Recipe

Find a vector's coordinates relative to a basis

Use when: you have a basis $\mathcal B=\{\mathbf b_1,\dots,\mathbf b_n\}$ for a space $V$ and one specific vector $\mathbf x\in V$ , and need $[\mathbf x]_{\mathcal B}$

Goal: the unique list of scalars $c_1,\dots,c_n$ with $\mathbf x=c_1\mathbf b_1+\cdots+c_n\mathbf b_n$ , packaged as a column

Taught in Coordinates relative to a basis

Quick recipe

Set up the combination equation
Turn the combination into a system of scalar equations
Solve the system for $c_1,\dots,c_n$
Verify, then write the coordinate vector

1
Write $c_1\mathbf b_1+c_2\mathbf b_2+\cdots+c_n\mathbf b_n=\mathbf x$ , one unknown scalar per basis vector — this is the equation whose solution IS the coordinate vector.

Example

For $\mathcal B=\{\mathbf b_1,\mathbf b_2\}$ with $\mathbf b_1=(2,1)$ , $\mathbf b_2=(-1,1)$ , find $[\mathbf x]_{\mathcal B}$ for $\mathbf x=(3,6)$ : $c_1(2,1)+c_2(-1,1)=(3,6)$

Why: $[\mathbf x]_{\mathcal B}$ is defined as exactly this equation's solution, and a basis guarantees the solution exists and is the only one.
2
Match corresponding pieces on both sides: entry-by-entry if $V=\mathbb R^n$ , coefficient-by-coefficient of like powers if $V$ is a polynomial space, entry-by-entry of the matrix if $V$ is a space of matrices. Whatever $V$ actually is, matching the same kind of piece turns one vector equation into $n$ ordinary scalar equations.

Example

Matching the two coordinates of $(3,6)$ : $2c_1 - c_2 = 3, \qquad c_1+c_2=6$
- $V=\mathbb R^n$ — match components directly, as above.
- $V$ is an abstract space (polynomials, matrices, …) — expand the combination algebraically first, then match coefficients of like terms (same power of $x$ , same matrix position) — see the worked example below using $P_2$ .
Why: a single equation between two elements of $V$ carries exactly as much information as $V$ has dimensions — matching pieces is how that information gets extracted into solvable scalar equations.
3
Use substitution or elimination, exactly as for solving any linear system.

Example

From the second equation, $c_1=6-c_2$ . Substitute into the first: $2(6-c_2)-c_2=3 \implies 12-3c_2=3 \implies c_2=3 \implies c_1=3$

Why: this is an ordinary linear system in disguise — the unknowns happen to be coordinate coefficients rather than $x,y,z$ , but the solving technique doesn't change.
4
Substitute the solved $c_i$ back into the original combination and confirm it reconstructs $\mathbf x$ exactly, then package them as $[\mathbf x]_{\mathcal B}$ in $\mathcal B$ 's own stated order.

Example

$3(2,1)+3(-1,1) = (6,3)+(-3,3) = (3,6) \ \checkmark$ $[\mathbf x]_{\mathcal B} = \boxed{\begin{bmatrix}3\\3\end{bmatrix}}$

Why: this catches an arithmetic slip before it propagates into anything built on $[\mathbf x]_{\mathcal B}$ later, such as a change-of-basis matrix or a transformation's matrix relative to $\mathcal B$ .

Common mistakes

Forgetting that $[\mathbf x]_{\mathcal B}$ always lives in $\mathbb R^n$ even when $V$ itself isn't $\mathbb R^n$ — writing the answer back as a polynomial or a matrix instead of a plain column of numbers.
Matching coefficients before fully expanding the combination in an abstract space, e.g. forgetting to expand $(x-1)^2$ before comparing powers of $x$ .
Assuming $[\mathbf x]_{\mathcal B}=\mathbf x$ for a basis $\mathcal B$ that ISN'T the standard basis — that shortcut only ever applies to $\mathbb R^n$ 's own standard basis.
Listing the $c_i$ in a different order than $\mathcal B$ 's own stated order — $[\mathbf x]_{\mathcal B}$ 's entries must match which basis vector is $\mathbf b_1$ , which is $\mathbf b_2$ , and so on.