CODEWITHSUNDEEP
mathwolfram-mathematica
CHM
CHM

Reputation: 337

Mathematica matrix diagonalization

I'm considering a matrix A such that A=PDP^-1.

The way I solve this using Mathematica is:

a={{0, -1}, {-1, 0}}
d = DiagonalMatrix[Eigenvalues[a]]
{{-1,0}, {0,1}}
p = Transpose[Eigenvectors[a]]

p.d.Inverse[p]
{{0, -1}, {-1, 0}}

Which is correct.

Problem is, the P matrix is not what I expected it to be. The matrix that Mathematica generates is

p={{1, -1}, {1, 1}}

But I am looking for

p2={{1/Sqrt[2], 1/Sqrt[2]}, {1/Sqrt[2], -(1/Sqrt[2])}}
p2.d.Inverse[p2]
{{0,-1}, {-1,0}}

Which also solves the equation. Is there a way for me to force Mathematica to show me different answers when executing Transpose[Eigenvectors[a]]?

Upvotes: 6

Views: 38060

Answers (4)

Mark McClure
Mark McClure

Reputation: 4964

Perhaps you should consider a built in matrix factorization tool? Of course, there are many matrix factorizations, many of which are built in to Mathematica. The one that likely corresponds most closely to your situation is JordanDecomposition. Here it is in action

SeedRandom[1];
a = RandomReal[{-2, 2}, {4, 4}];
{s, j} = JordanDecomposition[a];
Column[MatrixForm /@ {s, j}]

enter image description here

Chop[s.j.Inverse[s] - a]

enter image description here

Of course, many matrices are not diagonalizable but, if it is, then the Jordan decomposition will give you what you want. The vectors are normalized in the same way that the Eigen functions normalize them, but I fail to see why this is a problem.

Upvotes: 2

C. E.
C. E.

Reputation: 10627

What you need to do is normalize the answer you get. There is a function called Normalize, which can be used like this:

Normalize /@ {{1, -1}, {1, 1}}

Mathematica graphics

Upvotes: 11

David Z
David Z

Reputation: 131710

Eigenvectors can be freely rescaled by a constant, which means there are an infinite number of possible eigenvectors. Naturally, Mathematica cannot and will not show you all of them. So you'll need to normalize the eigenvectors in some way.

One option is to convert your matrix to numeric form using N. Mathematica returns normalized eigenvectors for numeric matrices.

p2 = Transpose[Eigenvectors[N[a]]]

This is risky, though, because computing the inverse of a numeric matrix can often fail spectacularly due to various numerical errors.

The other, better option is to manually normalize the eigenvectors using Normalize. You'll have to apply it to each eigenvector in the list returned by the Eigenvectors function:

p2 = Transpose[Normalize/@Eigenvectors[a]]

Upvotes: 4

acl
acl

Reputation: 6520

You could normalize your eigenvectors:

a = {{0, -1}, {-1, 0}};
d = DiagonalMatrix[Eigenvalues[a]];
p = Transpose[Normalize /@ Eigenvectors[a]];

so p is what you want: 

{{1/Sqrt[2], -(1/Sqrt[2])}, {1/Sqrt[2], 1/Sqrt[2]}}

Upvotes: 5

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