Let \(\lambda\) be an eigenvalue of \(A\).
Recall that the eigenvectors of \(A\) for \(\lambda\)
are the nonzero vectors in the nullspace of
\(A-\lambda I\).
We call the nullspace \(A-\lambda I\) the **eigenspace** of
\(A\) for \(\lambda\) denoted by \({\cal E}_A(\lambda)\).
In other words, \({\cal E}_A(\lambda)\) consists of all the
eigenvectors of \(A\) for \(\lambda\) and the zero vector.

Let \(A = \begin{bmatrix} 1 & 2 \\ 1 & 0\end{bmatrix}\). Note that \(-1\) is an eigenvalue of \(A\). Then \(A - (-1)I_2= \begin{bmatrix} 2 & 2 \\ 1 & 1\end{bmatrix}.\) The nullspace of this matrix is spanned by the single vector \(\begin{bmatrix} -1 \\ 1 \end{bmatrix}\). Hence, \({\cal E}_A(-1)\) is the span of \(\begin{bmatrix} -1 \\ 1 \end{bmatrix}\).

The **geometric multiplicity** of an eigenvalue \(\lambda\) of \(A\)
is the dimension of \({\cal E}_A(\lambda)\). In the example above,
the geometric multiplicity of \(-1\) is \(1\) as the eigenspace is spanned
by one nonzero vector.

In general, determining the geometric multiplicity of an eigenvalue requires no new technique because one is simply looking for the dimension of the nullspace of \(A - \lambda I\).

The **algebraic multiplicity** of an eigenvalue \(\lambda\) of \(A\)
is the number of times \(\lambda\) appears as a root of \(p_A\).
For the example above, one can check that \(-1\) appears only once as
a root. Let us now look at an example in which an eigenvalue has multiplicity
higher than \(1\).

Let \(A = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\).

Then \(p_A = \det(A- \lambda I_2) = \left| \begin{matrix} 1-\lambda & 2 \\ 0 & 1-\lambda\end{matrix} \right| = (1-\lambda)^2.\) Note that \(1\) is a root and it appears twice. Hence, the algebraic multiplicity of \(1\) is \(2\). But what is the geometric multiplicity?

Consider \(A-\lambda I = \begin{bmatrix} 0 & 2 \\ 0 & 0\end{bmatrix}\). What is the dimension of its nullspace? Clearly, the rank of this matrix is \(1\). By the rank-nullity formula, we get that the nullspace has dimension \(1\). Hence, the geometric multiplicity is \(1\). This is different from the algebraic multiplicity!

In general, **the algebraic multiplicity and geometric multiplicity
of an eigenvalue can differ. However, the geometric multiplicity can
never exceed the algebraic multiplicity**.

It is a fact that summing
up the algebraic multiplicities of all the eigenvalues
of an \(n \times n\) matrix \(A\) gives exactly \(n\).
**If for every eigenvalue of \(A\),
the geometric multiplicity equals the algebraic multiplicity, then
\(A\) is said to be diagonalizable**. As we will see,
it is relatively easy to compute powers of a diagonalizable matrix.

Let \(A = \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix}\).

Find all eigenvalues of \(A\).

For each eigenvalue of \(A\), determine its algebraic multiplicity and geometric multiplicity.

Let \(A = \begin{bmatrix} 8 & -9 \\ 4 & -4 \end{bmatrix}\).

Find all eigenvalues of \(A\).

For each eigenvalue of \(A\), determine its algebraic multiplicity and geometric multiplicity.

Is \(\begin{bmatrix} 2 & 2 & 2 \\ 0 & 2 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) diagonalizable?

Let \(A, B \in \mathbb{C}^{n\times n}\). \(A\) and \(B\) are said to be

*similar*if \(B= S^{-1}AS\) for some invertible matrix \(S\in\mathbb{C}^{n\times n}\).Show that similar matrices have the same eigenvalues, including multiple appearances.

Let \(\lambda\) be an eigenvalue of \(A\). Prove that the geometric multiplicity of \(\lambda\) is at most the algebraic multiplicity of \(\lambda\).

(Hint: Let \(\lambda\) be an eigenvalue of \(A\). Let \(\{v_1,\ldots,v_k\}\) be a basis for \({\cal E}_A(\lambda)\). Extend \(\{v_1,\ldots,v_k\}\) to a basis \(\{v_1,\ldots,v_n\}\) for \(\mathbb{C}^n\). Form the matrix \(S = [v_1 \cdots v_n]\). What can you say about the first \(k\) columns of the matrix \(B = S^{-1} A S\)?.)