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


  • 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.

  • $\begingroup$ 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 $n \times n$)? $\endgroup$ Commented Mar 7, 2017 at 5:03
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    $\begingroup$ 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 $\lambda = \dim \ker (T - \lambda I)^{\dim V}$. $\endgroup$ Commented Mar 7, 2017 at 6:18

1 Answer 1


Here's one approach. Whether it's helpful depends on your answer to my comment above.

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 $A$.

See also theorem 8.10 in the text.

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    $\begingroup$ Why without loss of generality? Is it possible change the positions of the diagonal entries between each other? How? $\endgroup$ Commented Mar 7, 2017 at 17:01
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    $\begingroup$ You're right, it's not so clear. I'll clarify when I get the chance. Also, you should look at that theorem $\endgroup$ Commented Mar 7, 2017 at 17:47
  • $\begingroup$ I already looked and indeed it's equivalent to the statement of the problem. I didn't know it, because the theorem 8.10 to which you refer is from the 2nd edition and I study by the 3rd edition. $\endgroup$ Commented Mar 7, 2017 at 17:52
  • $\begingroup$ @RafaelDeiga aha! I missed that. $\endgroup$ Commented Mar 7, 2017 at 18:03
  • $\begingroup$ @BenGrossmann The last sentence "So, $\dim \ker (A - \lambda I)^n$", is that supposed to equal something? $\endgroup$ Commented Mar 29, 2021 at 6:05

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