Projections and Their Two Subspaces
A projection keeps one component of a vector and removes another. To describe that action efficiently, we choose coordinates that separate the vectors it keeps from those it removes. The relevant subspaces are the image, consisting of all outputs of the map, and the kernel, consisting of all inputs sent to zero. In a suitable basis, the projection’s matrix records these two actions directly: identity on the image and zero on the kernel.
Throughout, let \(V\) be a finite-dimensional vector space over \(\mathbb{R}\). A linear map \(T:V\to V\) is a projection if applying it twice has the same effect as applying it once.
The equation \(T^2=T\) says that every output of \(T\) is fixed by another application of \(T\). Thus \(T\) acts as the identity on its image. By definition, it acts as zero on its kernel. These facts explain the two blocks in the adapted matrix, but first we establish that every vector can be split into one part from each subspace.
Proof. Let \(\mathbf{v}\in V\). Write \(\mathbf{v}=T(\mathbf{v})+(\mathbf{v}-T(\mathbf{v}))\). The first term belongs to \(\operatorname{im}T\). For the second term, linearity and \(T^2=T\) give \(T(\mathbf{v}-T(\mathbf{v}))=T(\mathbf{v})-T^2(\mathbf{v})=\mathbf{0}\), so it belongs to \(\ker T\). This proves that every vector is a sum of the required kind. To prove uniqueness, suppose \(\mathbf{y}\in\operatorname{im}T\cap\ker T\). Since \(\mathbf{y}\in\operatorname{im}T\), there is a \(\mathbf{z}\in V\) with \(\mathbf{y}=T(\mathbf{z})\). Then \(T(\mathbf{y})=T^2(\mathbf{z})=T(\mathbf{z})=\mathbf{y}\). Since \(\mathbf{y}\in\ker T\), we also have \(T(\mathbf{y})=\mathbf{0}\). Therefore \(\mathbf{y}=\mathbf{0}\). The intersection is just \(\{\mathbf{0}\}\), so a sum from the two subspaces is unique: if \(\mathbf{u}_1+\mathbf{k}_1=\mathbf{u}_2+\mathbf{k}_2\), then \(\mathbf{u}_1-\mathbf{u}_2=\mathbf{k}_2-\mathbf{k}_1\) belongs to both subspaces and must be zero. Hence both components agree. \(\square\)
Building the Adapted Basis
Let \(\mathcal{U}=(\mathbf{u}_1,\dots,\mathbf{u}_r)\) be a basis of \(\operatorname{im}T\), and let \(\mathcal{K}=(\mathbf{k}_1,\dots,\mathbf{k}_s)\) be a basis of \(\ker T\). The theorem says that the combined list, with the image basis first, spans \(V\). It also says no nonzero vector belongs to both subspaces, which ensures that the combined list is independent.
Proof. For spanning, take any \(\mathbf{v}\in V\). The direct-sum decomposition gives \(\mathbf{v}=\mathbf{u}+\mathbf{k}\) with \(\mathbf{u}\in\operatorname{im}T\) and \(\mathbf{k}\in\ker T\). The basis \(\mathcal{U}\) expresses \(\mathbf{u}\) as a linear combination of its listed vectors, and \(\mathcal{K}\) expresses \(\mathbf{k}\) as a linear combination of its listed vectors. Thus \(\mathbf{v}\) is a linear combination of the combined list. For independence, suppose \(\sum_{i=1}^{r}a_i\mathbf{u}_i+\sum_{j=1}^{s}b_j\mathbf{k}_j=\mathbf{0}\). Then \(\sum_{i=1}^{r}a_i\mathbf{u}_i=-\sum_{j=1}^{s}b_j\mathbf{k}_j\) belongs to both the image and kernel. Their intersection is \(\{\mathbf{0}\}\), so each side is zero. Independence of each subspace basis forces every \(a_i\) and \(b_j\) to be zero. The combined list is a basis. To find the matrix, recall that its \(j\)-th column is the coordinate vector of \(T\) applied to the \(j\)-th basis vector. For each image basis vector, \(T(\mathbf{u}_i)=\mathbf{u}_i\), since if \(\mathbf{u}_i=T(\mathbf{z})\), then \(T(\mathbf{u}_i)=T^2(\mathbf{z})=T(\mathbf{z})\). Its coordinate column is therefore the \(i\)-th identity column in the image coordinates. For each kernel basis vector, \(T(\mathbf{k}_j)=\mathbf{0}\), whose coordinate column is zero. These columns give exactly the displayed block matrix. \(\square\)
If \(r=0\), the image is \(\{\mathbf{0}\}\) and \(T\) is the zero map; the adapted matrix has only its zero block. If \(s=0\), the kernel is \(\{\mathbf{0}\}\) and the adapted matrix is the identity matrix. In the general case, the block form is a coordinate description of the action, not a claim that the matrix in every basis has this form.
Worked Examples
Worked Example: A Projection in \(\mathbb{R}^3\)
Let \(T:\mathbb{R}^3\to\mathbb{R}^3\) be represented in the standard basis by \(\mathbf{A}=\begin{bmatrix}1&0&-1\\0&1&-1\\0&0&0\end{bmatrix}\). Find bases for its image and kernel, then write its matrix in an adapted basis.
For \(\mathbf{x}=(x,y,z)^{\top}\), matrix multiplication gives \(T(\mathbf{x})=(x-z,y-z,0)^{\top}\). Every output lies in the span of \(\mathbf{e}_1=(1,0,0)^{\top}\) and \(\mathbf{e}_2=(0,1,0)^{\top}\), and both vectors are outputs: \(T(\mathbf{e}_1)=\mathbf{e}_1\) and \(T(\mathbf{e}_2)=\mathbf{e}_2\). Hence \(\mathcal{U}=(\mathbf{e}_1,\mathbf{e}_2)\) is a basis of the image. The kernel equation gives \(x-z=0\) and \(y-z=0\), so \(x=y=z\). Thus the kernel has basis \(\mathcal{K}=((1,1,1)^{\top})\).
Take \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2,(1,1,1)^{\top})\). This list spans because \((x,y,z)^{\top}=(x-z)\mathbf{e}_1+(y-z)\mathbf{e}_2+z(1,1,1)^{\top}\). It is independent because if \(a\mathbf{e}_1+b\mathbf{e}_2+c(1,1,1)^{\top}=\mathbf{0}\), the third coordinate gives \(c=0\), and the first two then give \(a=b=0\).
Applying \(T\) to the ordered basis vectors gives \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{0}\), respectively. Therefore \([T]_{\mathcal{B}}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\). For a direct matrix check, the change-of-coordinate matrix with these basis vectors as columns is \(\mathbf{P}=\begin{bmatrix}1&0&1\\0&1&1\\0&0&1\end{bmatrix}\). The columns of \(\mathbf{A}\mathbf{P}\) are \(T(\mathbf{e}_1)=\mathbf{e}_1\), \(T(\mathbf{e}_2)=\mathbf{e}_2\), and \(T(1,1,1)^{\top}=\mathbf{0}\), so \(\mathbf{A}\mathbf{P}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix} =\mathbf{P}[T]_{\mathcal{B}}\). This verifies the adapted-coordinate matrix.
Worked Example: A Projection That Is Not Orthogonal
Consider the map on \(\mathbb{R}^2\) given by \(T(x,y)=(x+y,0)\), whose standard matrix is \(\mathbf{A}=\begin{bmatrix}1&1\\0&0\end{bmatrix}\). Find an adapted basis and check idempotence.
The output is \((x+y,0)=(x+y)(1,0)\), and every multiple of \((1,0)\) occurs, so the image basis is \(\mathcal{U}=((1,0)^{\top})\). The equation \(T(x,y)=\mathbf{0}\) is \(x+y=0\), so a kernel basis is \(\mathcal{K}=((-1,1)^{\top})\). Every \((x,y)^{\top}\) has the decomposition \((x,y)^{\top}=(x+y)(1,0)^{\top}+y(-1,1)^{\top}\). The coefficients are unique: if \(a(1,0)+b(-1,1)=(0,0)\), the second coordinate gives \(b=0\), and the first gives \(a=0\). Thus \(\mathcal{B}=((1,0)^{\top},(-1,1)^{\top})\) is a basis.
The map fixes the first basis vector and kills the second, so \([T]_{\mathcal{B}}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\). Idempotence can also be checked in standard coordinates: \(\mathbf{A}^2=\begin{bmatrix}1&1\\0&0\end{bmatrix} \begin{bmatrix}1&1\\0&0\end{bmatrix} =\begin{bmatrix}1&1\\0&0\end{bmatrix}=\mathbf{A}\). This projection is not orthogonal with respect to the usual dot product: its image direction \((1,0)\) is not perpendicular to its kernel direction \((-1,1)\), since \((1,0)\cdot(-1,1)=-1\). Idempotence alone does not require perpendicular subspaces.
Worked Example: Constructing the Projection from Its Subspaces
In \(\mathbb{R}^3\), let the image subspace have basis \(\mathbf{u}_1=(1,0,1)^{\top}\), \(\mathbf{u}_2=(0,1,1)^{\top}\), and let the kernel have basis \(\mathbf{k}=(1,1,0)^{\top}\). Construct the projection onto the first subspace along the second.
First check that the three vectors form a basis. If \(a\mathbf{u}_1+b\mathbf{u}_2+c\mathbf{k}=\mathbf{0}\), the coordinates give \(a+c=0\), \(b+c=0\), and \(a+b=0\). The first two equations give \(a=b=-c\); the third gives \(-2c=0\), hence \(c=a=b=0\). For any \((x,y,z)^{\top}\), write \((x,y,z)^{\top}=a\mathbf{u}_1+b\mathbf{u}_2+c\mathbf{k}\). Matching coordinates gives \(x=a+c\), \(y=b+c\), \(z=a+b\). Solving, \(a=(x-y+z)/2\), \(b=(-x+y+z)/2\), and \(c=(x+y-z)/2\).
The projection keeps the image component and discards the kernel component, so \(T(x,y,z)^{\top}=a\mathbf{u}_1+b\mathbf{u}_2 =((x-y+z)/2,(-x+y+z)/2,z)^{\top}\). Its standard matrix is therefore \(\mathbf{A}=\begin{bmatrix}1/2&-1/2&1/2\\-1/2&1/2&1/2\\0&0&1\end{bmatrix}\). In the adapted basis \(\mathcal{B}=(\mathbf{u}_1,\mathbf{u}_2,\mathbf{k})\), the map sends \(\mathbf{u}_1\) to \(\mathbf{u}_1\), \(\mathbf{u}_2\) to \(\mathbf{u}_2\), and \(\mathbf{k}\) to \(\mathbf{0}\). Hence \([T]_{\mathcal{B}}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\).
Check the actions using the standard matrix. Its first column combination gives \(\mathbf{A}\mathbf{u}_1=(1,0,1)^{\top}=\mathbf{u}_1\), and its second gives \(\mathbf{A}\mathbf{u}_2=(0,1,1)^{\top}=\mathbf{u}_2\). Also \(\mathbf{A}\mathbf{k}=(0,0,0)^{\top}\). Since these vectors form a basis, these checks verify that the matrix fixes the image basis and kills the kernel basis, as required.
Every Direct-Sum Decomposition Defines a Projection
The construction works in the reverse direction too. Suppose a finite-dimensional space is already written as a direct sum \(V=U\oplus K\). Each vector then has a unique representation \(\mathbf{u}+\mathbf{k}\), with \(\mathbf{u}\in U\) and \(\mathbf{k}\in K\). Keeping \(\mathbf{u}\) and discarding \(\mathbf{k}\) defines a projection.
Proof. The direct-sum condition guarantees that each vector has a unique decomposition, so the rule assigns exactly one output to each input. For \(\mathbf{v}_1=\mathbf{u}_1+\mathbf{k}_1\), \(\mathbf{v}_2=\mathbf{u}_2+\mathbf{k}_2\), and scalars \(a,b\in\mathbb{R}\), their linear combination has decomposition \(a\mathbf{v}_1+b\mathbf{v}_2=(a\mathbf{u}_1+b\mathbf{u}_2)+(a\mathbf{k}_1+b\mathbf{k}_2)\). The two terms belong to \(U\) and \(K\), respectively. Applying the rule gives \(T(a\mathbf{v}_1+b\mathbf{v}_2)=a\mathbf{u}_1+b\mathbf{u}_2=aT(\mathbf{v}_1)+bT(\mathbf{v}_2)\), so \(T\) is linear. Every output belongs to \(U\), and every \(\mathbf{u}\in U\) is an output because \(T(\mathbf{u}+\mathbf{0})=\mathbf{u}\); thus \(\operatorname{im}T=U\). Finally, \(T(\mathbf{u}+\mathbf{k})=\mathbf{0}\) exactly when \(\mathbf{u}=\mathbf{0}\), so the kernel is precisely \(K\). Applying \(T\) twice to \(\mathbf{u}+\mathbf{k}\) gives \(T(\mathbf{u})=\mathbf{u}\), so \(T^2=T\). \(\square\)
Common Pitfall: The Matrix Depends on the Basis
A projection does not usually have an identity-and-zero matrix in the standard basis. For the map \(T(x,y)=(x+y,0)\), the standard matrix is \(\mathbf{A}=\begin{bmatrix}1&1\\0&0\end{bmatrix}\), not \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\). A tempting but incorrect calculation replaces the standard matrix by the adapted form without changing the basis. The correction is to check what the map does to the actual standard basis vectors: \(T(1,0)=(1,0)\) and \(T(0,1)=(1,0)\). Thus the standard-coordinate columns are \((1,0)^{\top}\) and \((1,0)^{\top}\), which gives the displayed matrix. The zero column belongs to the kernel basis vector \((-1,1)^{\top}\), not to the standard vector \((0,1)^{\top}\).
The practical procedure is therefore to identify the image and kernel, choose a basis for each, and put the image basis first. In that ordered basis the columns follow directly from the action: each image vector is fixed, and each kernel vector is sent to zero. No inner product is needed for this argument. If an inner product is present, an orthogonal projection is a special case in which the kernel is perpendicular to the image; a general projection need not have that property.
Check Your Understanding
Use the image-and-kernel description to justify each answer.
- For \(T(x,y)=(x-y,0)\), find a basis for the image and a basis for the kernel. Write the matrix in the basis with the image vector first.
- Explain why \(T^2=T\) implies that \(T\) fixes every vector in \(\operatorname{im}T\).
- Suppose a projection on a four-dimensional space has a two-dimensional image. What is the dimension of its kernel, and what block sizes appear in its adapted matrix?
- Given \(V=U\oplus K\), describe the projection onto \(U\) along \(K\) and state its image and kernel.
- For the matrix \(\begin{bmatrix}1&1\\0&0\end{bmatrix}\), check idempotence and explain why its standard matrix is not its adapted-basis matrix.
