Dimension and Rank | Real Analysis

Watch a grid get squeezed flat by a matrix. Count the dimensions that vanish and those that survive.

Dimension counts independent directions. A matrix can squash directions together. Directions crushed to nothing form the kernel's dimension. Directions still spanning room form the rank. The two add up to the original dimension.

For example, sunlight presses a solid object into a flat shadow. The height direction vanishes in the shadow. Only the flat directions survive. Lost plus surviving equals the original dimension.

Examples & Exercises

Show that $\{(1, 0), (0, 1)\}$ is a basis of $\mathbb{R}^2$ .

(x, y) = x(1, 0) + y(0, 1)

The identity matrix — it preserves the standard basis vectors.

Show proof

Independence: if $c_1(1, 0) + c_2(0, 1) = (0, 0)$ , coordinates give $c_1 = 0$ and $c_2 = 0$ , so only the trivial combination yields the zero vector.
Span: for any $(x, y) \in \mathbb{R}^2$ , $(x, y) = x(1, 0) + y(0, 1)$ , so the two vectors linearly span all of $\mathbb{R}^2$ .
Independent and spanning, hence $\{(1, 0), (0, 1)\}$ is a basis of $\mathbb{R}^2$ , and $\dim \mathbb{R}^2 = 2$ .

View $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ as $T: \mathbb{R}^3 \to \mathbb{R}^2$ ; find its rank and nullity and verify the rank–nullity theorem.

\operatorname{rank} T + \operatorname{nullity} T = 2 + 1 = 3 = \dim \mathbb{R}^3

A rank-1 matrix — the one-dimensional kernel collapses to the origin, leaving a one-dimensional image.

Show proof

Both rows $(1, 2, 3)$ and $(4, 5, 6)$ are nonzero — this check is needed: a zero row IS a scalar multiple ( $0$ times) of the other, so the "not a scalar multiple" criterion yields independence only for nonzero rows.
Neither row is a scalar multiple of the other (e.g. $4 \cdot (1, 2, 3) = (4, 8, 12) \ne (4, 5, 6)$ : the multiple $4$ forced by the first coordinate already fails at the second), so the rows are linearly independent and the row rank is $2$ .
Verify the column rank directly too (without citing row rank $=$ column rank): for the first two columns $(1, 4)^T$ , $(2, 5)^T$ , $\det \begin{pmatrix} 1 & 2 \\ 4 & 5 \end{pmatrix} = 1 \cdot 5 - 2 \cdot 4 = -3 \ne 0$ , so these two columns are linearly independent, the column space is all of $\mathbb{R}^2$ , and the column rank is also $2$ .
Hence $\operatorname{rank} T = 2$ .
Substituting $(-1, 2, -1)^T$ gives $1(-1) + 2(2) + 3(-1) = -1 + 4 - 3 = 0$ and $4(-1) + 5(2) + 6(-1) = -4 + 10 - 6 = 0$ , so it is a nonzero element of the kernel. Thus $\dim \ker T \ge 1$ .
Now take any $(x, y, z)^T \in \ker T$ .
From the two equations $x + 2y + 3z = 0$ and $4x + 5y + 6z = 0$ , subtracting 4 times the first from the second gives $-3y - 6z = 0$ , i.e. $y = -2z$ .
Substituting into the first equation, $x = -2y - 3z = 4z - 3z = z$ , so $(x, y, z) = (z, -2z, z) = z \cdot (1, -2, 1) = (-z) \cdot (-1, 2, -1)$ .
Every kernel vector is a scalar multiple of $(-1, 2, -1)^T$ , so $\dim \ker T \le 1$ .
The previous step gave $\dim \ker T \ge 1$ , so the nullity is exactly $1$ , and $\operatorname{rank} T + \operatorname{nullity} T = 2 + 1 = 3 = \dim \mathbb{R}^3$ , verifying the rank–nullity theorem — both sides computed directly, with no appeal to the theorem itself.

Find the rank of $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ by row reduction, and exhibit a basis of its column space.

\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \xrightarrow{R_2 - 2R_1} \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}

A projection matrix — another rank-1 example with a one-dimensional kernel and image.

Show proof

Applying $R_2 \to R_2 - 2R_1$ gives $\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$ . There is exactly one nonzero row, so the dimension of the row space — the rank — is $1$ .
The columns satisfy $(2, 4)^T = 2(1, 2)^T$ , so one column is a scalar multiple of the other.
The single column $(1, 2)^T$ spans the whole column space (spanning).
Independence is also required: since $(1, 2)^T \ne (0, 0)^T$ , the singleton $\{(1, 2)^T\}$ is linearly independent (if $c(1, 2)^T = 0$ then $c = 0$ ).
Spanning and independent, hence a basis of the column space is $\{(1, 2)^T\}$ , and $\dim$ (column space) $= 1 =$ the rank.