Normality Meets Upper Triangularity
An upper triangular matrix has a built-in asymmetry: entries below its main diagonal are zero, while entries above it may be nonzero. Normality imposes a balance between rows and columns. When these two conditions hold at once, the balance is so restrictive that every entry above the diagonal must vanish as well. The result is useful when a matrix has been put into triangular form by a change of orthonormal coordinates: if the matrix is normal, that triangular form is already diagonal.
We work over \(\mathbb{C}\). For a complex matrix \(\mathbf{T}\), its adjoint \(\mathbf{T}^{*}\) is its conjugate transpose: its rows and columns are interchanged, and each entry is replaced by its complex conjugate. For example, if \(t_{ij}\) is the entry in row \(i\), column \(j\), then the entry in row \(i\), column \(j\) of \(\mathbf{T}^{*}\) is \(\overline{t_{ji}}\).
The equality defining normality is an equality of whole matrices, so corresponding entries must agree. In particular, their diagonal entries agree. Those entries have a useful interpretation as sums of squared absolute values.
Proof. By matrix multiplication, the entry in position \((j,j)\) of \(\mathbf{T}^{*}\mathbf{T}\) is \((\mathbf{T}^{*}\mathbf{T})_{jj}=\sum_{i=1}^{n}\overline{t_{ij}}t_{ij}=\sum_{i=1}^{n}|t_{ij}|^2\). The entry in position \((j,j)\) of \(\mathbf{T}\mathbf{T}^{*}\) is \((\mathbf{T}\mathbf{T}^{*})_{jj}=\sum_{k=1}^{n}t_{jk}\overline{t_{jk}}=\sum_{k=1}^{n}|t_{jk}|^2\). Normality says these two matrices are equal, so the corresponding diagonal entries are equal. \(\square\)
This lemma is a necessary condition for normality, not a replacement for the full matrix identity: equality of the diagonal entries alone does not assert equality of all entries. For an upper triangular matrix, however, the condition gives a strong starting point. Its first column has no entries below the diagonal, so it contains only its diagonal entry. The first row may have additional entries. Equality of the first row and column norms therefore forces those additional entries to be zero.
Worked Low-Dimensional Cases
Worked Example: A Two-by-Two Upper Triangular Matrix
Let \(\mathbf{T}=\begin{bmatrix}a&b\\0&d\end{bmatrix}\), where \(a,b,d\in\mathbb{C}\). Determine what normality requires by computing both products.
First take the conjugate transpose and multiply:
In the opposite order, the product is
If \(\mathbf{T}\) is normal, equality of the \((1,1)\) entries gives \(|a|^2=|a|^2+|b|^2\), hence \(|b|^2=0\). Since a squared absolute value is nonnegative and vanishes only when its entry is zero, \(b=0\). Conversely, if \(b=0\), both products equal \(\begin{bmatrix}|a|^2&0\\0&|d|^2\end{bmatrix}\), so \(\mathbf{T}\) is normal. Thus this two-by-two upper triangular matrix is normal exactly when it is diagonal.
Check: Setting \(b=0\) in each displayed product gives the same diagonal matrix. If \(b\ne0\), then \(|b|^2>0\), so the two \((1,1)\) entries differ and normality fails.
Worked Example: Forcing Entries to Vanish in Three Dimensions
Consider the general upper triangular matrix \(\mathbf{T}=\begin{bmatrix}a&b&c\\0&d&e\\0&0&f\end{bmatrix}\), with complex entries. The diagonal entries of its two normality products are \((|a|^2,\ |b|^2+|d|^2,\ |c|^2+|e|^2+|f|^2)\) and \((|a|^2+|b|^2+|c|^2,\ |d|^2+|e|^2,\ |f|^2)\), respectively. If \(\mathbf{T}\) is normal, equality in position \((1,1)\) gives \(|b|^2+|c|^2=0\). Both summands are nonnegative, so \(b=c=0\). With these entries zero, equality in position \((2,2)\) gives \(|d|^2=|d|^2+|e|^2\), so \(e=0\).
Thus normality forces this three-by-three upper triangular matrix to be diagonal. To check the result in the full identity, set \(b=c=e=0\). Then \(\mathbf{T}=\operatorname{diag}(a,d,f)\), and direct multiplication gives \(\mathbf{T}^{*}\mathbf{T}=\mathbf{T}\mathbf{T}^{*}=\operatorname{diag}(|a|^2,|d|^2,|f|^2)\).
Worked Example: An Upper Triangular Matrix That Is Not Normal
Upper triangularity alone does not imply diagonality. Take \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&3&0\\0&0&4\end{bmatrix}\). Its entries are real, so \(\mathbf{A}^{*}=\mathbf{A}^{\top}\). Multiplying the columns against one another gives \(\mathbf{A}^{*}\mathbf{A}=\begin{bmatrix}1&2&0\\2&13&0\\0&0&16\end{bmatrix}\): the second diagonal entry is \(2^2+3^2=13\). Multiplying the rows against one another gives \(\mathbf{A}\mathbf{A}^{*}=\begin{bmatrix}5&6&0\\6&9&0\\0&0&16\end{bmatrix}\): the first diagonal entry is \(1^2+2^2=5\).
Since the products differ, \(\mathbf{A}\) is not normal. The first diagonal entries already show the row-column imbalance: the first column has squared norm \(1\), while the first row has squared norm \(5\).
Check: The \((1,1)\) entries are \(1\) and \(5\), so the products cannot be equal. This agrees with the computed matrices and confirms that upper triangularity without normality does not force the off-diagonal entry \(2\) to vanish.
The General Result
Proof. We use induction on \(n\), the size of the matrix. If \(n=1\), the matrix has no off-diagonal entries and is diagonal. Now suppose \(n\ge2\), and assume the theorem holds for upper triangular normal matrices of size \(n-1\). Write \(\mathbf{T}\) in block form as \(\mathbf{T}=\begin{bmatrix}a&\mathbf{r}\\\mathbf{0}&\mathbf{B}\end{bmatrix}\), where \(a\in\mathbb{C}\), \(\mathbf{r}\) is a row with \(n-1\) entries, and \(\mathbf{B}\in\mathbb{C}^{(n-1)\times(n-1)}\) is upper triangular.
The first column of \(\mathbf{T}\) has first entry \(a\) and all remaining entries zero, so its squared norm is \(|a|^2\). Its first row has entries \(a\) followed by the entries of \(\mathbf{r}\), so its squared norm is \(|a|^2+\sum_{j=1}^{n-1}|r_j|^2\). The lemma and normality equate these norms. Therefore \(\sum_{j=1}^{n-1}|r_j|^2=0\), which forces every \(r_j=0\). Hence \(\mathbf{r}=\mathbf{0}\) and \(\mathbf{T}=\begin{bmatrix}a&\mathbf{0}\\\mathbf{0}&\mathbf{B}\end{bmatrix}\).
For this block diagonal matrix, direct block multiplication gives \(\mathbf{T}^{*}\mathbf{T}=\begin{bmatrix}|a|^2&\mathbf{0}\\\mathbf{0}&\mathbf{B}^{*}\mathbf{B}\end{bmatrix}\) and \(\mathbf{T}\mathbf{T}^{*}=\begin{bmatrix}|a|^2&\mathbf{0}\\\mathbf{0}&\mathbf{B}\mathbf{B}^{*}\end{bmatrix}\). Since \(\mathbf{T}\) is normal, these products are equal, so \(\mathbf{B}^{*}\mathbf{B}=\mathbf{B}\mathbf{B}^{*}\). Thus \(\mathbf{B}\) is normal as well as upper triangular. The induction hypothesis implies that \(\mathbf{B}\) is diagonal. The block form of \(\mathbf{T}\), with \(\mathbf{r}=\mathbf{0}\), now shows that \(\mathbf{T}\) is diagonal. This completes the induction. \(\square\)
The proof does not require that the diagonal entries be distinct, nonzero, or real. It uses only upper triangularity, normality, and the fact that a sum of squared absolute values can be zero only when each term is zero.
Common Pitfall: Transpose Is Not the Adjoint
Over \(\mathbb{C}\), replacing \(\mathbf{T}^{*}\) with \(\mathbf{T}^{\top}\) can give a false test for normality. For a concrete example, let \(\mathbf{C}=\begin{bmatrix}1&i\\i&0\end{bmatrix}\). It is symmetric, so the incorrect transpose-based test gives \(\mathbf{C}^{\top}\mathbf{C}=\mathbf{C}\mathbf{C}^{\top}=\mathbf{C}^{2}=\begin{bmatrix}0&i\\i&-1\end{bmatrix}\). That calculation would incorrectly declare \(\mathbf{C}\) normal.
The correction is to conjugate the entries as well as transpose. Here \(\mathbf{C}^{*}=\begin{bmatrix}1&-i\\-i&0\end{bmatrix}\), and the actual products are \(\mathbf{C}^{*}\mathbf{C}=\begin{bmatrix}2&i\\-i&1\end{bmatrix}\) and \(\mathbf{C}\mathbf{C}^{*}=\begin{bmatrix}2&-i\\i&1\end{bmatrix}\). They differ, so \(\mathbf{C}\) is not normal. This example is not triangular; it illustrates why the conjugate transpose in the definition must be used throughout, including when applying the row-column norm argument.
Why the Result Matters
Schur triangularization gives a complex matrix an upper triangular representation in suitable orthonormal coordinates. This theorem explains what happens when the matrix is also normal: that triangular representation cannot have any nonzero entries above its diagonal. Indeed, if \(\mathbf{Q}\) is unitary and \(\mathbf{S}=\mathbf{Q}^{*}\mathbf{A}\mathbf{Q}\), then \(\mathbf{S}^{*}\mathbf{S}=\mathbf{Q}^{*}\mathbf{A}^{*}\mathbf{A}\mathbf{Q}\) and \(\mathbf{S}\mathbf{S}^{*}=\mathbf{Q}^{*}\mathbf{A}\mathbf{A}^{*}\mathbf{Q}\). So when \(\mathbf{A}\) is normal, \(\mathbf{S}\) is normal too. If \(\mathbf{S}\) is upper triangular, the theorem makes it diagonal. This is a key step toward understanding the structure of normal operators.
The practical reasoning is short once its justification is clear: compare the first row and first column norms to eliminate the first row’s off-diagonal entries; then apply the same argument to the remaining block. Comparing only one row and column establishes the first step, not the whole theorem. The induction is what ensures that every off-diagonal entry is eliminated.
Check Your Understanding
Use the conjugate transpose when working over \(\mathbb{C}\), and justify each conclusion from the normality identity.
- For \(\mathbf{T}=\begin{bmatrix}a&b\\0&d\end{bmatrix}\), which pair of entries forces \(b=0\) when \(\mathbf{T}\) is normal?
- For an upper triangular normal three-by-three matrix, explain why the first row’s second and third entries must vanish.
- State the diagonal-entry equality that holds in position \((j,j)\) for any normal matrix.
- Why does the proof need an induction step after the first row’s off-diagonal entries have vanished?
- Compute \(\mathbf{C}^{*}\mathbf{C}\) for \(\mathbf{C}=\begin{bmatrix}1&i\\i&0\end{bmatrix}\) and explain why \(\mathbf{C}\) is not normal.
