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# Consider the eigenvalue decomposition of a symmetric matrix A. Prove that two eigenvectors Vị and V; associated with two distinct eigenvalues li and l; of A are mutually orthogonal; that is, v. Vj = 0

Accepted Solution

A:
Lets consider the symmetric matrix $$A$$, and the two eigenvectors $$\vec{v}_i$$ and $$\vec{v}_j$$ such as:$$A \vec{v} _i = \lambda_i \vec{v} _i$$$$A \vec{v} _j = \lambda_j \vec{v} _j$$with$$\lambda_i \ne \lambda_j$$.The dot product between $$\vec{v}_i$$ and $$\vec{v}_j$$ can be obtained with:$$\vec{v}_i \cdot \vec{v}_j = (\vec{v}_i )^t \vec{v}_j$$Using the first eigenvector equation we can find:$$\vec{v}_i = \frac{1}{\lambda_i} A \vec{v} _i$$Lets transpose it$$(\vec{v}_i)^t = (\frac{1}{\lambda_i} A \vec{v} _i)^t$$$$(\vec{v}_i)^t = (\vec{v} _i)^t A^t ((\frac{1}{\lambda_i})^t$$as $$\lambda_i$$ is an scalar$$(\vec{v}_i)^t = (\vec{v} _i)^t A^t (\frac{1}{\lambda_i})$$Now, as A is symmetric:$$A^t = A$$so$$(\vec{v}_i)^t = (\vec{v} _i)^t A (\frac{1}{\lambda_i})$$Lets take the dot product again:$$\vec{v}_i \cdot \vec{v}_j = (\vec{v}_i )^t \vec{v}_j$$but this is :$$\vec{v}_i \cdot \vec{v}_j = (\vec{v} _i)^t A (\frac{1}{\lambda_i}) \vec{v}_j$$$$\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t A \vec{v}_j$$$$\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t ( A \vec{v}_j )$$But, the parenthesis is equal to $$A \vec{v} _j = \lambda_j \vec{v} _j$$so$$\vec{v}_i \cdot \vec{v}_j = (\frac{1}{\lambda_i}) (\vec{v} _i)^t \lambda_j \vec{v} _j$$$$\vec{v}_i \cdot \vec{v}_j = (\frac{\lambda_j }{\lambda_i}) (\vec{v} _i)^t \vec{v}_j$$Now, subtracting the dot product$$\vec{v}_i \cdot \vec{v}_j - \vec{v}_i \cdot \vec{v}_j = (\frac{\lambda_j }{\lambda_i}) (\vec{v} _i)^t \vec{v}_j - (\vec{v} _i)^t \vec{v}_j = 0$$$$( \frac{\lambda_j }{\lambda_i} - 1 ) (\vec{v} _i)^t \vec{v}_j= 0$$As the eigenvalues are distinct,  $$\frac{\lambda_j }{\lambda_i}$$ can't be 1, so $$( \frac{\lambda_j }{\lambda_i} - 1 ) \ne 0$$this implies $$(\vec{v} _i)^t \vec{v}_j= 0$$so the eigenvectors are orthogonal.