# Which eigenvalue has multiplicity greater than 1 ?

1. Apr 23, 2012

### sid9221

http://dl.dropbox.com/u/33103477/question.png [Broken]

I have determined the eigenvalues which are -2, 2 and 1 respectively.

I'm pretty sure that the one with multiplicity of 2 is the, the eigenvalue = 2 as it occur's twice in the diagnol. But I don't think thats a concrete enough reason.

Any idea's how to proceed ?

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2. Apr 23, 2012

### Staff: Mentor

I don't think it's even a correct reason, let alone being concrete enough.

Since you now have the eigenvalues, row reduce the matrix A - λI, substituting for λ each of your eigenvalues. For one value, the row-reduced matrix will have a solution space that is two-dimensional. For the other two values, the solution space will be one-dimensional.

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3. Apr 23, 2012

### sid9221

Thanks, worked it out and it makes sense. Turns out 2 is the one with multiplicity equal to 2.

The other part of the question ask's for a 4x4 invertible matrix P and an upper triangular matrix T such that T=P^-1 A P.

Now the problem with that is that any combination of the eigenvectors to form P would involve 2identical columns(as the multiplicity is 2), hence the matrix would not be invertible. So am I missing something ?

4. Apr 23, 2012

### Staff: Mentor

Check your work. In my work, λ = 2 has multiplicity 1, and the associated eigenvector is <1, 0, -1, 1>T.

5. Apr 23, 2012

### sid9221

Are you sure ? Too recheck I plugged the values in to an applet to find eigen values and vectors and here is the result...

Characteristic polynomial:
x^4 - 3x^3 - 2x^2 + 12x - 8

Real eigenvalues:
{-2, 1, 2, 2}

Eigenvector of eigenvalue -2:
(1, 1, 0, 0)
Eigenvector of eigenvalue 1:
(0, 0, -1, 1)
Eigenvector of eigenvalue 2:
(1, 0, -1, 1)
Eigenvector of eigenvalue 2:
(1, 0, -1, 1)

So thats multiplicity 2 for the eigenvalue 2 ?

Also any matrix formed with these would be not invertible(2 identical columns) so how do I work around that when creating an invertible matrix ?

6. Apr 23, 2012

### Staff: Mentor

There are two kinds of multiplicities: algebraic and geometric. The algebraic multiplicity of the eigenvalue 2 is two, because the λ - 2 factor in the characteristic equation occurs to power two. The geometric multiplicity of this eigenvalue is one, because there is only a single eigenvector for λ = 2.

The upshot is that I don't think you can diagonalize matrix A because there aren't enough eigenvectors.

7. Apr 23, 2012

### HallsofIvy

Staff Emeritus
Right. If one is an eigenvalue with both algeraic and geometric multiplicity 1 and 2 is an eigenvalue with algebraic multiplicity 2 but its geometric multiplicity is only 1, then it is similar to the "Jordan Normal Form"
$$\begin{bmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
but cannot be diagonalized.

8. Apr 23, 2012

### Ray Vickson

In general, the method you try (viz., knowing the eigenvectors and then determining the eigenvalues) will only tell you about multiplicity in case some of your eigenvector equations have more than one linearly-independent solution. For some matrices, that happens, but for others it does not. When the dimensionality of the eigenvector space is less than the n (for an nxn matrix) the only sure way to proceed is to determine the characteristic polynomial of the matrix, to determine the multiplicity of its roots.

RGV

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