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

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