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Consider the matrix $A=\begin{bmatrix}
50 &70 \\
70 & 80
\end{bmatrix}$ whose eigenvectors corresponding to eigenvalues $\lambda _{1}$ and $\lambda _{2}$ are $x_{1}=\begin{bmatrix}
70 \\
\lambda_{1}-50
\end{bmatrix}$ and $x_{2}=\begin{bmatrix}
\lambda _{2}-80\\
70
\end{bmatrix}$, respectively. The value of $x^{T}_{1} x_{2}$ is _________.
in Linear Algebra 24.6k points
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1 vote
Answer should be $0.$

“Eigen-vectors of a real symmetric matrix are orthogonal(perpendicular).”

So, here matrix $A$ is symmetric because $A^T = A.$

So, $x_{1} \perp x_{2}$ (or) $x_{1}^{T} x_{2}=0$ (or) $\vec{x_{1}}.\vec{x_{_{2}}} = 0$ i.e. $(70\hat{i}+(\lambda_{1}-50 )\hat{j}).((\lambda_{2}-80 )\hat{i}+70\hat{j})=0$ i.e $70(\lambda _{2}-80) + 70(\lambda _{1}-50)=0$
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