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Proof of Theorem 6].

We are given n linearly independent eigenvectors p_{1}, p_{2}, ... , p_{n} of n-square matrix A and the
corresponding eigenvalues λ_{1}, λ_{2}, ... , λ_{n}. An eigenvector is a vector X that a matrix A carries
into a multiple of itself according to the equation AX = λX where λ is the corresponding
eigenvalue. Thus the eigenvectors p_{1}, p_{2}, ... , p_{n} and eigenvalues λ_{1}, λ_{2}, ... , λ_{n} must satisfy the
following set of equations:

Ap_{1} = λ_{1}p_{1}

1) Ap_{2} = λ_{2}p_{2}

............

Ap_{n} = λ_{n}p_{n} .

Now system 1) is equivalent to

2) A[p_{1} p_{2} ... p_{n}] = [λ_{1}p_{1} λ_{2}p_{2} ... λ_{n}p_{n}] .

Why? Because of the following theorem:

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Theorem 1. Suppose that the matrix A carries the vector X_{1} into the vector Y_{1}, the vector X_{2 }
into the vector Y_{2} , etc. i.e.

AX_{1} = Y_{1}

AX_{2} = Y_{2}

......

......

AX_{n} = Y_{n}

Then

A [ X_{1} X_{2} .... X_{n}] = [Y_{1} Y_{2} ... Y_{n}]

where [ X_{1} X_{2} .... X_{n}] is a matrix whose columns are X_{1}, X_{2}, .... , X_{n} and [Y_{1} Y_{2} ... Y_{n}] is a
matrix whose columns are Y_{1},Y_{2}, ... ,Y_{n} .

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Furthermore,

Why? Because of the following theorem:

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Theorem 2. The effect of post-multiplying a matrix

by a diagonal matrix

is that of multiplying the i-th column of matrix A by the factor i.e. the successive columns of the original matrix are simply multiplied by successive diagonal elements of the diagonal matrix. Explicitly:

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From equations 2) and 3) we get

AP = PD

or, equivalently,

4) A = PDP^{-1}

where we have substituted P for [p_{1} p_{2} ... p_{n}] and D for

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