Let $A$ be a $7\times 7$ matrix over $\mathbb C$ with minimal polynomial $(t-2)^3$. I need to prove $\dim \ker (A-2)\geq 3$.

The handwavy argument I have is that $\deg m$ is the size of the greatest Jordan block while $\dim \ker (A-2)$ is the number of blocks, and since $2\cdot 3<7$, the dimension must be at least $3$. However, I realized I don't know how to formally prove this, i.e without taking the sentences I said as facts.

  • $\begingroup$ if you are allowed to use the existence of the Jordan form then your handwavy argument seems close to complete in my estimation. Perhaps missing is the comment that the sum of the dimensions of the generalized e-spaces is $7$ and since you cannot have a $4$-dimensional gen. e-space it follows you have at worst $3,3,1$ as the dimension of the gen. e-spaces. But, to each space there is an e-vector and so the worst case scenario is $dim(ker(A-2))=3$. $\endgroup$ – James S. Cook Jul 7 '16 at 15:18
  • $\begingroup$ @JamesS.Cook why can't there by a 4d gen eigenspace? $\endgroup$ – linalg Jul 7 '16 at 15:35
  • $\begingroup$ because the degree of each factor in the minimal polynomial is the size of the largest Jordan block corresponding to the e-value in question. If there was a $4 \times 4$ block then the minimal polynomial would necessarily be at least $4$-th order (but, you have just order $3$) $\endgroup$ – James S. Cook Jul 8 '16 at 0:53

I wouldn’t call your argument handwavy, but we can replace the use of the Jordan normal form by the underlying calculations:

Lemma: Let $V$ be a finite-dimensional vector space and $f, g \colon V \to V$ be two endomorphisms. Then $\dim \ker (fg) \leq \dim \ker f + \dim \ker g$.

Proof: We have $g( \ker(fg) ) \subseteq \ker f$ and thus by the dimensional formula that $$ \dim \ker (fg) = \dim \operatorname{im} g|_{\ker(fg)} + \dim \ker g|_{\ker(fg)} \leq \dim \ker f + \dim \ker g. $$

Thus we get the following by induction:

Corollary: If $V$ is a finite-dimensional vector space and $f_1, \dotsc, f_n \colon V \to V$ are endomorphisms, then $$ \dim \ker(f_1 \dotsm f_n) \leq \dim \ker f_1 + \dotsb + \dim \ker f_n. $$ We have in particular that $\dim \ker f^k \leq k \dim \ker f$ for every endomorphism $f \colon V \to V$ and every $k \in \mathbb{N}$.

If we apply this corollary to the endomorphism $f \colon \mathbb{C}^7 \to \mathbb{C}^7$, $x \mapsto (A-2) x$ we find that $$ 7 = \dim \mathbb{C}^7 = \dim \ker f^3 \leq 3 \dim \ker f = 3 \dim \ker (A-2), $$ and thus $\dim \ker (A-2) \geq 3$.


Note that we're not really interested in $A$, we're interested in $B = A - 2$, which has minimal polynomial $t^3$.

If $\dim\ker B \leq 2$, then the kernel of $B^2$ would have dimension at most $4$, and the kernel of $B^3$ would have dimension at most $6$. But $B$ is a $7\times 7$ matrix with minimal polynomial $t^3$, so $\dim\ker B^3 = 7$. This is a contradicion.

Some might say that this is essentially the same argument. But this is more elementary, and thus it is perhaps easier to formalise.

  • $\begingroup$ How do you show that $\mathrm{dim}\;\mathrm{ker}(CD)\leq\mathrm{dim}\;\mathrm{ker}(C)+ \mathrm{dim}\;\mathrm{ker}(D)$? I feel like it should be obvious but I can't find a proof. $\endgroup$ – Oscar Cunningham Jul 7 '16 at 14:55
  • $\begingroup$ @OscarCunningham Since $(\operatorname{Im} D \cap \ker C) \subseteq \ker C$, we see that $C$ cannot collapse more dimensions of the image of $D$ than it does to its whole domain. $\endgroup$ – Arthur Jul 7 '16 at 15:07

Foremost, if $\lambda$ is an eigenvalue of $A$, then $(A-\lambda)$ will be a factor of the minimal polynomial (I hope this isn't too hand-wavey, but it is true, so I don't think it is).

We know that $V$ decomposes into the direct sum of the generalized eigenspaces since your field is algebraically closed. Since we know $A$ only has 1 eigenvalue by the minimal polynomial argument,

$$V = G(A, 2) = \ker(A-2I)^7$$ (Where 7 is $\dim V$)

Since the root 2 has multiplicity 3 in the minimal polynomial, we in fact know that $V = \ker(A-2I)^3$, which means that $\dim \ker(A-2I)^3 = 7$, and we also know that $dim\ker(A-2I)^2 \leq 6$, since if it were 7, the minimal polynomial would instead be $(2-t)^2$.

As we know the exact form of the basis, we can discuss the rest in terms of the Jordan chains. We know that the Jordan basis corresponding to generalized eigenvalues of $V$ must all be corresponding to eigenvalue $2$ due to the decomposition of $V$ into its generalized eigenspaces discussed above. We know of a length 3 Jordan chain because of the multiplicity in the minimal polynomial, so there is some chain in our basis that is $v_1, (A-2I)v_1, (A-2I)^2v_1$, where $(A-2I)^2v_1$ is an eigenvalue by definition of the Jordan basis.

The other chains in our basis could be at most of length 3, which would minimize the number of vectors strictly in the eigenspace. Due to dimension counting, we couldd only have one chain of length 3, and then another of length 1.

So the other two chains, at their longest, are: $v_2, (A-2I)v_2, (A-2I)^2v_2, v_3$, where $(A-2I)^2v_2$ and $v_3$ are the eigenvalues by definition of the Jordan basis.

We could also have two chains of length 2, which also results in 3 total eigenvectors, one for each chain, 1 chain of length 2 and 2 chains of length 1 (4 total eigenvectors), or 4 chains of length 1 (5 total eigenvectors), all which produce a dimension of the eigenspace greater than or equal to 3.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.