# Number of times that a eigenvalue appears in the diagonal of a upper-triangular matrix is equal to the multiplicity

Problem from Linear Algebra Done Right 3rd edition, page 260.

Notation:

• F (field) denotes R or C
• $V$ denotes a finite-dimensional nonzero vector space over F
• $\mathcal{L}(V)$ is the set of all linear transformations from $V$ to $V$

Suppose $T \in \mathcal{L}(V)$ and $\lambda \in$ F. Prove that for every basis of $V$ with respect to which $T$ has an upper-triangular matrix, the number of times that $\lambda$ appears on the diagonal of the matrix of $T$ equals the multiplicity of $\lambda$ as an eigenvalue of $T$.

Here multiplicity is the algebraic multiplicity.

• What is Axler's definition of algebraic multiplicity? I know he doesn't like to do things in terms of determinants. Has he said anything about $\dim \ker (A - \lambda I)^n$ at this point (where $A$ is $n \times n$)? Commented Mar 7, 2017 at 5:03
• The definition in my book of the algebraic multiplicity of an eigenvalue $\lambda$ for an operator $T$ on $V$ is the dimension of the generalized eigenspace corresponding to $\lambda$. In other words, algebraic multiplicity of $\lambda = \dim \ker (T - \lambda I)^{\dim V}$. Commented Mar 7, 2017 at 6:18

Without loss of generality, suppose that $$\lambda$$ appears on the first few diagonal entries. Thus, the matrix $$A$$ with respect to this basis is such that $$A - \lambda I = \pmatrix{T& B\\0 & C}$$ where $$C$$ is invertible and $$T = \pmatrix{0&*&\cdots&*\\ &\ddots&\ddots&\vdots\\ &&&*\\&&&0}$$ Note that $$T^n = 0$$. So, $$\dim \ker (A - \lambda I)^n$$ is the size of $$A$$.
• @BenGrossmann The last sentence "So, $\dim \ker (A - \lambda I)^n$", is that supposed to equal something? Commented Mar 29, 2021 at 6:05