Symmetric matrices exhibit equality between geometric and algebraic multiplicities

For the matrix $B=begin{bmatrix} 5 & 0 \ 0 & 5end{bmatrix}$, the eigenvalue $5$ has algebraic multiplicity $2$, which matches its geometric multiplicity of $2$ since $dim ker (5I-B) = 2$. On the other hand, for the matrix $C=begin{bmatrix} 5 &1 \ 0 & 5end{bmatrix}$, the eigenvalue $5$ also has algebraic multiplicity $2$ with $p_C(x) = (x-5)^2$, but its geometric multiplicity is only $1$ since $dim ker (5I-C) = 1$. It can be said that a matrix is ‘deficient’ when its algebraic and geometric multiplicities do not match.
If there is a linear transformation represented by a matrix A and an eigenvalue $lambda$, a vector $v$ is considered a generalized eigenvector if there exists an integer n such that $(A-lambda I)^nv =0$. The set of all such generalized eigenvectors is called the generalized eigenspace, and its dimension is equal to the algebraic multiplicity of $lambda$. A more detailed explanation on the intuition behind generalized eigenvectors can be found here.


Solution 1:

The matrix’s characteristic polynomial can be expressed as $p_A(x) = det (xI-A)$. For instance, if we take $A$ to be $begin{bmatrix} 1 & 4 \ 2 & 3end{bmatrix}$, then $p_A(x) = (x+1)(x-5)$, indicating the existence of two unique eigenvalues, each with an algebraic multiplicity of one.

Let $B=begin{bmatrix} 5 & 0 \ 0 & 5end{bmatrix}$. This implies that the polynomial $p_B(x) = (x-5)^2$, indicating that the eigenvalue $5$ has an algebraic multiplicity of $2$. Additionally, since the dimension of the kernel of $(5I-B)$ is $2$, the geometric multiplicity is also $2$.

Let $C$ be a matrix given by $C=begin{bmatrix} 5 &1 \ 0 & 5end{bmatrix}$ and $p_C$ be its characteristic polynomial. It follows that $p_C(x) = (x-5)^2$ and hence the eigenvalue $5$ has an algebraic multiplicity of $2$. However, the geometric multiplicity of the eigenvalue $5$ is $1$ since $dim ker (5I-C) = 1$.

In a broad sense, when the multiplicities are not equivalent, the matrix could be considered to be lacking in some way.

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

The number of eigenvectors that are linearly independent and can be obtained for an eigenvalue is referred to as the geometric multiplicity.


Solution 2:


To explain the multiplicities related to eigen-values of matrices, there are two approaches. The first is to define the eigen-value of a matrix A as the zeros of the polynomial $det(A-xI)$, and the multiplicities of the eigen-values are determined by the number of occurrences of these zeros in the polynomial. The second approach is to compute the kernel of the linear transformation defined by a singular matrix $A-lambda I$, where $lambda$ is an eigen-value of A, and the geometric multiplicity of the eigen-value is the dimension of this kernel.
Therefore, to find the multiplicities of eigen-values, one can either compute a polynomial or a transformation and determine the dimension of its kernel. For a $2 times 2$ matrix with two eigen-values, the multiplicities must be $1$, indicating that the two approaches coincide and the matrix is diagonalisable.

Frequently Asked Questions