Encoding a Vector List as Matrix Columns

From a Vector List to a Matrix

When a list of vectors is involved in a linear combination, its coefficients can be collected into one column vector. Placing the original vectors in the columns of a matrix makes the matrix-vector product perform exactly that combination. In particular, setting the product equal to zero records precisely the coefficient choices that make a combination of the list equal to the zero vector.

Work over \(\mathbb{R}\). Let \(m,k\in\mathbb{N}\), and let \(\mathbf{v}_1,\dots,\mathbf{v}_k\) be an ordered list of vectors in \(\mathbb{R}^m\). “Ordered list” matters: each position has its own coefficient, even if two vectors happen to be equal. For coefficients \(x_1,\dots,x_k\in\mathbb{R}\), write their coefficient vector as the column \(\mathbf{x}=(x_1,\dots,x_k)^\top\).

The dimensions keep track of the roles: \(\mathbf{A}\) has \(m\) rows because each vector has \(m\) coordinates, and \(k\) columns because the list has \(k\) positions. Thus \(\mathbf{A}\mathbf{x}\) is a vector in \(\mathbb{R}^m\), the same space as the original vectors. We now verify exactly what that product contains.

The Product Is the Corresponding Linear Combination

Proof. Write the \(i\)th coordinate of column \(\mathbf{v}_j\) as \(v_{ij}\), where \(1\le i\le m\) and \(1\le j\le k\). By the definition of matrix multiplication, coordinate \(i\) of \(\mathbf{A}\mathbf{x}\) is \(\sum_{j=1}^k v_{ij}x_j\). Coordinate \(i\) of \(x_1\mathbf{v}_1+\cdots+x_k\mathbf{v}_k\) is also \(\sum_{j=1}^k x_jv_{ij}\). These coordinates agree for every \(i\), since real-number multiplication is commutative. Therefore the two vectors are equal. \(\square\)

This formula explains the construction: the coefficient \(x_j\) multiplies column \(j\), and adding the resulting vectors gives the product. A coefficient of zero simply contributes the zero vector; a negative coefficient reverses the direction of that column’s contribution. No column is discarded or rearranged when forming the product.

Why the Zero Equation Records Every Relation

A linear combination of the list is equal to zero exactly when the product formed from its coefficient vector is the zero vector. This is an equivalence: each coefficient choice gives one matrix-vector product, and each column vector of coefficients specifies one combination. The equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) therefore keeps all possible coefficient choices in view at once.

Proof. By the column-combination formula, \(\mathbf{A}\mathbf{x}=x_1\mathbf{v}_1+\cdots+x_k\mathbf{v}_k\). If the combination equals \(\mathbf{0}\), this equality gives \(\mathbf{A}\mathbf{x}=\mathbf{0}\). Conversely, if \(\mathbf{A}\mathbf{x}=\mathbf{0}\), the same equality says that the combination equals \(\mathbf{0}\). Both implications hold for every coefficient vector \(\mathbf{x}\), including \(\mathbf{x}=\mathbf{0}\). \(\square\)

Thus a solution of the homogeneous equation is not just a yes-or-no report. It is a specific coefficient vector, and its entries directly state how the original vectors combine. A nonzero solution supplies a nontrivial zero combination; the all-zero coefficient vector supplies the trivial one. The equation does not change the list or its vectors: it expresses the combination question in matrix notation.

The Matrix as a Rule on Coefficient Vectors

Once the columns are fixed, multiplication by \(\mathbf{A}\) turns every coefficient vector in \(\mathbb{R}^k\) into the corresponding combination in \(\mathbb{R}^m\). In other words, \(\mathbf{A}\) defines a function \(T_{\mathbf{A}}:\mathbb{R}^k\to\mathbb{R}^m\) by \(T_{\mathbf{A}}(\mathbf{x})=\mathbf{A}\mathbf{x}\). The matrix stores the input-output rule in its columns.

Proof. Write \(\mathbf{x}=(x_1,\dots,x_k)^\top\) and \(\mathbf{y}=(y_1,\dots,y_k)^\top\). The \(j\)th entry of \(a\mathbf{x}+b\mathbf{y}\) is \(ax_j+by_j\). Applying the column-combination formula gives \(T_{\mathbf{A}}(a\mathbf{x}+b\mathbf{y})=\sum_{j=1}^k(ax_j+by_j)\mathbf{v}_j\). Distributing each scalar and grouping the sums gives \(a\sum_{j=1}^k x_j\mathbf{v}_j+b\sum_{j=1}^k y_j\mathbf{v}_j\). Applying the formula again, this is \(aT_{\mathbf{A}}(\mathbf{x})+bT_{\mathbf{A}}(\mathbf{y})\), as required. \(\square\)

The zero equation asks which coefficient vectors this rule sends to the zero vector. The rule applies to all coefficient vectors, not only ones that produce zero. If a target \(\mathbf{b}\in\mathbb{R}^m\) is used instead, then \(\mathbf{A}\mathbf{x}=\mathbf{b}\) says that the columns combine to give \(\mathbf{b}\). The zero case is the same column-combination rule with target \(\mathbf{0}\).

A Common Mistake: Placing the Vectors as Rows

To encode a list for the product \(\mathbf{A}\mathbf{x}\), the vectors must be columns. A tempting but incorrect construction is to place the coordinates of each vector across a row and then multiply that matrix by the coefficient column. That product generally combines rows instead of the given vectors.

For instance, let \(\mathbf{v}_1=(1,2)^\top\), \(\mathbf{v}_2=(3,4)^\top\), and \(\mathbf{x}=(2,-1)^\top\). The correct column matrix is \(\mathbf{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix}\), so \(\mathbf{A}\mathbf{x}=(1(2)+3(-1),\;2(2)+4(-1))^\top=(-1,0)^\top\). Directly checking gives \(2\mathbf{v}_1-\mathbf{v}_2=(2,4)^\top-(3,4)^\top=(-1,0)^\top\).

The wrong row arrangement would be \(\begin{bmatrix}1&2\\3&4\end{bmatrix}\). Multiplying this by the same \(\mathbf{x}\) gives \((2-2,\;6-4)^\top=(0,2)^\top\), which is not \(2\mathbf{v}_1-\mathbf{v}_2\). The correction is to place each vector vertically as a column before multiplying by its coefficient vector.

Keep the roles distinct: the ordered list supplies the columns, the coefficient vector supplies one scalar per column, and the product is the resulting linear combination. With this setup, \(\mathbf{A}\mathbf{x}=\mathbf{0}\) means exactly that the chosen coefficients produce the zero vector; the equation itself neither rearranges the list nor changes the meaning of its entries.