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I don't understand how to find the multiplicity for an eigenvalue. To be honest, I am not sure what the books means by multiplicity.

For instance, finding the multiplicty of each eigenvalue for the given matrix: $$\begin{bmatrix}1 & 4\\2 & 3\end{bmatrix}$$

I found the eigenvalues of this matrix are -1 and 5, but what are the multiplicities of these? Thanks in advance

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Do you mean geometric multiplicity or do you mean algebraic multiplicity? Thanks. –  awllower Mar 8 '13 at 6:13
What is the difference b/w the 2? I know the answer for both of their multiplicities is 1 if that helps - i just don't know why –  Johnathon Svenkat Mar 8 '13 at 6:15
Algebraic multiplicity is the number of times an eigenvalue appears in a characteristic polynomial of a matrix. The geometric one is the nullity of $A-kI$ where $k$ is an eigenvalue of $A$. When the two coincide, and only when so, the matrix is diagonalisable. –  awllower Mar 8 '13 at 6:17
What is your matrix? It seems to have 8 values, I would expect a square number of values... –  copper.hat Mar 8 '13 at 6:20
There is no such thing as the multiplicity of an eigenvector. (And for eigenvalues more then one notion of multiplicity exist, one that is often forgotten is the multiplicity as root of the minimal polynomial.) I'll edit the title to avoid the nonsensical term. –  Marc van Leeuwen Mar 8 '13 at 7:31

2 Answers 2

up vote 20 down vote accepted

The characteristic polynomial of the matrix is $p_A(x) = \det (xI-A)$. In your case, $A = \begin{bmatrix} 1 & 4 \\ 2 & 3\end{bmatrix}$, so $p_A(x) = (x+1)(x-5)$. Hence it has two distinct eigenvalues and each occurs only once, so the algebraic multiplicity of both is one.

If $B=\begin{bmatrix} 5 & 0 \\ 0 & 5\end{bmatrix}$, then $p_B(x) = (x-5)^2$, hence the eigenvalue $5$ has algebraic multiplicity $2$. Since $\dim \ker (5I-B) = 2$, the geometric multiplicity is also $2$.

If $C=\begin{bmatrix} 5 &1 \\ 0 & 5\end{bmatrix}$, then $p_C(x) = (x-5)^2$ (same as $p_C$), hence the eigenvalue $5$ has algebraic multiplicity $2$. However, $\dim \ker (5I-C) = 1$, the geometric multiplicity is $1$.

Very loosely speaking, the matrix is 'deficient' in some sense when the two multiplicities do not match.

The algebraic multiplicity of an eigenvalue $\lambda$ is the power $m$ of the term $(x-\lambda)^m$ in the characteristic polynomial.

The geometric multiplicity is the number of linearly independent eigenvectors you can find for an eigenvalue.

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Wow that sounds a lot simpler when you put it like that. What about the geometric multiplicity? Can you find it with the information given? –  Johnathon Svenkat Mar 8 '13 at 6:28
I have noticed you answering many of my linear algebra questions lately Copper.hat, and I just wanted to let you know how thankful I have been for assistance. Definitely cleared a lot of things up for me. (tried to pm this but could not find any option to) –  Johnathon Svenkat Mar 8 '13 at 6:38
Glad to help! ${}{}{}$ –  copper.hat Mar 8 '13 at 6:39
@copper.hat why do dim ker$(5I−C)=1$? –  gbox Aug 27 at 12:22
@gbox: Look at solutions of $(5I-C)x =0$. –  copper.hat Aug 27 at 15:40

Let me explain the two multiplicities that I know are related to eigen-values of matrices:
Firstly, what is the eigenvalue of a matrix $A$? By definition it consists of the zeros of the polynomial: $\det(A-xI)$. So the muliplicities that they occur in this polynomial are defined to be the multiplicities of the eigen-values.
Secondly, since, for an eigen-value $\lambda$, we have $\det(A-\lambda I)=0$, i.e. $A-\lambda I$ is a singular matrix, and the linear transformation it defines has a non-trivial kernel. The dimension of this kernel is then said to be the geometric multiplicity of the eigen-value.
Hence, in one case, one has to compute some polynomial; while, on the other hand, one has to compute some transformations, to find its kernel, and to determine the dimension of the kernel, to find the multiplicites of eigen-values. Notice here that you have a $2\times 2$ matrix with $2$ eigen-values, and hence they must be of multiplicities $1$. That is to say, the two notions coincide, and the matrix in question must be diagonalisable.

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Thanks for your help! –  Johnathon Svenkat Mar 8 '13 at 6:39
It is my pleasure. –  awllower Mar 8 '13 at 6:40

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