Maps and Their Coordinate Matrices
An identity map leaves each vector unchanged, and a zero map sends every vector to zero. Their actions are simple, but their coordinate matrices depend on the bases used to describe inputs and outputs. In particular, the matrix of an identity map need not be an identity matrix: it must convert coordinates in the domain basis into coordinates in the codomain basis while representing the same vector.
Let \(V\) and \(W\) be finite-dimensional vector spaces over \(\mathbb{R}\), 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 \(W\). The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) records the coefficients of \(\mathbf{v}\) in the basis \(\mathcal{B}\). A representing matrix \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\) has \(m\) rows and \(n\) columns, and acts by \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Its columns are the coordinate vectors of the images of the domain basis vectors, written in the codomain basis.
The identity map has the same space as its domain and codomain, though the bases used for its input and output coordinates may differ. A zero map may have different domain and codomain spaces. If \(\dim V=n\) and \(\dim W=m\), its matrix therefore has \(m\) rows and \(n\) columns, just like any other map from \(V\) to \(W\).
The Identity Map in the Same or Different Bases
Write \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) for the domain basis of \(\operatorname{id}_V\), and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_n)\) for its codomain basis. These bases have the same number of vectors because both are bases of \(V\). Since \(\operatorname{id}_V(\mathbf{b}_j)=\mathbf{b}_j\), column \(j\) of the representing matrix is \([\mathbf{b}_j]_{\mathcal{C}}\).
Proof. Take any \(\mathbf{v}\in V\) and write \([\mathbf{v}]_{\mathcal{B}}=(x_1,\dots,x_n)^{\top}\). By the definition of coordinates, \(\mathbf{v}=\sum_{j=1}^{n}x_j\mathbf{b}_j\). The identity map leaves this vector unchanged, so its output coordinates in \(\mathcal{C}\) are \([\operatorname{id}_V(\mathbf{v})]_{\mathcal{C}}=\sum_{j=1}^{n}x_j[\mathbf{b}_j]_{\mathcal{C}}\). This is the product of the displayed matrix with \([\mathbf{v}]_{\mathcal{B}}\), proving the formula for every input. If \(\mathcal{B}=\mathcal{C}\), then \([\mathbf{b}_j]_{\mathcal{C}}=\mathbf{e}_j\) for each \(j\), so the matrix is \(\mathbf{I}_n\). Conversely, if the matrix is \(\mathbf{I}_n\), its \(j\)th column is \(\mathbf{e}_j\), so \([\mathbf{b}_j]_{\mathcal{C}}=\mathbf{e}_j=[\mathbf{c}_j]_{\mathcal{C}}\). Uniqueness of coordinates in the basis \(\mathcal{C}\) gives \(\mathbf{b}_j=\mathbf{c}_j\) for every \(j\). Thus the ordered bases agree. \(\square\)
This result separates the map from its matrix. The map \(\operatorname{id}_V\) always sends a vector to itself. Its matrix is \(\mathbf{I}_n\) only when the coordinates before and after applying the map are taken in the same ordered basis. With different bases, the matrix records how to express the unchanged vector in the output basis.
Worked Example: Identity Map with the Same Basis
Let \(V=\mathbb{R}^2\) and let \(\mathcal{B}=((1,1),(1,-1))\). Find the matrix of \(\operatorname{id}_V\) when \(\mathcal{B}\) is used for both input and output. Then apply it to the coordinate vector \((2,-1)^{\top}\).
Each domain basis vector has its own standard coordinate vector in \(\mathcal{B}\): \([\mathbf{b}_1]_{\mathcal{B}}=(1,0)^{\top}\) and \([\mathbf{b}_2]_{\mathcal{B}}=(0,1)^{\top}\). Therefore \([ \operatorname{id}_V ]_{\mathcal{B}\leftarrow\mathcal{B}}=\mathbf{I}_2\). Applying it gives
Check: The input represents \(2(1,1)-1(1,-1)=(1,3)\). The output coordinates represent the same vector, so the identity map has indeed left the vector unchanged.
Worked Example: Identity Map with Different Bases
Use \(\mathcal{B}=((1,1),(1,-1))\) for the input coordinates and the standard basis \(\mathcal{C}=((1,0),(0,1))\) for the output coordinates. Find the matrix of \(\operatorname{id}_{\mathbb{R}^2}\), then apply it to \((2,-1)^{\top}\).
The identity map sends \(\mathbf{b}_1=(1,1)\) to itself, whose standard coordinates are \((1,1)^{\top}\). It sends \(\mathbf{b}_2=(1,-1)\) to itself, whose standard coordinates are \((1,-1)^{\top}\). Placing these output coordinate vectors in the columns gives
For the given input coordinates, \(\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix} =\begin{bmatrix}2-1\\2+1\end{bmatrix} =\begin{bmatrix}1\\3\end{bmatrix}\).
Check: The input coordinates \((2,-1)^{\top}\) represent \(2(1,1)-(1,-1)=(1,3)\). The output coordinates are standard coordinates, and \((1,3)^{\top}\) represents exactly \((1,3)\). The matrix is not \(\mathbf{I}_2\), but the map still leaves the vector itself unchanged.
The Zero Map in Coordinates
For a zero map \(Z:V\to W\), every domain basis vector is sent to the zero vector in \(W\). Its coordinate vector in any codomain basis is the zero column. Thus all columns of its representing matrix are zero. The matrix shape still reflects both spaces: if the domain has dimension \(n\) and the codomain has dimension \(m\), there are \(n\) zero columns, each with \(m\) entries.
Proof. For every domain basis vector \(\mathbf{b}_j\), the definition of the zero map gives \(Z(\mathbf{b}_j)=\mathbf{0}\). The \(j\)th column of the representing matrix is therefore \([Z(\mathbf{b}_j)]_{\mathcal{C}}=[\mathbf{0}]_{\mathcal{C}}\), the zero column of length \(m\). This holds for all \(n\) columns, so the representing matrix is \(\mathbf{O}\) of size \(m\times n\). For an arbitrary \(\mathbf{v}\in V\), write \([\mathbf{v}]_{\mathcal{B}}=\mathbf{x}\). Then \(\mathbf{O}\mathbf{x}=\mathbf{0}\), and \(\mathbf{0}=[\mathbf{0}]_{\mathcal{C}}\). Hence the matrix action gives the coordinates of \(Z(\mathbf{v})=\mathbf{0}\), as claimed. \(\square\)
Worked Example: A Zero Map from R³ to R²
Let \(Z:\mathbb{R}^3\to\mathbb{R}^2\) send every vector to \((0,0)\). Use the standard bases in both spaces. Find its matrix and calculate its action on the arbitrary input coordinate vector \((2,-1,4)^{\top}\).
The domain has three basis vectors and the codomain has two. Each domain basis vector maps to the zero vector, whose standard coordinates are \((0,0)^{\top}\). Therefore the matrix has two rows and three columns:
Check: The input represents \((2,-1,4)\), and the defining rule for \(Z\) sends that vector to \((0,0)\). The computed output coordinates represent \((0,0)\), as required.
Reading the Matrices Correctly
For these maps, the columns give a direct check of the matrix. In an identity-map matrix, column \(j\) is the output coordinate vector of the \(j\)th domain basis vector, which is that same vector expressed in the codomain basis. In a zero-map matrix, every column is the coordinate vector of the zero output. In both cases, multiplying by an input coordinate vector confirms the effect on an arbitrary vector, not just on the basis vectors.
| Map | Domain and codomain dimensions | Matrix in the chosen bases | Effect on coordinates |
|---|---|---|---|
| \(\operatorname{id}_V\), same basis in and out | \(n\) and \(n\) | \(\mathbf{I}_n\) | Leaves the coordinate vector unchanged |
| \(\operatorname{id}_V\), different bases | \(n\) and \(n\) | Columns \([\mathbf{b}_j]_{\mathcal{C}}\) | Re-expresses the same vector in the output basis |
| Zero map \(V\to W\) | \(n\) and \(m\) | \(m\times n\) zero matrix | Sends every input coordinate vector to zero |
Common pitfall: using \(\mathbf{I}_n\) for an identity map regardless of basis. For the mixed bases in the second worked example, writing \(\mathbf{I}_2\) would give \(\mathbf{I}_2(2,-1)^{\top}=(2,-1)^{\top}\), which are not the standard coordinates of the vector represented by the input. The input represents \((1,3)\), so the correct output coordinates are \((1,3)^{\top}\). The correction is to use the images of the domain basis vectors as columns, giving \(\begin{bmatrix}1&1\\1&-1\end{bmatrix}\).
Common pitfall: assuming a zero-map matrix must be square. A \(3\times3\) zero matrix cannot represent a map from \(\mathbb{R}^3\) to \(\mathbb{R}^2\), because multiplying it by a three-entry input gives a three-entry output. The correct matrix is \(2\times3\), so its product has the two coordinates required for an output in \(\mathbb{R}^2\). Always count rows from the codomain dimension and columns from the domain dimension.
When a space has dimension zero, its basis is the empty list; the same coordinate rules still apply. An identity map on that space has the \(0\times0\) identity matrix, and a zero map with a zero-dimensional domain has a matrix with zero columns. These boundary cases follow the same principle: the matrix must act between coordinate spaces of the dimensions of the codomain and domain.
Check Your Understanding
In each question, state the bases used and check that the matrix action has the claimed coordinate interpretation.
- For the identity map on \(\mathbb{R}^2\), use \(\mathcal{B}=((1,0),(1,1))\) for input coordinates and the standard basis for output coordinates. Find the representing matrix.
- Using \(\mathcal{B}=((1,0),(1,1))\) for both input and output, find the identity-map matrix and apply it to \((3,-2)^{\top}\).
- What is the shape of the matrix of a zero map from a four-dimensional space to a two-dimensional space?
- Explain why every column of a zero-map matrix is zero, using the rule for columns of a representing matrix.
- A student uses \(\mathbf{I}_3\) for an identity map from \(\mathbb{R}^3\) to itself with different input and output bases. State what condition would make that choice correct.
