Change of Basis

Coordinates depend on the basis

In the standard basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ of $\mathbb{R}^2$, a vector $\mathbf{v} = (5, 3)$ means $5\mathbf{e}_1 + 3\mathbf{e}_2$. The numbers $(5, 3)$ are the coordinates of $\mathbf{v}$ with respect to the standard basis.

In a different basis $B = \{\mathbf{b}_1, \mathbf{b}_2\}$, the same geometric vector has different coordinates: $\mathbf{v} = c_1 \mathbf{b}_1 + c_2 \mathbf{b}_2$, and the pair $(c_1, c_2)$ is called $[\mathbf{v}]_B$.

The change-of-basis formulas

Let $P = [\mathbf{b}_1 \mid \mathbf{b}_2 \mid \cdots \mid \mathbf{b}_n]$ be the matrix with basis vectors as columns. Then $\mathbf{v} = P [\mathbf{v}]_B \quad \Longleftrightarrow \quad [\mathbf{v}]_B = P^{-1} \mathbf{v}$

$P$ converts $B$-coordinates back to standard; $P^{-1}$ converts standard to $B$-coordinates.

Step-by-step

$\mathbf{v} = (5, 3)$ in standard coords; $B = \{(1, 1), (1, -1)\}$.

Step 1 — Build $P$.

$P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

Step 2 — Compute $P^{-1}$.

$\det P = (1)(-1) - (1)(1) = -2$.

$P^{-1} = \dfrac{1}{-2} \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

Step 3 — Multiply: $[\mathbf{v}]_B = P^{-1} \mathbf{v}$.

$[\mathbf{v}]_B = \dfrac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 5 \\ 3 \end{pmatrix} = \dfrac{1}{2} \begin{pmatrix} 8 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$

Verification: $4 \mathbf{b}_1 + 1 \mathbf{b}_2 = 4(1, 1) + (1, -1) = (5, 3) = \mathbf{v} ✓$

So in the basis $B$, the vector $\mathbf{v} = (5, 3)_{\text{std}}$ has coordinates $\boxed{(4, 1)_B}$.

Change of basis for linear transformations

Given a linear map $T$ with matrix $A$ in the standard basis, its matrix $A_B$ in a new basis $B$ is $A_B = P^{-1} A P$

Same map, different coordinate system. Diagonalization $A = P D P^{-1}$ is exactly this: pick $P$ = eigenvectors, and the map becomes diagonal — stretching along the new axes.

Why change basis?

• Simpler representations. Rotations simplify in aligned axes; covariance is diagonal in PCA coordinates; linear ODEs decouple in eigenbases.
• Computational speedup. Diagonal matrices are trivial to exponentiate, invert, multiply.
• Conceptual clarity. Principal components, symmetry axes, natural frequencies — all are bases adapted to the problem.

When is $P^{-1}$ painless?

• Orthogonal basis ($P$ is orthogonal): $P^{-1} = P^T$ — cheap.
• Orthonormal basis: even better; $[\mathbf{v}]_B$ entries are dot products $\mathbf{v} \cdot \mathbf{b}_i$.
• Upper/lower triangular bases: $P^{-1}$ comes from back-substitution.

Our $B$ has orthogonal (not orthonormal) columns: $(1,1) \cdot (1,-1) = 0$. Their lengths are both $\sqrt{2}$, not 1. For dot-product-based formulas, scale the basis to orthonormal first.

Orthonormal variant

$\mathbf{q}_1 = (1, 1)/\sqrt{2}, \quad \mathbf{q}_2 = (1, -1)/\sqrt{2}$. Then $[\mathbf{v}]_{Q} = \begin{pmatrix} \mathbf{v} \cdot \mathbf{q}_1 \\ \mathbf{v} \cdot \mathbf{q}_2 \end{pmatrix} = \begin{pmatrix} 8/\sqrt{2} \\ 2/\sqrt{2} \end{pmatrix} = \begin{pmatrix} 4\sqrt{2} \\ \sqrt{2} \end{pmatrix}$

Different numbers for the same vector — coordinates always depend on basis.

Common mistakes

• Inverting the wrong direction. $P$ converts $B$-coords to standard; $P^{-1}$ converts standard to $B$.
• Forgetting the basis needs to be linearly independent. If columns of $P$ are dependent, $P^{-1}$ doesn't exist.
• Conflating "vector" with "coordinates." A geometric vector is basis-independent; its coordinates change with the basis. Keep them conceptually separate.

Try it in the visualization

Two coordinate grids are overlaid: the standard $x$–$y$ axes (gray) and the new $B$-basis axes (gold). A single vector is drawn; its coordinates are read in both grids and displayed.