Coordinates as Coefficients to Be Found
A basis gives a unique way to build each vector from a fixed ordered list. In practice, the vector may be given while its coefficients in that list are unknown. Finding coordinates is then a system-solving problem: express the target as a linear combination of the basis vectors, and solve for the coefficients. Row reduction is a direct method for doing so, provided the basis vectors are placed in the correct order as the columns of a matrix.
Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis of \(\mathbb{R}^n\), and let \(\mathbf{v}\in\mathbb{R}^n\). The basis matrix \(\mathbf{B}\) has the vectors \(\mathbf{b}_1,\dots,\mathbf{b}_n\) as its columns. The coordinate vector records the coefficients in that same order. Thus finding coordinates means solving \(\mathbf{B}\mathbf{c}=\mathbf{v}\), where \(\mathbf{c}\) is the unknown column of coefficients.
The displayed reconstruction equation follows from matrix multiplication: multiplying \(\mathbf{B}\) by a column of coefficients forms the corresponding linear combination of its columns. Since a basis spans \(\mathbb{R}^n\), such coefficients exist for every \(\mathbf{v}\). Since a basis is linearly independent, they are unique. These two parts of the definition explain why the coordinate problem has exactly one answer.
The Augmented-Matrix Method
To solve \(\mathbf{B}\mathbf{c}=\mathbf{v}\), append the target vector to the basis matrix and row reduce the augmented matrix \(\begin{bmatrix}\mathbf{B}&\mathbf{v}\end{bmatrix}\). Use elementary row operations on the entire augmented matrix, including its final column. The resulting equations have the same solutions for the coefficients as the original equations.
Proof. Because \(\mathcal{B}\) spans \(\mathbb{R}^n\), there are coefficients \(c_1,\dots,c_n\) such that \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{b}_j\). By the definition of the basis matrix, this equality is \(\mathbf{B}\mathbf{c}=\mathbf{v}\), where \(\mathbf{c}=(c_1,\dots,c_n)^{\top}\). If another coefficient vector \(\mathbf{d}\) also solves the equation, then \(\mathbf{B}(\mathbf{c}-\mathbf{d})=\mathbf{0}\). This says that the linear combination of the basis vectors with coefficients given by \(\mathbf{c}-\mathbf{d}\) is zero. Their linear independence implies \(\mathbf{c}-\mathbf{d}=\mathbf{0}\), so \(\mathbf{c}=\mathbf{d}\). Thus the system has exactly one solution. Each elementary row operation preserves the solution set: swapping equations does not change which equations must hold; multiplying an equation by a nonzero scalar is reversible; and adding a multiple of one equation to another is reversible by subtracting that same multiple. Therefore row reduction preserves the unique solution, and solving the reduced system recovers \([\mathbf{v}]_{\mathcal{B}}\). Substitution into \(\mathbf{B}\mathbf{c}=\mathbf{v}\) checks the result. \(\square\)
The theorem gives both the method and its reason. Coordinates are not read from the target vector by inspection unless the basis happens to make that convenient; they are the solution to a system determined by the basis columns. The final multiplication is useful even when every row-reduction step is correct: it confirms that the computed coefficients reconstruct the given vector.
Worked Coordinate Calculations
Worked Example: A Nonsymmetric Basis in \(\mathbb{R}^3\)
Find the coordinates of \(\mathbf{v}=(1,2,3)^{\top}\) in the ordered basis \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3)\), where \(\mathbf{b}_1=(1,0,0)^{\top}\), \(\mathbf{b}_2=(1,1,0)^{\top}\), and \(\mathbf{b}_3=(1,1,1)^{\top}\). These vectors form a basis: if \(a\mathbf{b}_1+b\mathbf{b}_2+c\mathbf{b}_3=\mathbf{0}\), the third coordinate gives \(c=0\), the second then gives \(b=0\), and the first gives \(a=0\); and the three vectors span \(\mathbb{R}^3\), as the row reduction below also confirms.
Put the basis vectors in columns and append the target:
This matrix is already in echelon form. Read the equations from the bottom row upward: \(c_3=3\), \(c_2+c_3=2\), and \(c_1+c_2+c_3=1\). Thus \(c_2=-1\) and \(c_1=-1\), giving
Check: multiply the basis matrix by this coordinate vector: \(\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\begin{bmatrix}-1\\-1\\3\end{bmatrix} =\begin{bmatrix}-1-1+3\\-1+3\\3\end{bmatrix} =\begin{bmatrix}1\\2\\3\end{bmatrix}=\mathbf{v}\). Equivalently, \(-\mathbf{b}_1-\mathbf{b}_2+3\mathbf{b}_3=(1,2,3)^{\top}\).
Worked Example: Row Reducing a Two-Vector Basis
Find the coordinates of \(\mathbf{v}=(7,4)^{\top}\) in the ordered list \(\mathcal{B}=( (2,1)^{\top}, (1,1)^{\top})\). Its basis matrix is \(\mathbf{B}=\begin{bmatrix}2&1\\1&1\end{bmatrix}\). The columns are independent: \(2a+b=0\) and \(a+b=0\) imply \(a=0\), then \(b=0\). They span \(\mathbb{R}^2\), since the system below has a solution for the target and, more generally, the row reduction has a pivot in each coefficient column.
Reduce the augmented matrix. First interchange the two rows, then subtract twice the first row from the second, multiply the second row by \(-1\), and subtract the second row from the first:
The final equations give \(c_1=3\) and \(c_2=1\), so \([\mathbf{v}]_{\mathcal{B}}=(3,1)^{\top}\). Check: \(3(2,1)^{\top}+1(1,1)^{\top}=(6+1,3+1)^{\top}=(7,4)^{\top}\), as required.
Worked Example: Coordinates of a Polynomial
Find the coordinates of \(p(x)=3+2x+x^2\) in the ordered basis \(\mathcal{B}=(1+x,\;x+x^2,\;1+x^2)\) of \(P_2\), the vector space of real polynomials of degree at most \(2\). Use the standard basis \((1,x,x^2)\) to write coefficient columns. The columns of the basis matrix are therefore \((1,1,0)^{\top}\), \((0,1,1)^{\top}\), and \((1,0,1)^{\top}\). They are independent: if \(a+c=0\), \(a+b=0\), and \(b+c=0\), then \(b=-a\), \(c=a\), and the last equation gives \(a=0\), hence \(b=c=0\). Thus this list is a basis of \(P_2\).
The target polynomial has standard-coordinate column \((3,2,1)^{\top}\). Row reduce the augmented matrix:
The operations were \(R_2\to R_2-R_1\), then \(R_3\to R_3-R_2\), then \(R_3\to \tfrac12R_3\). Now apply \(R_2\to R_2+R_3\), followed by \(R_1\to R_1-R_3\):
Hence \([p]_{\mathcal{B}}=(2,0,1)^{\top}\). Check: \(2(1+x)+0(x+x^2)+1(1+x^2)=2+2x+1+x^2=3+2x+x^2\). In polynomial spaces, the same method applies after representing every polynomial in the same standard basis and using those coefficient columns consistently.
Why the Vectors Must Be Columns
The order and orientation in the basis matrix are part of the method. If \(\mathbf{B}=[\mathbf{b}_1\ \cdots\ \mathbf{b}_n]\), then \(\mathbf{B}\mathbf{c}\) is exactly \(c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). Writing the vectors as rows instead creates \(\mathbf{B}^{\top}\), which generally describes a different system. The solution of that different system need not be the coordinates of the target in the stated basis.
For the first worked example, the correct basis matrix is \(\begin{bmatrix}1&1&1\\0&1&1\\0&0&1\end{bmatrix}\). If the vectors are incorrectly arranged as rows, the matrix becomes \(\begin{bmatrix}1&0&0\\1&1&0\\1&1&1\end{bmatrix}\). Solving the resulting, incorrect equation \(\mathbf{B}^{\top}\mathbf{c}=(1,2,3)^{\top}\) gives \(c_1=1\), \(c_1+c_2=2\), and \(c_1+c_2+c_3=3\), hence the wrong candidate \(\mathbf{c}=(1,1,1)^{\top}\). Substituting it into the actual basis matrix gives
The check catches the error: the candidate coefficients reconstruct a different vector. The correct coordinates, already found by using the basis vectors as columns, are \((-1,-1,3)^{\top}\). This example also shows why one must check the answer against the original basis matrix, not against the incorrectly arranged matrix.
A Reliable Routine and Its Limits
For a stated basis in \(\mathbb{R}^n\), a dependable calculation follows this sequence:
- Write the basis as an ordered list, preserving the given order.
- Place its vectors as the columns of \(\mathbf{B}\).
- Append the target vector to form \(\begin{bmatrix}\mathbf{B}&\mathbf{v}\end{bmatrix}\).
- Row reduce the entire augmented matrix and solve for the coefficient column.
- Multiply \(\mathbf{B}\) by the proposed coordinates and compare the result with \(\mathbf{v}\).
The basis hypothesis matters. If a list is not a basis, the system may have no solution for some target vectors or more than one solution for others. For instance, a dependent list can have two different coefficient columns that reconstruct the same vector: any nonzero dependence relation can be added to one solution to make another. Thus a unique coordinate vector is guaranteed by a basis, not merely by having a list of vectors. When the problem states that the list is a basis, the coordinate theorem ensures that row reduction will lead to a consistent system with a unique solution.
A final practical point is that the coordinate vector depends on the ordered basis. Changing the order of the basis vectors changes the order of the coefficients, even though the reconstructed vector remains the same when each coefficient stays attached to its corresponding basis vector. Record the ordered list before forming the matrix, and keep that order through the row reduction and the final check.
Check Your Understanding
For each problem, keep the basis vectors as columns and verify your proposed coordinate vector by reconstruction.
- Find the coordinates of \((5,3)^{\top}\) in the ordered basis \(((1,1)^{\top},(2,1)^{\top})\) of \(\mathbb{R}^2\).
- For the basis \(((1,0,0)^{\top},(1,1,0)^{\top},(1,1,1)^{\top})\), set up the augmented matrix for the target \((2,5,4)^{\top}\).
- Explain why elementary row operations on the augmented matrix preserve the solutions for the coordinates.
- Write the coefficient columns of \(1+x\), \(x+x^2\), and \(1+x^2\) in the standard basis \((1,x,x^2)\).
- Using the basis from the first worked example, explain why \((1,1,1)^{\top}\) is not the coordinate vector of \((1,2,3)^{\top}\).
