Building a Matrix from Prescribed Basis Images

From Basis Images to Matrix Columns

A linear map is determined by what it does to a basis: once the images of the basis vectors are known, linearity determines the image of every vector. A matrix records this information in a particular way. Its \(j\)th column is the coordinate vector of the image of the \(j\)th domain basis vector, expressed in the chosen codomain basis. This tutorial develops that construction and uses it to calculate images reliably.

Work over \(\mathbb{R}\). Let \(V\) and \(W\) be finite-dimensional vector spaces, with ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) for \(V\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\) for \(W\). Because \(\mathcal{C}\) is a basis, each vector in \(W\) has a unique expression as a linear combination of its vectors. For \(\mathbf{w}\in W\), the column of coefficients in that expression is its coordinate vector \([\mathbf{w}]_{\mathcal{C}}\).

The arrow in the notation indicates the direction of the coordinate calculation: start with coordinates in \(\mathcal{B}\), and obtain coordinates in \(\mathcal{C}\). The columns are coefficient lists, not the images written in some unrelated coordinate system. If an image is given directly as a linear combination of the \(\mathbf{c}_i\), its coefficients can be placed into the corresponding column immediately.

Why the Prescribed Images Determine a Map

Proof. Since \(\mathcal{B}\) is a basis, each \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=x_1\mathbf{b}_1+\cdots+x_n\mathbf{b}_n\). Define \(T(\mathbf{v})=x_1\mathbf{w}_1+\cdots+x_n\mathbf{w}_n\). This definition gives \(T(\mathbf{b}_j)=\mathbf{w}_j\), because the coordinates of \(\mathbf{b}_j\) in \(\mathcal{B}\) are \(1\) in position \(j\) and \(0\) elsewhere. To check linearity, write \(\mathbf{u}=\sum_{j=1}^n x_j\mathbf{b}_j\), \(\mathbf{v}=\sum_{j=1}^n y_j\mathbf{b}_j\), and let \(a,b\in\mathbb{R}\). Then \(a\mathbf{u}+b\mathbf{v}=\sum_{j=1}^n(ax_j+by_j)\mathbf{b}_j\), so the definition gives \(T(a\mathbf{u}+b\mathbf{v})=\sum_{j=1}^n(ax_j+by_j)\mathbf{w}_j=aT(\mathbf{u})+bT(\mathbf{v})\). Thus \(T\) is linear. If \(S\) is another linear map with \(S(\mathbf{b}_j)=\mathbf{w}_j\) for all \(j\), then for each \(\mathbf{v}=\sum_{j=1}^n x_j\mathbf{b}_j\), linearity gives \(S(\mathbf{v})=\sum_{j=1}^n x_j\mathbf{w}_j=T(\mathbf{v})\). Hence \(S=T\), proving uniqueness. By the definition of the matrix of a linear map, its \(j\)th column is \([T(\mathbf{b}_j)]_{\mathcal{C}}=[\mathbf{w}_j]_{\mathcal{C}}\), as claimed. \(\square\)

This theorem means that any list of proposed images, including a list containing \(\mathbf{0}\), gives a valid and unique linear map. No extra compatibility condition is needed: the domain basis vectors can be assigned images freely, and every other input is handled by its unique basis expansion.

Using the Matrix for Any Input

Proof. Write \(\mathbf{v}=x_1\mathbf{b}_1+\cdots+x_n\mathbf{b}_n\), so \([\mathbf{v}]_{\mathcal{B}}=(x_1,\dots,x_n)^{\top}\). By linearity, \(T(\mathbf{v})=\sum_{j=1}^n x_jT(\mathbf{b}_j)\). Taking \(\mathcal{C}\)-coordinates gives \([T(\mathbf{v})]_{\mathcal{C}}=\sum_{j=1}^n x_j[T(\mathbf{b}_j)]_{\mathcal{C}}\). The vectors \([T(\mathbf{b}_j)]_{\mathcal{C}}\) are exactly the columns of \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\), so this sum is the matrix product \([T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). \(\square\)

Use the rule in order: find the input coordinates in the domain basis, multiply by the matrix, then interpret the resulting coefficients in the codomain basis. The result of the multiplication is an output coordinate vector; it is not necessarily the output vector written in standard coordinates.

When the Images Are Given as Ordinary Vectors

Sometimes a problem states \(T(\mathbf{b}_j)\) as a vector in familiar coordinates rather than as a combination of the codomain basis. The matrix column must still contain the coefficients relative to \(\mathcal{C}\). Find those coefficients by solving the basis expansion equation for each image before assembling the matrix.

Common Pitfall: Putting Images in Rows

Suppose the first worked example’s two basis images are written as the rows of a matrix instead of its columns. This produces the incorrect array \(\begin{bmatrix}2&-1&1\\-1&3&2\end{bmatrix}\). Multiplying that array by \(\begin{bmatrix}3\\1\end{bmatrix}\) is not even the required coordinate calculation: it does not place the two input coefficients against the two basis-image columns. The correct matrix is \(\begin{bmatrix}2&-1\\-1&3\\1&2\end{bmatrix}\), because column 1 stores the \(\mathcal{C}\)-coordinates of \(T(\mathbf{b}_1)\) and column 2 stores those of \(T(\mathbf{b}_2)\).

A second common error is to confuse coordinates with the vector they describe. In the third example, \(\begin{bmatrix}4\\-1\end{bmatrix}\) is the coordinate vector of the output in \(\mathcal{C}\), not the output in standard coordinates. The conversion \(4\mathbf{c}_1-\mathbf{c}_2=(3,5)\) is necessary to state the actual vector in \(\mathbb{R}^2\).

The construction is useful because it separates a map into two manageable tasks: encode its basis images as columns, then use matrix multiplication on input coordinates. Keeping the basis order fixed is essential throughout: changing the order of the domain basis changes the column order, and changing the codomain basis changes the coefficients in every column.