Choosing a Basis That Reveals Invariance
A linear map can look complicated in one basis and structured in another. When a subspace is invariant under the map, a useful choice is to list a basis of that subspace first, then complete the list to a basis of the full space. In this adapted basis, the matrix has a zero block in its lower-left corner. The other diagonal blocks are not arbitrary: they represent the map restricted to the subspace and the map induced on the quotient by that subspace.
Let \(V\) be a finite-dimensional vector space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(T:V\to V\) be linear. A subspace \(W\) of \(V\) is invariant under \(T\) if \(T(\mathbf{w})\in W\) for every \(\mathbf{w}\in W\). Invariance says that applying \(T\) to a vector in \(W\) never produces a component outside \(W\).
Such an extension always exists in a finite-dimensional space. To see why, begin with the independent list \(\mathcal{B}_W\) and take any basis \((\mathbf{v}_1,\dots,\mathbf{v}_n)\) of \(V\). Examine these vectors one at a time. Append \(\mathbf{v}_j\) if it is not in the span of the list already assembled; otherwise do not append it. Appending a vector outside the current span preserves independence. At the end, every \(\mathbf{v}_j\) is in the span of the assembled list, so that list spans \(V\). It is therefore a basis extending \(\mathcal{B}_W\).
The Block Form and Its Meaning
Use an adapted basis \(\mathcal{B}=(\mathbf{w}_1,\dots,\mathbf{w}_r,\mathbf{u}_1,\dots,\mathbf{u}_s)\), and write the matrix of \(T\) in this basis with columns giving the coordinates of the images of basis vectors. The first \(r\) images belong to \(W\), by invariance. Thus their coordinates in the final \(s\) basis directions are zero. The matrix must have the form shown below.
Here \(\mathbf{A}\) is \(r\times r\), \(\mathbf{B}\) is \(r\times s\), \(\mathbf{O}\) is the \(s\times r\) zero matrix, and \(\mathbf{C}\) is \(s\times s\). The block \(\mathbf{B}\) records the components in \(W\) of the images of the complement basis vectors. It need not be zero: the complement vectors themselves need not map into the complement.
Proof. For each \(j\) with \(1\le j\le r\), invariance gives \(T(\mathbf{w}_j)\in W\). Its expression in the adapted basis therefore uses only \(\mathbf{w}_1,\dots,\mathbf{w}_r\), so the last \(s\) entries of column \(j\) of \([T]_{\mathcal{B}}\) are zero. This proves the displayed block form. The first \(r\) entries of these columns are exactly the coordinates of \(T(\mathbf{w}_j)\) in the basis \(\mathcal{B}_W\). Since the restriction \(T|_W\) sends each \(\mathbf{w}_j\) to this vector, its matrix in \(\mathcal{B}_W\) is \(\mathbf{A}\).
For the quotient statement, define \(V/W\) to be the vector space of cosets \(\mathbf{v}+W=\{\mathbf{v}+\mathbf{w}:\mathbf{w}\in W\}\). Define \(\overline{T}:V/W\to V/W\) by \(\overline{T}(\mathbf{v}+W)=T(\mathbf{v})+W\). This is well-defined: if \(\mathbf{v}+W=\mathbf{v}'+W\), then \(\mathbf{v}-\mathbf{v}'\in W\), so invariance gives \(T(\mathbf{v})-T(\mathbf{v}')=T(\mathbf{v}-\mathbf{v}')\in W\), and hence the two output cosets agree. The cosets \(\mathbf{u}_1+W,\dots,\mathbf{u}_s+W\) form a basis of \(V/W\): they span because every vector of \(V\) is a combination of the adapted basis, and they are independent because a combination of the \(\mathbf{u}_j\) lying in \(W\) must have all its \(\mathbf{u}_j\)-coefficients zero by independence of the full adapted basis.
For each \(j\) with \(1\le j\le s\), write \(T(\mathbf{u}_j)\) in adapted-basis coordinates. Its last \(s\) entries are column \(j\) of \(\mathbf{C}\). Passing to the quotient discards the part in \(W\), so \(\overline{T}(\mathbf{u}_j+W)\) has exactly those last \(s\) coordinates in the coset basis. Therefore the matrix of \(\overline{T}\) is \(\mathbf{C}\), as claimed. \(\square\)
Reading the Blocks in Examples
Worked Example: A Two-Dimensional Invariant Subspace
Let \(T:\mathbb{R}^3\to\mathbb{R}^3\) have standard matrix
The first two columns give \(T(\mathbf{e}_1)=\mathbf{e}_1\) and \(T(\mathbf{e}_2)=2\mathbf{e}_1+3\mathbf{e}_2\), both in \(W\), so \(W\) is invariant. The ordered basis \((\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\) is adapted, and the blocks are
The restriction sends \(\mathbf{e}_1\) to \(\mathbf{e}_1\) and \(\mathbf{e}_2\) to \(2\mathbf{e}_1+3\mathbf{e}_2\), which verifies the block \(\mathbf{A}\). In the quotient, \(\mathbf{e}_1+W\) and \(\mathbf{e}_2+W\) are zero, so \(T(\mathbf{e}_3)+W=4\mathbf{e}_1+5\mathbf{e}_2+6\mathbf{e}_3+W=6(\mathbf{e}_3+W)\). Thus the quotient matrix is \([6]\), as indicated by \(\mathbf{C}\).
Check: the columns of the full matrix give the three images just stated, so the block decomposition reproduces \(\mathbf{M}\) exactly.
Worked Example: An Adapted Basis That Is Not the Standard Basis
Define \(T:\mathbb{R}^3\to\mathbb{R}^3\) by the standard matrix
Take \(\mathbf{w}_1=(1,1,0)\), \(\mathbf{u}_1=(0,1,0)\), and \(\mathbf{u}_2=(0,0,1)\). The matrix with these columns is \(\mathbf{P}=\begin{bmatrix}1&0&0\\1&1&0\\0&0&1\end{bmatrix}\), whose columns form a basis because its determinant is \(1\). Direct multiplication gives
Thus \(W=\operatorname{span}\{\mathbf{w}_1\}\) is invariant, and \((\mathbf{w}_1,\mathbf{u}_1,\mathbf{u}_2)\) is adapted. The displayed image computations give its matrix column by column:
The first block is \([2]\), the action on \(W\). The quotient basis is \((\mathbf{u}_1+W,\mathbf{u}_2+W)\); after terms in \(W\) are discarded, the two images are \(3(\mathbf{u}_1+W)\) and \(5(\mathbf{u}_1+W)+6(\mathbf{u}_2+W)\). Hence the quotient block is \(\begin{bmatrix}3&5\\0&6\end{bmatrix}\).
Check: the three computed images are \((2,2,0)\), \((1,4,0)\), and \((4,9,6)\). Multiplying the displayed adapted-basis matrix by the corresponding coordinate columns \((1,0,0)^{\top}\), \((0,1,0)^{\top}\), and \((0,0,1)^{\top}\) gives exactly those coordinates in \(\mathcal{B}\).
Worked Example: A Two-Dimensional Quotient Block
Let \(T:\mathbb{R}^4\to\mathbb{R}^4\) have standard matrix
With \(W=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\), the first two columns show that \(W\) is invariant. In the standard basis, which is adapted, the decomposition is
In the quotient basis \((\mathbf{e}_3+W,\mathbf{e}_4+W)\), the first image is \(T(\mathbf{e}_3)+W=2(\mathbf{e}_3+W)+(\mathbf{e}_4+W)\), while the second is \(T(\mathbf{e}_4)+W=7(\mathbf{e}_3+W)+3(\mathbf{e}_4+W)\). These are exactly the columns of \(\mathbf{C}\).
Check: multiplying the full matrix by each standard basis vector returns its corresponding column: \((1,0,0,0)^{\top}\), \((2,1,0,0)^{\top}\), \((3,5,2,1)^{\top}\), and \((4,6,7,3)^{\top}\). The first two images lie in \(W\); the last two yield the stated quotient coordinates after the first two components are discarded.
Why the Lower-Left Block Must Vanish
The zero block is the matrix expression of invariance, not a general feature of a chosen basis. The same reasoning also gives the converse: if a basis begins with a basis of \(W\) and the lower-left block is zero, then the images of the first basis vectors have no coordinates outside \(W\). Their images lie in \(W\), and linearity then shows that every vector in \(W\) maps into \(W\).
Proof. If \(W\) is invariant, then each \(T(\mathbf{w}_j)\) lies in \(W\), so its coordinates in every \(\mathbf{u}_i\) direction are zero. These coordinates are precisely the entries of the lower-left block. Conversely, suppose that block is zero. Then for each \(j\le r\), the coordinate column of \(T(\mathbf{w}_j)\) has zero entries in all complement directions, so \(T(\mathbf{w}_j)\in W\). Any \(\mathbf{w}\in W\) has the form \(\sum_{j=1}^r c_j\mathbf{w}_j\). By linearity, \(T(\mathbf{w})=\sum_{j=1}^r c_jT(\mathbf{w}_j)\), a linear combination of vectors in \(W\), and therefore belongs to \(W\). This proves invariance. \(\square\)
Common Pitfall: Reversing the Block Direction
A frequent error is to treat coordinate vectors as rows when reading the matrix. With the column convention, the \(j\)-th column is the coordinate vector of \(T\) applied to the \(j\)-th basis vector. For example, consider \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2)\) and \(W=\operatorname{span}\{\mathbf{e}_1\}\). A mistaken lower-triangular candidate is
Its first column says \(T(\mathbf{e}_1)=2\mathbf{e}_1+\mathbf{e}_2\), which is not in \(W\). Thus this matrix does not describe an invariant first-coordinate subspace. For invariance, the first column must have zero second entry; for instance, \(\begin{bmatrix}2&1\\0&3\end{bmatrix}\) has the required zero lower-left block. It sends \(\mathbf{e}_1\) to \(2\mathbf{e}_1\), while it may send \(\mathbf{e}_2\) partly into \(W\).
The adapted form separates two actions: \(\mathbf{A}\) describes how vectors already in \(W\) move within \(W\), while \(\mathbf{C}\) describes how cosets move after contributions in \(W\) are ignored. The off-diagonal block records the part of the motion from the complement directions into \(W\). This makes the block form useful whenever a problem singles out an invariant subspace and asks how the map behaves both on that subspace and beyond it.
Check Your Understanding
Use the column convention throughout. When a matrix is requested, state the basis in which it is written.
- For the matrix \(\begin{bmatrix}2&1&0\\0&3&4\\0&0&5\end{bmatrix}\), verify that the span of the first two standard basis vectors is invariant, and identify the restriction and quotient blocks.
- Explain why the images of the complement basis vectors may have nonzero components in the invariant subspace.
- For an adapted basis with three vectors in \(W\) and two additional vectors, give the sizes of the four blocks in the adapted-basis matrix.
- Show that the cosets of the added basis vectors span \(V/W\).
- For the matrix \(\begin{bmatrix}1&0\\2&3\end{bmatrix}\), decide whether the span of the first standard basis vector is invariant, and justify the answer from the first column.
