Assume $V$ to be a finite dimensional vector space. Define the algebraic multiplicity $am(\lambda)$ of an eigenvalue $\lambda$ of a linear operator $T:V\to V$ as the maximum index of the factor $(t-\lambda)$ appearing in the characteristic polynomial of $T$. Also define $G_\lambda=\{v\in V:(T-\lambda I)^kv=0\}$. I want to show that $\dim(G_\lambda)=am(\lambda)$ without using Jordan Form.

Sheldon Axler in "Linear Algebra Done Right" specifically defined the "multiplicity" of $\lambda$ as $\dim(G_\lambda)$, hence I could not get any help from it. I am not very conversant with the properties of the Jordan form, hence I would like a more elementary proof. Please note that I cannot use the decomposition of $V$ into a direct sum of generalized eigenspaces because I will need to prove that indeed, $am(\lambda)=\dim(G_\lambda)$ to prove this.

I started by assuming that $f(t)=(t-\lambda)^kp(t)$ where $f$ is the characteristic polynomial of $T$, $p$ is any other polynomial not containing the factor $(t-\lambda)$. So I will have to show that $\dim(G_\lambda)=k$.

By Cayley Hamilton Theorem, $f(T)=0\implies (T-\lambda I)^kp(T)=0$ hence $p(T)v\in G_\lambda \forall v\in V$. Now consider the collection $\{p(T)v,(T-\lambda I)p(T)v,...,(T-\lambda I)^{k-1}p(T)v\}$ for a nonzero $v\in V$ which I know is linearly independent (based on the previous exercise) and hence $\dim(G_\lambda)\geq k$.

How will the other direction follow?

  • $\begingroup$ You don't need to know anything about dimensions to show that any finite dimensional space decomposes as a direct sum of generalised eigenspaces. This depends only on the fact that the minimal polynomial splits, as it does over$~\Bbb C$, after which the primary decomposition theorem can be applied. $\endgroup$ Commented Apr 24, 2015 at 11:51
  • $\begingroup$ Okay then. But could you kindly provide some hints for the part I am stuck at? Please note that $G_\lambda$ is the generalized eigenspace w.r.t. $\lambda$. I hope there is no confusion. $\endgroup$ Commented Apr 24, 2015 at 12:47
  • $\begingroup$ @Landon Carter: What you have written in the last line how do you get such a $v$ ? What happens if $(X-\lambda)P(X)$ is the minimal polynomial for $T$ while $k > 1.$ $\endgroup$
    – user371231
    Commented Dec 20, 2018 at 18:44

2 Answers 2


Don't use the Cayley-Hamilton theorem; it is less elementary than what you need. And in any case in Axler's book it (8.20) follows results to the effect you are asking about (8.10, 8.18). In fact Axler defines the (algebraic) multiplicity of $\lambda$ as $\dim(G_\lambda)$, and then goes on to define the characteristic polynomial to be the product over eigenvalues$~\lambda$ of $(X-\lambda)^{\dim(G_\lambda)}$ (which is a crazy thing to do, born of irrational fear of determinants, but) which makes the question you ask void of content in the context of that book.

I suppose you know that given a direct sum decomposition into invariant subspaces, the characteristic polynomial of $T$ is the product of those of its restrictions to those subspaces. I will also assume you know the characteristic polynomial of the restriction of $T$ to $G_\lambda$ is $(X-\lambda)^{\dim(G_\lambda)}$. Both things are quite obvious if you define the characteristic polynomial using determinants (for the second part use that the restriction has a triangular matrix on an appropriate basis). Now you will be done if you can show that $G_\lambda$ is a factor in a direct sum decomposition into two invariant subspaces, where (the restriction of $T$ to) the other factor does not have $\lambda$ as an eigenvalue.

There are two approaches to proving that fact. The one related to the primary decomposition theorem is to write the minimal polynomial~$\mu$ of$~T$ (or any polynomial annihilating $T$) as product $\mu=(X-\lambda)^dQ$ of a power of $X-\lambda$ and a factor$~Q$ relatively prime to it; then using Bézout coefficients $B,C$ of these two factors (so $1=B(X-\lambda)^d+CQ$) (one can find certain polynomials of $T$ (namely $(CQ)[X:=T]$ and $\def\Id{\mathrm{id}}B[X:=T](T-\lambda\Id)^d$) that give projections onto the kernels of $(T-\lambda\Id)^d$ respectively $Q[X:=T]$, and which kernels therefore form a direct sum decomposition. The kernel associated the factor $(X-\lambda)^d$ is in fact $G_\lambda$ (that approach does not even explicitly depend on the space being finite dimensional, though having an annihilating polynomial in the first place does depend on that).

But there is more elementary: if $G_\lambda$ is the kernel of $(T-\lambda\Id)^d$ with $d=\dim(G_\lambda)$, then the image of $(T-\lambda\Id)^d$ provides an invariant complementary factor. The intersection has dimension zero, since one would otherwise have vectors that are not annihilated by $(T-\lambda\Id)^d$, but which are annihilated by a higher power of $T-\lambda\Id$, which contradicts what you ought to know of generalised eigenspaces. But then the two subspaces are complementary by the rank-nullity theorem, and form a direct sum decomposition.

  • $\begingroup$ From the last paragraph how that follows ? $\endgroup$
    – user371231
    Commented Dec 20, 2018 at 19:26
  • $\begingroup$ @MarianoSuárez-Álvarez: Yes, the order of definitions is with page numbers in Linear Algebra Abridged in brackets: generalized eigenvector/eigenspace [99/100]. multiplicity of eigenvalue is dimension of its generalized eigenspace [102], characteristic polynomial is product over eigenvalues $\lambda_i$ of $(x-\lambda_i)^{d_i}$ where $d_i$ is multiplicity of $\lambda_i$ [111]. And the feared determinant is the product of the eigenvalues, each repeated according to its multiplicity. Don't forget this is a didactic masterpiece (according to zbMATH). $\endgroup$ Commented Nov 29, 2022 at 15:49

I would like to complement Marc van Leeuwen's answer. I am going to use some results from the PDF that Marc linked (Axler's paper). First, we know that $V$ can be written as the direct sum of the generalized eigenspaces, $G(\lambda_i)$, $V = G(\lambda_1) \oplus \cdots \oplus G(\lambda_r)$, because generalized eigenvectors corresponding to different eigenvalues are linearly independent, and because the generalized eigenvectors span $V$. What is the shape of the matrix associated with $T$ under this basis? The direct sum translates into a block diagonal matrix $$ T = \left[ \begin{array}{cccc} M_1 & 0 & 0 & 0 \\ 0 & M_2 & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & M_r \\ \end{array} \right], $$ where $M_i$ is the operator $T$ restricted to $G(\lambda_i)$. What is the shape of each $M_i$? Because we are working on $G(\lambda_i)=\{v\in V:(T-\lambda_i I)^kv=0 \text{ for some }k \in \mathbb{Z}^+\}$, we'll choose a basis constructed in the following way. If $(T-\lambda_i I)^kv=0$ for $k=1$, then $Tv=\lambda_iv$ and this translates into a diagonal element $\lambda_i$ in the matrix $M_i$ if we choose $v$ as an element of our basis. Now extract as many linearly independent vectors $v$ as you can that satisfy $(T-\lambda_i I)v=0$. Call them $B_i(k=1) = \{x_1,\ldots,x_p\}$. The $i$ makes reference to the $i$-th block $M_i$ and the $k$ indicates that we are working on a subspace of $G(\lambda_i)$ associated with the equation $(T-\lambda_i I)^kv=0$. So far, the matrix $M_i$ has the following structure with respect to $B_i(k=1)$. $$ M_i = \left[ \begin{array}{cccc} \lambda_i & 0 & 0 & ? & ? \\ 0 & \ddots & 0 & ? & ? \\ 0 & 0 & \lambda_i & ? & ? \\ 0 & 0 & 0 & ? & ? \\ 0 & 0 & 0 & ? & ? \\ \end{array} \right], $$

Now repeat the same process for $(T-\lambda_i I)^kv=0$ for $k=2$. Note that $(T-\lambda_i I)^2v=0$ implies that the vector $w=(T-\lambda_i I)v$ satisfies $(T-\lambda_i I)w=0$. Hence, $w$ can be written as a linear combination of the $x_1,\ldots,x_p$, which implies that there are scalars $c_1,\ldots,c_p$, such that $$ c_1x_1 + \cdots c_px_p = w = Tv - \lambda_i v, $$ and consequently that $$ Tv = \lambda_i v + c_1x_1 + \cdots c_px_p. $$ This means that $T$ acting on $v$ produces a multiple of $v$ plus a combination of vectors in $B_i(k=1)$. Thus, the matrix $M_i$ now looks like $$ M_i = \left[ \begin{array}{cccc} \lambda_i & 0 & 0 & \times & ? \\ 0 & \ddots & 0 & \times & ? \\ 0 & 0 & \lambda_i & \times & ? \\ 0 & 0 & 0 & \lambda_i & ? \\ 0 & 0 & 0 & 0 & ? \\ \end{array} \right], $$ with respect to the basis $\{x_1,\ldots,x_p,v\}$. The symbol $\times$ indicates a potentially nonzero entry. Continue building and appending the bases $B_i(k)$, $k=2,3,\ldots$, until you span the whole $G(\lambda_i)$. Finally, note that the resulting matrix $M_i$ is upper-triangular with the eigenvalues in the main diagonal. Hence, the multiplicity of $\lambda_i$, $m(\lambda_i)$, corresponds to the number of diagonal entries in $M_i$, i.e., the size of the block, and by construction, this same number tells us the dimension of $G(\lambda_i)$.


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