Schur decomposition to show matrix has $n$ orthonormal eigenvectors

Question

From Gilbert Strang's "Introduction to Linear Algebra." We are trying to show by Schur decomposition that all symmetric matrices are diagonalizable. We write down the Schur decomposition as $A=QTQ^{-1}$ where $A$ is square, $T$ is upper triangular and $\bar Q^{T}=Q^{-1}$, $\bar Q$ is the complex conjugate of $Q$. The text looks for $AQ=QT$ and argues that the first column of $Q$ has to be an eigenvector. Since $T$ is triangular and not necessarily diagonal, the first proposed step is to use the first column of $Q$ and supplement it with $n-1$ columns to complete an orthonormal matrix $Q_{1}$. Then we write: $\bar Q^{T}_{1}AQ_{1}= \begin{bmatrix} \bar q^{T}_{1} \\ \vdots \\ \bar q^{T}_{n} \\ \end{bmatrix} \begin{bmatrix} Aq_{1} & \cdots & Aq_{n} \\ \end{bmatrix} = \begin{bmatrix} t_{11} & \cdots \\ 0 & A_{2} \\ \end{bmatrix} $

(1) Why do we get the right hand side? I am able to reach it by multiplication and from the properties of a complex conjugate. At the same time, shouldn't the right hand side simplify to a triangular matrix? What am I missing?

Going on, the book makes and argument by "induction" (not a formal one). It assumes a Schur factorization $A_2=Q_{2}T_{2}Q_{2}^{-1}$ is possible for $A_{2}$ of size $n-1$. Then it "puts" $Q_2$ and $T_2$ into $Q$ and $T$: $Q=Q_{1} \begin{bmatrix} 1 & 0 \\ 0 & Q_{2} \\ \end{bmatrix}$ and $T= \begin{bmatrix} t_{11} & \cdots \\ 0 & T_{2} \\ \end{bmatrix}$ and $AQ=QT$

(2) Where do we get this transformation from? In particular how can we see where $Q_{1}, T_{1}, Q_{}2, T_{2}$ fit into $Q$ and $T$?

Once this is cleared, I am able to show that for a symmetric matrix $T$ is diagonal and the matrix has the requisite number of eigenvectors.

Just to be clear, this is the first time the Schur decomposition is presented in the material so the answer might be obvious if you have mastered the decomposition but it definitely is not at this point in the book.

Answer 1

Regarding the first step:

All we know about the matrix $Q_1$ is that its first column $q_1$ is an eigenvector of $A$. We are trying to prove the existence of a matrix $Q$ such that $Q^*AQ$ is upper triangular, but we're not quite there yet.

Now, consider the product $$ Q_1^*AQ_1 = \pmatrix{q_1^*\\ \vdots \\ q_n^*}\pmatrix{Aq_1 & \cdots & Aq_n} = \\ \pmatrix{q_1^*Aq_1 & q_1^*Aq_2 & \cdots & q_1^*Aq_n\\ q_2^*Aq_1 & q_2^*Aq_2 & \cdots & q_2^*Aq_n\\ \vdots & \vdots & \ddots & \vdots\\ q_n^*Aq_1 & q_n^*Aq_2 & \cdots & q_n^*Aq_n} $$ Note, however, that for complex vectors $u$ and $v$, $u^*v$ is the inner product (Hermitian inner product) of the vectors $u$ and $v$. So, since $q_1$ is an eigenvector, and since $q_1$ is orthogonal to the other $q_i$, we may rewrite the above matrix as $$ Q_1^*AQ_1 = \pmatrix{q_1^*Aq_1 & q_1^*Aq_2 & \cdots & q_1^*Aq_n\\ 0 & q_2^*Aq_2 & \cdots & q_2^*Aq_n\\ \vdots & \vdots & \ddots & \vdots\\ 0 & q_n^*Aq_2 & \cdots & q_n^*Aq_n} $$ which is precisely the desired form, if we set $t_{11} = q_1^*Aq_1$.

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Regarding the second:

Using block-matrix multiplication, note that $$ \tilde Q A \tilde Q^{-1} = \pmatrix{I_{k}&0\\0&Q} \pmatrix{A_{11}&A_{12}\\A_{21}&A_{22}} \pmatrix{I_{k}&0\\0&Q}^{-1} = \pmatrix{A_{11}&A_{12}Q^{-1}\\QA_{21}&QA_{22}Q^{-1}} $$ Note that $I_k$ is the size-$k$ identity matrix, and that each of these block matrices are partitioned in the same way.

In a more abstract proof, we could build this matrix in a more immediately intuitive fashion. In particular, to build the linear transformation $\tilde Q$, first consider its restriction $Q$ to a certain invariant subspace of $A$.