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The real Schur decomposition theorem states that for any matrix $A\in\mathbb R^{n\times n}$, there exists an orthogonal matrix $Q$ and a "quasitriangular" matrix $T$ such that $A=QTQ^T$. Here, "quasitriangular" means that $T$ has the form

$$T=\begin{pmatrix}B_1&&&*\\&B_2&&\\&&\ddots&\\0&&&B_n\end{pmatrix},$$ where all $B_i$ are either $1\times1$ or $2\times2$ matrices. These matrices form the "quasidiagonal".

I want to prove that in the specific case of real Schur decomposition of an orthogonal matrix, $T$ must be a "quasidiagonal matrix", i.e. all entries above the quasidiagonal are zero. This is claimed without proof in this answer.

It's easy to see that $T$ must be orthogonal. In the special case where $T$ is upper triangular, it's simple to prove that the norm of the $i$-th column equals that of the $i$-th row, which then yields the desired result by a simple induction. However, my attempt at adapting this proof to the quasidiagonal case failed.

This result seems to me like it should be well-known. A reference would also suffice.

Rodrigo de Azevedo
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ViHdzP
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3 Answers3

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Suppose $A, T, Q \in \mathbb{R}^{n \times n}$, $A=Q T Q^T$, $A$ and $Q$ are orthogonal. Then $T$ is orthogonal. We want to prove that if $T$ is quasitriangular, then it's quasidiagonal. This follows from the following theorem.

Theorem

Suppose $$T=\begin{bmatrix} B_{1,1} & B_{1,2} & \dots & B_{1,m} \\ & B_{2,2} & \dots & \vdots \\ & & \ddots & \vdots \\ & & & B_{m,m} \end{bmatrix}$$ is a real orthogonal or complex unitary block matrix, where each $B_{i,i}$ is a $1 \times 1$ or a $2 \times 2$ matrix. Then for each $i$ for each $j>i$ we have $B_{i,j}=0$.

Proof of the theorem

Since T is orthogonal or unitary, we have $T^* T = I$, which can be visualized as $$\begin{bmatrix} B_{1,1}^* & & & \\ B_{1,2}^* & B_{2,2}^* & & \\ \vdots & \vdots & \ddots & \\ B_{1,m}^* & \dots & \dots & B_{m,m}^* \end{bmatrix} \cdot \begin{bmatrix} B_{1,1} & B_{1,2} & \dots & B_{1,m} \\ & B_{2,2} & \dots & \vdots \\ & & \ddots & \vdots \\ & & & B_{m,m} \end{bmatrix} = \begin{bmatrix} I&&& \\ &I&& \\ &&\ddots& \\ &&&I \end{bmatrix}. $$

Now we will prove by induction on $i$ that for each $i$, $B_{i,i}^*$ is invertible and for each $j > i$ we have $B_{i,j}=0$.

Base of induction: it can be seen from the structure of matrices in the equation above that $B_{1,1}^* B_{1,1} = I$ and thus $B_{1,1}^*$ is invertible, and that for each $j>1$, $B_{1,1}^* B_{1,j}=0$, which by invertibility of $B_{1,1}^*$ gives us $B_{1,j}=0$.

Inductive step. Suppose for each $i$ up to and including $k$, for each $j > i$ we have $B_{i,j}=0$. So, we have $$\begin{bmatrix} B_{1,1}^*&&&&&& \\ & \ddots &&&&& \\ && B_{k,k}^* &&&& \\ &&&B_{k+1,k+1}^* & & & \\ &&&B_{k+1,k+2}^* & B_{k+2,2}^* & & \\ &&&\vdots & \vdots & \ddots & \\ &&&B_{k+1,m}^* & \dots & \dots & B_{m,m}^* \end{bmatrix} \cdot \begin{bmatrix} B_{1,1}&&&&&& \\ & \ddots &&&&& \\ && B_{k,k} &&&& \\ &&&B_{k+1,k+1} & B_{k+1,k+2} & \dots & B_{k+1,m} \\ &&&& B_{k+2,k+2} & \dots & \vdots \\ &&&& & \ddots & \vdots \\ &&&& & & B_{m,m} \end{bmatrix} = \begin{bmatrix} I&&&&&& \\ &\ddots&&&&& \\ &&I&&&& \\ &&&I&&& \\ &&&&I&& \\ &&&&&\ddots& \\ &&&&&&I \end{bmatrix}. $$ From the structure of matrices in this equation we can see that $B_{k+1,k+1}^* B_{k+1,k+1} = I$, which means that $B_{k+1,k+1}^*$ is invertible, and that for each $j > k+1$ we have $B_{k+1,k+1}^* B_{k+1,j} = 0$, which by invertibility of $B_{k+1,k+1}^*$ gives us $B_{k+1,j}=0$. QED

ViHdzP
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CrabMan
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Have you tried to use the fact that $B_i$ are also orthogonal? For proving this, using $T^tT=TT^t=I$ on the diagonal (using block matrix).

Then you can easily prove that $*$ can only be zero, using $T^tT=TT^t=I$ besides diagonal step by step, from top row to bottom row.

DreamAR
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It will depend on background knowledge, but the simplest algebraic proof comes from knowing that

(i) Invertible Block upper triangular matrices (with specified dimensions for each diagonal block) form a group.
(ii) Inverses of orthogonal matrices are given by their transpose.

$T=Q^TAQ$
and $Q,A \in O_n\big(\mathbb R\big)\implies T \in O_n\big(\mathbb R\big)$
applying (i) we know $T^{-1}$ is block upper triangular and applying (ii) we know that $T^{-1}=T^T$ which must be block lower triangular since the transpose of a block upper triangular matrix is block lower triangular.

Thus $T^T$ is both block upper triangular and block lower triangular, i.e. $T^T$ is block diagonal, and so is $T$.

This sort of argument runs identically when you encounter an uglier group, like $Q,A \in O_n\big(\mathbb C\big)$
(the complex orthogonal group which is not the unitary group).

user8675309
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  • to clarify on (i), i.e. ref e.g. ex 2.4.12 of Artin's Algebra 1st edition where you show $\begin{pmatrix}A &* \ \mathbf 0&D\end{pmatrix}$ for $A \in GL_r(\mathbb R)$ and $D\in \mathbb GL_{n-r}(\mathbb R)$ forms a subgroup of $GL_n(\mathbb R)$-- once this is known, (i) follows almost immediately by strong induction. – user8675309 Dec 01 '21 at 19:46