From Unitary Change of Basis to Triangular Form
Diagonalization represents a matrix using a basis of eigenvectors. Such a basis need not exist, even over \(\mathbb{C}\). Schur triangularization asks for less: instead of requiring every basis vector to be an eigenvector, it finds an orthonormal basis in which the matrix is upper triangular. That weaker requirement always succeeds for a complex square matrix. It preserves the eigenvalues on the diagonal while allowing entries above the diagonal to record interactions between basis vectors.
For a complex matrix \(\mathbf{U}\), its conjugate transpose \(\mathbf{U}^{*}\) is obtained by transposing and then taking the complex conjugate of every entry. A square matrix \(\mathbf{U}\) is unitary if \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}\); equivalently, \(\mathbf{U}\mathbf{U}^{*}=\mathbf{I}\). Its columns therefore form an orthonormal basis of \(\mathbb{C}^n\) under the standard inner product. Two square matrices \(\mathbf{A}\) and \(\mathbf{T}\) are unitarily similar if \(\mathbf{T}=\mathbf{U}^{*}\mathbf{A}\mathbf{U}\) for some unitary \(\mathbf{U}\).
Here, upper triangular means \(t_{ij}=0\) whenever \(i>j\). The columns of \(\mathbf{U}\) specify the orthonormal basis used to express the transformation: the \(j\)th column of \(\mathbf{A}\mathbf{U}\) is \(\mathbf{A}\mathbf{u}_j\), while the \(j\)th column of \(\mathbf{U}\mathbf{T}\) expresses that same vector in the columns of \(\mathbf{U}\). Thus the equivalent equation \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{T}\) says that the matrix of the transformation in this basis is \(\mathbf{T}\).
Proof. If \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\), multiply on the left by \(\mathbf{U}\). Since \(\mathbf{U}\mathbf{U}^{*}=\mathbf{I}\), this gives \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{T}\). Conversely, if \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{T}\), multiply on the left by \(\mathbf{U}^{*}\). Since \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}\), this gives \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\). Finally, multiplying this last equation on the left by \(\mathbf{U}\) and on the right by \(\mathbf{U}^{*}\) gives \(\mathbf{A}=\mathbf{U}\mathbf{T}\mathbf{U}^{*}\). Reversing those multiplications proves the converse. \(\square\)
What the Diagonal Entries Tell Us
The diagonal of a Schur form is not arbitrary: it lists the eigenvalues of the original matrix, including repetitions according to algebraic multiplicity. Recall that an eigenvalue \(\lambda\) is a scalar for which \(\mathbf{A}-\lambda\mathbf{I}\) is singular, and its algebraic multiplicity is its multiplicity as a root of the characteristic polynomial \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\).
Proof. Since \(\mathbf{A}-\lambda\mathbf{I}=\mathbf{U}(\mathbf{T}-\lambda\mathbf{I})\mathbf{U}^{*}\), multiplicativity of the determinant and \(\det(\mathbf{U})\det(\mathbf{U}^{*})=\det(\mathbf{U}\mathbf{U}^{*})=1\) give $$ p_{\mathbf{A}}(\lambda) =\det(\mathbf{A}-\lambda\mathbf{I}) =\det(\mathbf{T}-\lambda\mathbf{I}) =\prod_{j=1}^{n}(t_{jj}-\lambda). $$ The last equality holds because \(\mathbf{T}-\lambda\mathbf{I}\) is upper triangular, whose determinant is the product of its diagonal entries. The roots of this polynomial are exactly the \(t_{jj}\), with each repeated according to the number of times it occurs in the product. Those roots, with their multiplicities, are the eigenvalues of \(\mathbf{A}\). \(\square\)
This conclusion concerns the diagonal entries only; it does not say that \(\mathbf{T}\) is diagonal. An entry above the diagonal may be nonzero, and that possibility is exactly why triangularization is more general than unitary diagonalization.
Worked Matrix Forms
Worked Example: A Real Matrix with a Nonzero Triangular Entry
Let $$ \mathbf{A}=\begin{bmatrix}2&1\\ \tfrac12&\tfrac32\end{bmatrix}, \qquad \mathbf{U}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-1&1\end{bmatrix}. $$ Find and verify the corresponding triangular matrix.
The columns of \(\mathbf{U}\) are \(\tfrac{1}{\sqrt{2}}(1,-1)^{\top}\) and \(\tfrac{1}{\sqrt{2}}(1,1)^{\top}\). Each has norm \(1\), and their inner product is \(\tfrac12(1-1)=0\), so \(\mathbf{U}\) is unitary. Directly, \(\mathbf{U}^{*}=\mathbf{U}^{\top}\) and \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_2\). First calculate $$ \mathbf{A}\mathbf{U} =\frac{1}{\sqrt{2}} \begin{bmatrix} 2-1&2+1\\ \tfrac12-\tfrac32&\tfrac12+\tfrac32 \end{bmatrix} =\frac{1}{\sqrt{2}}\begin{bmatrix}1&3\\-1&2\end{bmatrix}. $$ Set \(\mathbf{T}=\begin{bmatrix}1&\tfrac12\\0&\tfrac52\end{bmatrix}\). Then $$ \mathbf{U}\mathbf{T} =\frac{1}{\sqrt{2}} \begin{bmatrix} 1&\tfrac12+\tfrac52\\ -1&-\tfrac12+\tfrac52 \end{bmatrix} =\frac{1}{\sqrt{2}}\begin{bmatrix}1&3\\-1&2\end{bmatrix} =\mathbf{A}\mathbf{U}. $$ Thus \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\). Check the reconstruction as well: $$ \mathbf{U}\mathbf{T}\mathbf{U}^{*} =\frac12 \begin{bmatrix}1&3\\-1&2\end{bmatrix} \begin{bmatrix}1&-1\\1&1\end{bmatrix} =\frac12 \begin{bmatrix}4&2\\1&3\end{bmatrix} =\mathbf{A}. $$ The diagonal entries are \(1\) and \(\tfrac52\). Indeed, $$ \det(\mathbf{A}-\lambda\mathbf{I}_2) =(2-\lambda)(\tfrac32-\lambda)-\tfrac12 =\lambda^2-\tfrac72\lambda+\tfrac52 =(1-\lambda)(\tfrac52-\lambda). $$ The characteristic polynomial confirms both eigenvalues.
Worked Example: A Complex Schur Form
Consider $$ \mathbf{T}=\begin{bmatrix}i&2\\0&1-i\end{bmatrix}, \qquad \mathbf{U}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-1&1\end{bmatrix}. $$ Construct the matrix \(\mathbf{A}=\mathbf{U}\mathbf{T}\mathbf{U}^{*}\), then verify its Schur form.
The same orthonormality calculation as in the previous example gives \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_2\). Since \(\mathbf{U}^{*}=\tfrac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\), multiply first to obtain $$ \mathbf{U}\mathbf{T} =\frac{1}{\sqrt{2}} \begin{bmatrix} i&2+1-i\\ -i&-2+1-i \end{bmatrix} =\frac{1}{\sqrt{2}} \begin{bmatrix}i&3-i\\-i&-1-i\end{bmatrix}. $$ Therefore $$ \begin{aligned} \mathbf{A} &=\frac12 \begin{bmatrix}i&3-i\\-i&-1-i\end{bmatrix} \begin{bmatrix}1&-1\\1&1\end{bmatrix}\\ &=\frac12 \begin{bmatrix}3&3-2i\\-1-2i&-1\end{bmatrix}. \end{aligned} $$ For the reverse check, multiply this \(\mathbf{A}\) by \(\mathbf{U}\): $$ \mathbf{A}\mathbf{U} =\frac{1}{\sqrt{2}} \begin{bmatrix}i&3-i\\-i&-1-i\end{bmatrix} =\mathbf{U}\mathbf{T}. $$ Multiplying by \(\mathbf{U}^{*}\) on the left now gives \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\). The diagonal entries are \(i\) and \(1-i\), so $$ \det(\mathbf{A}-\lambda\mathbf{I}_2) =(i-\lambda)(1-i-\lambda). $$ They are the two eigenvalues, each with algebraic multiplicity \(1\).
Worked Example: A Schur Form That Is Diagonal
Let \(\mathbf{A}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\) and $$ \mathbf{U}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-i&i\end{bmatrix}. $$ The columns of \(\mathbf{U}\) have norm \(1\). Their inner product, computed using conjugation in the second vector, is \(\tfrac12(1+i\cdot i)=0\), so \(\mathbf{U}\) is unitary. Set \(\mathbf{T}=\begin{bmatrix}i&0\\0&-i\end{bmatrix}\). Direct multiplication gives $$ \mathbf{A}\mathbf{U} =\frac{1}{\sqrt{2}}\begin{bmatrix}i&-i\\1&i\end{bmatrix} =\mathbf{U}\mathbf{T}. $$ Consequently \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\), and reconstruction gives \(\mathbf{U}\mathbf{T}\mathbf{U}^{*}=\mathbf{A}\). The characteristic polynomial is \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=\lambda^2+1=(i-\lambda)(-i-\lambda)\), in agreement with the diagonal. This example happens to be diagonal in an orthonormal basis; Schur's theorem guarantees triangular form in general, not diagonal form.
What the Form Does—and Does Not—Guarantee
A Schur form records eigenvalues in a convenient order along the diagonal, while the unitary matrix records the orthonormal basis that produces that order. Different choices of orthonormal basis can give different upper triangular matrices, so a Schur form need not be unique. The product identity is essential: the statement is not that \(\mathbf{A}\) itself is triangular, but that it becomes triangular after the change of basis.
Do not infer diagonalizability just from triangularizability. For instance, the upper triangular matrix \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\) has only one independent eigenvector: solving \((\mathbf{A}-\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\), so eigenvectors are multiples of \((1,0)^{\top}\). It is already in triangular form, but that fact alone does not provide a basis of eigenvectors. In particular, the Schur theorem is a statement about unitary similarity to an upper triangular matrix, not about unitary diagonalization.
Common pitfall: replacing the conjugate transpose by the transpose. In the rotation example, the columns of \(\mathbf{U}\) are complex. Using the ordinary transpose gives $$ \mathbf{U}^{\top}\mathbf{U} =\frac12\begin{bmatrix}1&-i\\1&i\end{bmatrix} \begin{bmatrix}1&1\\-i&i\end{bmatrix} =\begin{bmatrix}0&1\\1&0\end{bmatrix}, $$ which is not \(\mathbf{I}_2\). Thus \(\mathbf{U}\) is not orthogonal in the real-matrix sense. The correct calculation uses \(\mathbf{U}^{*}\), which conjugates the entries and satisfies \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_2\). Over \(\mathbb{C}\), writing \(\mathbf{U}^{\top}\mathbf{A}\mathbf{U}\) in place of \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}\) can invalidate both the change-of-basis claim and the eigenvalue interpretation.
Check Your Understanding
Use the matrix identity and the diagonal-entry interpretation to organize each answer.
- For \(\mathbf{U}^{*}\mathbf{A}\mathbf{U}=\mathbf{T}\), derive \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{T}\).
- If the diagonal of a \(3\times3\) Schur form is \(2,2,-1\), state the eigenvalues and their algebraic multiplicities.
- Explain why a nonzero entry above the diagonal does not contradict Schur triangularization.
- For the complex matrix in the rotation example, why is \(\mathbf{U}^{\top}\) not the appropriate inverse?
- Does Schur's theorem guarantee an orthonormal basis of eigenvectors? Explain the distinction between triangular and diagonal forms.
