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# Linear Maps

Module by: Thanos Antoulas, JP Slavinsky. E-mail the authors

Summary: (Blank Abstract)

A diagonal matrix is one whose elements not on the diagonal are equal to 0. The following matrix is one example. ( a000 0b00 00c0 000d ) a000 0b00 00c0 000d

A matrix AA is diagonalizable if there exists a matrix VRn×n V × n n , detV0 V 0 such that VAV-1=Λ V A V Λ is diagonal. In such a case, the diagonal entries of ΛΛ are the eigenvalues of AA.

Let's take an eigenvalue decomposition example to work backwards to this result.

Assume that the matrix AA has eigenvectors vv and ww and the respective eigenvalues λvλv and λwλw: Av=λvv A v λv v Aw=λww A w λw w

We can combine these two equations into an equation of matrices: A( vw )=( vw )( λv0 0λv ) A vw vw λv0 0λv

To simplify this equation, we can replace the eigenvector matrix with VV and the eigenvalue matrix with ΛΛ. AV=VΛ A V V Λ

Now, by multiplying both sides of the equation by V-1 V , we see the diagonalizability equation discussed above.

## the diagonalizability equation

A=VΛV-1 A V Λ V
(1)

When is such a diagonalization possible? The condition is that the algebraic multiplicity equal the geometric multiplicity for each eigenvalue, αi=γi αi γi . This makes sense; basically, we are saying that there are as many eigenvectors as there are eigenvalues. If it were not like this, then the V matrices would not be square, and therefore could not be inverted as is required by the diagonalizability equation. Remember that the eigenspace associated with a certain eigenvalue λ is given by kerAλI ker A λ I .

This concept of diagonalizability will come in handy in different linear algebra manipulations later. We can however, see a time-saving application of it now. If the matrix AA is diagonalizable, and we know its eigenvalues λiλi, then we can immediately find the eigenvalues of A2 A 2 : A2=(VΛV-1)(VΛV-1)=VΛ2V-1 A 2 V Λ V V Λ V V Λ 2 V

The eigenvalues of A2 A 2 are simply the eigenvalues of AA, squared.

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