Question

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

Answer 1

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 \\e \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 = (1)/(\lambda_i) A \vec{v} _i`

Lets transpose it

`(\vec{v}_i)^t = ((1)/(\lambda_i) A \vec{v} _i)^t`

`(\vec{v}_i)^t =   (\vec{v} _i)^t A^t (((1)/(\lambda_i))^t`

as
`\lambda_i` is an scalar

`(\vec{v}_i)^t =   (\vec{v} _i)^t A^t ((1)/(\lambda_i))`

Now, as A is symmetric:

`A^t = A`

so

`(\vec{v}_i)^t =   (\vec{v} _i)^t A ((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 ((1)/(\lambda_i)) \vec{v}_j`

`\vec{v}_i \cdot   \vec{v}_j =  ((1)/(\lambda_i))  (\vec{v} _i)^t A \vec{v}_j`

`\vec{v}_i \cdot   \vec{v}_j =  ((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 =  ((1)/(\lambda_i))  (\vec{v} _i)^t \lambda_j \vec{v} _j`

`\vec{v}_i \cdot   \vec{v}_j =  ((\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  =  ((\lambda_j )/(\lambda_i))  (\vec{v} _i)^t \vec{v}_j - (\vec{v} _i)^t \vec{v}_j = 0`

`( (\lambda_j )/(\lambda_i) - 1 ) (\vec{v} _i)^t \vec{v}_j= 0`

As the eigenvalues are distinct,
`(\lambda_j )/(\lambda_i)` can't be 1, so

`( (\lambda_j )/(\lambda_i) - 1 ) \\e 0`

this implies

`(\vec{v} _i)^t \vec{v}_j= 0`

so the eigenvectors are orthogonal.