A linear operator L on the vector space V can admit a set of vectors which, when transformed by L, result multiplied by a scalar.

Vectors of this set are said
*eigenvectors* l_{i} and the
corresponding scalars are said *eigenvalues* λ_{i}
of the linear operator L.

To compute eigenvectors and eigenvalues of L one need to solve the following equation said eigenvalue equation

From this equation we have

The last equation allows the calculation of the eigenvalues λ_{i}.
It is easy if the dimension *n* of the vector space is 2, but difficult if *n*
is greater, because it is equivalent to a polynomial equation of degree *n* and,
in general, it can have complex roots.

However if the matrix of L is diagonal, that is if all its entries are null except those with equal indices, the equation is equivalent to with a trivial set of solutions .

In this case, the eigenvalue equation is equivalent to the set of equations

which admit solution only if the components of the eigenvector |l_{i}> are null
except that with index *i*.

Example.

Given the diagonal matrix , one has that is therefore .

The eigenvector |d_{1}>, corresponding to the eigenvalue δ_{1}=*a*,
is obtained from the equation
,
that is
,
true for all *x* if *y*=0. So the calculation **doesn't find a determined vector,
but the set of all the vectors laying on the x-axis**. In order to determine the result
we select from this set the vector with unitary norm and with the same direction as
the x-axis. This is the **normalized eigenvector**.
So in the example we have
.

The analogous calculation with respect to the eigenvalue |d_{2}>,
corresponding to the eigenvalue δ_{2}=2, we obtain
.

The results shown in the above example can be generalized to n-dimensional diagonal matrices. Therefore, in general:

- the eigenvalues of a real diagonal matrix coincide with the not null entries of the matrix;
- the normalized eigenvectors are the unit vectors on the cartesian axes, so they result mutually orthogonal;
- the set of eigenvectors is an orthonormal basis for the space V.

From the eigenvalue equation we can deduce that, given the matrix L
and another matrix M with its inverse M^{-1}, the matrix M^{-1}LM
has the same eigenvalues as L. In fact

In particular this is true if M is a rotation.

If the matrix which undergoes the rotation M is a diagonal real matrix like the D of the
example, the transformed matrix S=M^{-1}DM is a
symmetric real matrix,
that is such that, if i≠j, S_{i,j}=S_{j,i}. In fact we have

Vice versa, using the opposite procedure, every real symmetric matrix can be transformed in a diagonal real matrix.

In fact, given the matrix , in order to find the rotation which diagonalizes S, we let

The term by term subtraction of the second equation from the third one gives

From this equation and from the first one we have

Therefore the rotation with angle -*θ* transforms S into D.

This procedure can be applied to any symmetric real matrix, so we can say that all the eigenvalues of a real symmetric matrix are real values.

Example.

Given the real symmetric matrix the eigenvalue equation is from which we have

The eigenvalues are σ_{1}=4 e σ_{2}=8.
In order to calculate the corresponding normalized eigenvectors one need to solve the equations
, where
the norm of the vectors |s_{i}> is 1, so they can be expressed as
with
.

Using the first eigenvalue we have

therefore .

Using the second eigenvalue we have

therefore .

The eigenvectors are orthonormal.

There exists a rotation that transforms S into a diagonal matrix with the same eigenvalues . The same rotation transforms the normalized eigenvectors of S into the unit vectors of the cartesian axes. In the proposed example the rotation angle is π/6.

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