Dimension and Rank | Real Analysis

If isomorphism means "same structure", dimension and rank measure its size. Watch the grid get squeezed by a matrix and verify

\dim\ker T + \mathrm{rank}\,T = \dim V

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

Neither row is a scalar multiple of the other (e.g. $4 \cdot (1, 2, 3) = (4, 8, 12) \ne (4, 5, 6)$ ), so the rows $(1, 2, 3)$ and $(4, 5, 6)$ are linearly independent. Hence the rank, the dimension of the row (column) space, is $\operatorname{rank} T = 2$ .
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.
Hence a basis of the column space is $\{(1, 2)^T\}$ , and $\dim$ (column space) $= 1 =$ the rank.