Goal: $Q^TAQ=D$ for any symmetric matrix $A$

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• By induction method.
• Find $A\sim B$ and $B\sim D\longrightarrow A\sim D$

ASSUME $Q_1=[v_1:v_2:\ldots:v_n]$ an orthogonal matrix for which $Av_1=\lambda_1v_1$.

$$\begin{aligned} Q_1^TAQ_1=&\begin{bmatrix}v_1^T\\v_2^T\\\vdots\\v_n^T\end{bmatrix}A[v_1:v_2:\ldots:v_n]\\ =&\begin{bmatrix}v_1^T\\v_2^T\\\vdots\\v_n^T\end{bmatrix}[Av_1:Av_2:\ldots:Av_n]\\ =&\begin{bmatrix}v_1^T\\v_2^T\\\vdots\\v_n^T\end{bmatrix}[\lambda_1v_1:Av_2:\ldots:Av_n]\\ =&\left [\begin{array}{c:c}\lambda_1&*\\\hdashline0&A_1 \end{array}\right]=B \end{aligned}$$ $$B^T=Q_1^TA^TQ_1=Q_1^TAQ_1=B$$ $$\Rightarrow B\ is\ symmertric$$ $$\Rightarrow\ A_1\ is \ symmetric$$ $$\therefore\ B=\left[\begin{array}{c:c}\lambda_1&0\\\hdashline0&A_1 \end{array}\right]$$

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$$\because\ A\sim B\\\Rightarrow c_A(\lambda)=c_B(\lambda)$$ $$\begin{aligned} \therefore\ c_A(\lambda)= c_B(\lambda)=&\mathrm{det}(B-\lambda I)=\mathrm{det}(\left[\begin{array}{c:c}\lambda_1&0\\\hdashline0&A_1\end{array}\right]-\lambda I)\\ =&\mathrm{det}(\left[\begin{array}{c:c}\lambda_1-\lambda&0\\\hdashline0&A_1-\lambda I'\end{array}\right])\\ =&(\lambda_1-\lambda)\mathrm{det}(A_1-\lambda I')\\ =&(\lambda_1-\lambda)c_{A_1}(\lambda) \end{aligned}$$

$\therefore$ The characteristic polynomial of $A_1$ divides the characteristic polynomial of $A$. It follows that the eigenvalues of $A_1$ are also eigenvalues of $A$.

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$${A_1}\ is \ a \ k\times k \ real\ symmertric \ matrix \\\Rightarrow\ Let\ P_2^TA_1P_2=D_1$$ $${Q_2}=\left[\begin{array}{c:c}1&0\\\hdashline0&P_2\end{array}\right]\\Q=Q_1Q_2$$ $$\begin{aligned}Q^TAQ=&(Q_1Q_2)^TA(Q_1Q_2)=(Q_2^TQ_1^T)A(Q_1Q_2)=Q_2^TBQ_2\\=&\left[\begin{array}{c:c}1&0\\\hdashline0&P_2^T\end{array}\right]\left[\begin{array}{c:c}\lambda_1&0\\\hdashline0&A_1\end{array}\right]\left[\begin{array}{c:c}1&0\\\hdashline0&P_2\end{array}\right]\\=&\left[\begin{array}{c:c}\lambda_1&0\\\hdashline0&P_2^TA_1P_2\end{array}\right]\\=&\left[\begin{array}{c:c}\lambda_1&0\\\hdashline0&D_1\end{array}\right]\end{aligned}$$