Matters

This connects the algebraic notion (independence) to the functional notion (injectivity) — they are the same thing.

Testing linear independence of measurements = testing injectivity of the measurement matrix
Removing dependent columns from a dataset = making the data matrix injective

Learning Resources

Independence and injectivity

Khan Academy

Linear independence and the null space connection.

13 min

Quiz

Question 1

If the columns of $A$ are linearly dependent, then $A$ is not injective.

Common Mistakes

Thinking column independence is a geometric property with no algebraic meaning — it is equivalent to injectivity.