Algebraic vs geometric multiplicity

For a square matrix $A$ and an eigenvalue $r$, two notions of “how many” the eigenvalue is:

• Algebraic multiplicity $m_a(r)$: the number of times $r$ appears as a root of the characteristic polynomial $\det (A-\lambda I)=0$.
• Geometric multiplicity $m_g(r)$: the dimension of the eigenspace, i.e., the number of linearly independent eigenvectors corresponding to $r$.

In general, $m_a(r)\ge m_g(r)$. They can be equal (the matrix is non-defective for $r$) or unequal ($m_a>m_g$, meaning $A$ is defective at $r$).

Why they can differ

The characteristic polynomial is a polynomial of degree $n$ for an $n\times n$ matrix. Its roots are eigenvalues; their multiplicities (algebraic) are how many times each root appears.

The eigenspace for a particular eigenvalue is the kernel of $A-rI$ — a subspace of ${\mathbb{R}}^n$. Its dimension is the geometric multiplicity.

These count different things. Algebraic multiplicity counts repetitions in the polynomial. Geometric multiplicity counts how many independent directions $A$ “stretches” by the same factor $r$.

Consider $A=\left[\begin{array}{cc}2& 1\\ 0& 2\end{array}\right]$. Characteristic polynomial: $(\lambda -2{)}^2$. So $\lambda =2$ has $m_a=2$.

But the eigenspace: $(A-2I)\mathbf{v}=0$ becomes $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\mathbf{v}=0$, giving $v_2=0$. So eigenspace is span of $(1,0)$, dimension 1: $m_g=1$.

So $m_a=2$ but $m_g=1$. The matrix is defective.

Diagonal vs defective matrices

A matrix is diagonalizable if and only if for every eigenvalue, $m_a=m_g$. In that case, you can find a basis of eigenvectors and the matrix decomposes as $A=PD{P}^{-1}$ with $D$ diagonal.

If any eigenvalue has $m_a>m_g$, the matrix is defective — you can’t fully diagonalize it. The closest you can get is the Jordan canonical form $A=PJ{P}^{-1}$ with $J$ block-diagonal of Jordan blocks. Each Jordan block of size $k$ corresponds to an eigenvalue with geometric multiplicity 1 contributing algebraic multiplicity $k$.

For an $n\times n$ matrix with $n$ distinct eigenvalues, $m_a=m_g=1$ for each — automatically diagonalizable.

For repeated eigenvalues, you might or might not be diagonalizable, depending on the eigenspace structure.

Implications for ODE systems

For ${\mathbf{x}}^{\prime }=A\mathbf{x}$, we want $n$ linearly independent solutions to span the solution space.

If $A$ is diagonalizable (every $m_a=m_g$), each eigenvalue $r$ with $m_a=k$ contributes $k$ linearly independent solutions of the form ${\mathbf{u}}_i{e}^{rt}$ (one per linearly independent eigenvector). Total: $n$ solutions.

If $A$ is defective at some eigenvalue $r$ ($m_a>m_g$), you don’t have enough eigenvectors. Need generalized eigenvectors to fill out the solution space. See Repeated eigenvalues case for the construction.

Inequality holds always

For any eigenvalue $r$ of an $n\times n$ matrix:

$1\le m_g(r)\le m_a(r)\le n$

Lower bound 1: every eigenvalue has at least one eigenvector.

Upper bound $n$: total of all $m_a$ equals $n$.

The middle inequality $m_g\le m_a$ comes from a counting argument: $A$ restricted to the generalized eigenspace acts as $rI+N$ where $N$ is nilpotent — so the dimension of the generalized eigenspace is $m_a$, and the eigenspace (kernel of $A-rI$, hence of $N$) is a subspace.

Computing them

For a given matrix:

1. Compute $\det (A-\lambda I)=0$ to find eigenvalues. Multiplicities of roots are algebraic multiplicities.
2. For each eigenvalue $r$, solve $(A-rI)\mathbf{v}=0$ — find the kernel. Dimension = geometric multiplicity.
3. Compare: if $m_g=m_a$, that eigenvalue’s eigenspace is “full.” If $m_g<m_a$, the matrix is defective at $r$.

Worked example

$A=\left[\begin{array}{cc}1& -1\\ 1& 3\end{array}\right]$.

Algebraic: $\det \left[\begin{array}{cc}1-\lambda & -1\\ 1& 3-\lambda \end{array}\right]=(1-\lambda )(3-\lambda )+1={\lambda }^2-4\lambda +4=(\lambda -2{)}^2$. So $\lambda =2$, $m_a=2$.

Geometric: $(A-2I)\mathbf{v}=0$ gives $\left[\begin{array}{cc}-1& -1\\ 1& 1\end{array}\right]\mathbf{v}=0$. The kernel: $v_1+v_2=0$, so $\mathbf{v}=(1,-1{)}^T$ (one independent eigenvector). $m_g=1$.

Since $m_a>m_g$, the matrix is defective. To solve ${\mathbf{x}}^{\prime }=A\mathbf{x}$, you need a generalized eigenvector — see Repeated eigenvalues case.

For more on the linear-algebra background, this concept connects to Jordan canonical form (not yet in this vault).