What does the rank-nullity theorem give for a 5x3 matrix of rank 2?

## The two spaces being counted An m x n matrix A is a map from R^n to R^m: it eats a vector with n entries and returns a vector with m entries. Two subspaces describe what that map does. - The **column space** (or image) is the span of A's columns: every vector of the form Ax. It lives in R^m. Its dimension is the **rank** of A. - The **nullspace** (or kernel) is the set of all x with Ax = 0. It lives in R^n. Its dimension is the **nullity**. The rank-nullity theorem ties them together: ``` rank(A) + dim(null(A)) = n = number of columns ``` ## Why it is the column count Both terms are budgeted out of the *input* side. Start with the n input directions. Some of them survive the map and produce independent output directions - that count is the rank. Every remaining input direction must be sent to zero, because there is nowhere else for it to go without adding to the rank. The two counts therefore partition n exactly. Nothing in this argument mentions m, which is why the row count never appears in the formula. The row count enters only through the ceiling rank <= m. This is the single most common slip in interviews: with a 5 x 3 matrix, candidates reach for 5 because it is the bigger number. The formula uses 3. ## The 5 x 3 example A is 5 x 3 with rank 2. - n = 3, so rank + nullity = 3. - nullity = 3 - 2 = 1. The nullspace is a **line** through the origin in R^3: one vector v (and all its multiples) satisfies Av = 0. - The column space has dimension 2. The three columns of A live in R^5 but span only a plane there - one of them is a combination of the other two. - Because the nullspace is non-trivial, A cannot be one-to-one: distinct inputs x and x + v produce the same output. ## The wide case: 3 x 5 Now A is 3 x 5, so n = 5 and rank <= min(3, 5) = 3. Rank-nullity gives ``` nullity = 5 - rank >= 5 - 3 = 2 ``` So the nullspace is at least a plane. This proves a fact worth carrying around: **a homogeneous system with more unknowns than equations always has non-zero solutions.** Five unknowns constrained by three equations cannot be pinned down; at least two directions of freedom remain, whatever the numbers are. ## Why rank <= min(m, n) Rank is the number of independent columns, so it cannot exceed n - there are only n columns to choose from. A theorem that surprises people the first time is that **row rank equals column rank**: the dimension of the span of the rows equals the dimension of the span of the columns. Since the rows are vectors in R^n and there are m of them, the same argument bounds the rank by m. Together: rank <= min(m, n). A matrix achieving that bound is called **full rank**; anything less is **rank deficient**. Row rank equalling column rank also means the row space and the column space always have the same dimension even though they usually live in different spaces (R^n versus R^m). ## A geometric reading Think of A as flattening R^n. The nullity counts how many directions get crushed to nothing; the rank counts how many survive as independent directions in the output. Flattening cannot create directions, only destroy them, and every input direction is either crushed or preserved. That accounting is the whole theorem. ## Using it as a sanity check Given only the shape of a matrix and its rank, you immediately know: whether the map can be one-to-one (only if nullity is 0, which needs rank = n), whether it can be onto R^m (only if rank = m), and how large the family of solutions to a consistent system will be (a set of dimension equal to the nullity). Those three readings cover most of what an interviewer is probing when they ask you to state the theorem and then hand you a shape and a rank.