Pairs, Coordinates, and the Direct Sum
When two vector spaces are combined as a direct sum, each vector records one component from each space. The inclusion maps place a vector into one component while setting the other to zero; the projection maps extract one component and discard the other. Their coordinate matrices are simple block matrices, but the blocks make sense only when the basis order is fixed and used consistently.
Throughout, the spaces are finite-dimensional vector spaces over \(\mathbb{R}\). For spaces \(V\) and \(W\), their direct sum \(V\oplus W\) is the vector space of ordered pairs \((\mathbf{v},\mathbf{w})\), where \(\mathbf{v}\in V\) and \(\mathbf{w}\in W\). Addition and scalar multiplication are defined componentwise: \((\mathbf{v},\mathbf{w})+(\mathbf{v}',\mathbf{w}')=(\mathbf{v}+\mathbf{v}',\mathbf{w}+\mathbf{w}')\) and \(a(\mathbf{v},\mathbf{w})=(a\mathbf{v},a\mathbf{w})\).
Each map is linear. For example, \(i_V(a\mathbf{v}+b\mathbf{v}')=(a\mathbf{v}+b\mathbf{v}',\mathbf{0})=a(\mathbf{v},\mathbf{0})+b(\mathbf{v}',\mathbf{0})\). For the projections, \(\pi_V(a(\mathbf{v},\mathbf{w})+b(\mathbf{v}',\mathbf{w}'))=a\mathbf{v}+b\mathbf{v}'=a\pi_V(\mathbf{v},\mathbf{w})+b\pi_V(\mathbf{v}',\mathbf{w}')\); the same componentwise calculation applies to \(i_W\) and \(\pi_W\).
Choose ordered bases \(\mathcal{B}=(\mathbf{b}_1,\ldots,\mathbf{b}_n)\) for \(V\) and \(\mathcal{C}=(\mathbf{c}_1,\ldots,\mathbf{c}_m)\) for \(W\). The basis order for the direct sum will be the concatenation \(\mathcal{D}=((\mathbf{b}_1,\mathbf{0}),\ldots,(\mathbf{b}_n,\mathbf{0}),(\mathbf{0},\mathbf{c}_1),\ldots,(\mathbf{0},\mathbf{c}_m))\). Thus all \(V\)-coordinates come first, followed by all \(W\)-coordinates.
Proof. The matrix of a linear map has, as its \(j\)th column, the coordinate vector of the image of the \(j\)th domain basis vector in the codomain basis. For \(i_V\), the \(j\)th domain basis vector is \(\mathbf{b}_j\), and its image is \((\mathbf{b}_j,\mathbf{0})\), which is the \(j\)th basis vector of \(\mathcal{D}\). Its coordinate column therefore has a \(1\) in position \(j\) and zeros elsewhere. These \(n\) columns form the block matrix with \(\mathbf{I}_n\) above \(\mathbf{O}_{m\times n}\). For \(i_W\), each \(\mathbf{c}_j\) maps to the \((n+j)\)th vector of \(\mathcal{D}\); its coordinate column has its \(1\) in position \(n+j\), giving the stated lower identity block. For \(\pi_V\), the first \(n\) basis vectors of \(\mathcal{D}\) map to \(\mathbf{b}_1,\ldots,\mathbf{b}_n\), and the last \(m\) map to \(\mathbf{0}\). The resulting columns are those of \(\mathbf{I}_n\) followed by \(m\) zero columns. For \(\pi_W\), the first \(n\) basis vectors map to \(\mathbf{0}\), and the last \(m\) map to \(\mathbf{c}_1,\ldots,\mathbf{c}_m\), giving \(n\) zero columns followed by \(\mathbf{I}_m\). These are exactly the four displayed matrices. \(\square\)
How the Blocks Act
The theorem gives a direct coordinate rule. If \(\mathbf{v}\) has coordinate column \(\mathbf{x}\in\mathbb{R}^n\) in \(\mathcal{B}\), and \(\mathbf{w}\) has coordinate column \(\mathbf{y}\in\mathbb{R}^m\) in \(\mathcal{C}\), then the pair \((\mathbf{v},\mathbf{w})\) has coordinate column \(\begin{bmatrix}\mathbf{x}\\\mathbf{y}\end{bmatrix}\) in \(\mathcal{D}\). The inclusions insert a zero block; the projections select one of the two blocks.
These rules also show why the direct-sum basis is a basis. Every pair has the displayed coordinate column, so the concatenated list spans \(V\oplus W\). If a linear combination of that list is zero, its \(V\)-component is a zero combination of \(\mathcal{B}\) and its \(W\)-component is a zero combination of \(\mathcal{C}\); independence of each basis forces all coefficients to vanish.
Proof. By the definitions and componentwise addition, \(i_V(\mathbf{v})+i_W(\mathbf{w})=(\mathbf{v},\mathbf{0})+(\mathbf{0},\mathbf{w})=(\mathbf{v},\mathbf{w})\), so a decomposition exists. Suppose also that \((\mathbf{v},\mathbf{w})=i_V(\mathbf{v}')+i_W(\mathbf{w}')\). The right-hand side is \((\mathbf{v}',\mathbf{w}')\). Equality of ordered pairs gives \(\mathbf{v}=\mathbf{v}'\) and \(\mathbf{w}=\mathbf{w}'\). Thus the decomposition is unique. \(\square\)
Worked Coordinate Calculations
Worked Example: Standard Bases in \(\mathbb{R}^2\oplus\mathbb{R}^3\)
Use the standard bases for \(V=\mathbb{R}^2\) and \(W=\mathbb{R}^3\). For \(\mathbf{x}=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}1\\0\\4\end{bmatrix}\), calculate both inclusions and both projections using the block matrices.
The concatenated basis has the two \(V\)-basis vectors first, then the three \(W\)-basis vectors. The inclusion matrices and their actions are
and
The projection matrices act on the pair coordinate column \(\begin{bmatrix}2&-1&1&0&4\end{bmatrix}^{\top}\):
Check: The first inclusion gives the coordinates of \((\mathbf{x},\mathbf{0})\), and the second gives those of \((\mathbf{0},\mathbf{y})\). The pair coordinates split as \(\begin{bmatrix}2\\-1\end{bmatrix}\) followed by \(\begin{bmatrix}1\\0\\4\end{bmatrix}\), which the two projections recover exactly.
Worked Example: Nonstandard Component Bases
Let \(V=W=\mathbb{R}^2\), with \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), where \(\mathbf{b}_1=(1,1)\), \(\mathbf{b}_2=(1,-1)\), and \(\mathcal{C}=(\mathbf{c}_1,\mathbf{c}_2)\), where \(\mathbf{c}_1=(1,0)\), \(\mathbf{c}_2=(1,2)\). Apply the inclusions and projections to component coordinates \(\mathbf{x}=\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}3\\-1\end{bmatrix}\).
Relative to the concatenated basis \(\mathcal{D}\), the matrices are
Multiplying the inclusion matrices gives
The pair has concatenated coordinates \(\begin{bmatrix}2&-1&3&-1\end{bmatrix}^{\top}\), so multiplication by the projection matrices gives
Check: The actual vectors represented are \(\mathbf{v}=2\mathbf{b}_1-\mathbf{b}_2=(2,2)-(1,-1)=(1,3)\) and \(\mathbf{w}=3\mathbf{c}_1-\mathbf{c}_2=(3,0)-(1,2)=(2,-2)\). Thus the pair is \(((1,3),(2,-2))\), and its two component coordinate columns are indeed \(\mathbf{x}\) and \(\mathbf{y}\).
Worked Example: Polynomial Components
Let \(V=P_1(\mathbb{R})\) with basis \(\mathcal{B}=(1+x,x)\), and let \(W=P_0(\mathbb{R})\) with basis \(\mathcal{C}=(1)\). For \(p(x)=3+2x\) and \(q(x)=-4\), find the pair coordinates and verify both projections.
Write \(p(x)=a(1+x)+bx=a+(a+b)x\). Matching coefficients gives \(a=3\) and \(a+b=2\), hence \(b=-1\). The coordinate column for \(p\) is \(\begin{bmatrix}3\\-1\end{bmatrix}\), while that for \(q\) is \(\begin{bmatrix}-4\end{bmatrix}\). The direct-sum coordinate column is therefore \(\begin{bmatrix}3&-1&-4\end{bmatrix}^{\top}\). The projection matrices give
For a further check, the inclusion of \(p\) gives \(\begin{bmatrix}3&-1&0\end{bmatrix}^{\top}\), representing \((3+2x,0)\), and the inclusion of \(q\) gives \(\begin{bmatrix}0&0&-4\end{bmatrix}^{\top}\), representing \((0,-4)\). Adding these coordinate columns gives \(\begin{bmatrix}3&-1&-4\end{bmatrix}^{\top}\), the coordinates of the original pair.
Common Pitfall: Changing the Basis Order
The block formulas depend on the concatenated order \(\mathcal{D}\). A frequent error is to use a matrix built for one order while reading coordinates in another. For example, with \(\dim V=2\) and \(\dim W=1\), suppose someone orders the basis as \(((\mathbf{b}_1,\mathbf{0}),(\mathbf{0},\mathbf{c}_1),(\mathbf{b}_2,\mathbf{0}))\), interleaving the components. A pair whose component coordinates are \((2,3)\) in \(V\) and \((-4)\) in \(W\) then has coordinate column \(\begin{bmatrix}2&-4&3\end{bmatrix}^{\top}\). Using the concatenated-order projection matrix \(\begin{bmatrix}1&0&0\\0&1&0\end{bmatrix}\) would give \(\begin{bmatrix}2&-4\end{bmatrix}^{\top}\), which incorrectly treats the \(W\)-coordinate as the second \(V\)-coordinate.
For the interleaved order, the correct matrix for the \(V\)-projection is \(\begin{bmatrix}1&0&0\\0&0&1\end{bmatrix}\), and it gives \(\begin{bmatrix}2&3\end{bmatrix}^{\top}\). The correction is not to memorize a different block formula, but to match each matrix column to the ordered basis vector it represents. With the stated concatenated order, the block formulas in the theorem apply without alteration.
Check Your Understanding
Use the concatenated basis order throughout unless a question explicitly gives a different order.
- Write the matrices of \(i_V\) and \(\pi_W\) when \(\dim V=1\) and \(\dim W=2\).
- For component coordinate columns \(\begin{bmatrix}4\\-2\end{bmatrix}\) and \(\begin{bmatrix}5\end{bmatrix}\), write the coordinate column of their pair in the concatenated basis.
- Explain why the columns of the matrix for \(i_W\) have their identity block below the zero block.
- If the pair coordinate column is \(\begin{bmatrix}1&3&-2&4\end{bmatrix}^{\top}\), with \(\dim V=2\) and \(\dim W=2\), find the coordinate columns returned by \(\pi_V\) and \(\pi_W\).
- In one sentence, describe how to correct a projection matrix if the ordered basis for the direct sum is changed.
