The question:

"Let $B$ be a $10 \times 10$ matrix and let $\lambda$ be a scalar. Suppose it is known that $$ \text{nullity}(B - \lambda I) = 5, \\ \text{nullity}(B - \lambda I)^2 = 8, \\ \text{nullity}(B - \lambda I)^3 = 9. $$ Find all possible Jordan forms of B."

The answer is $J = J_4(\lambda) \oplus J_2(\lambda) \oplus J_2(\lambda) \oplus J_1(\lambda) \oplus J_1(\lambda)$, where $\lambda$ is the only eigenvalue, or $J = J_3(\lambda) \oplus J_2(\lambda) \oplus J_2(\lambda) \oplus J_1(\lambda) \oplus J_1(\lambda) \oplus J_2(\mu)$, if the tenth eigenvalue is not $\lambda$.

I know in the question we have 9 out of the 10 eigenvalues for $B$ so we're basically just trying to find out where to place the last one. In the answers, it seems they have placed it in ker$(B - \lambda I)^4$.

But my question is, in the first case, why couldn't we have placed it in ker$(B - \lambda I)^3$ and thus gotten the Jordan form $J = J_3(\lambda) \oplus J_3(\lambda) \oplus J_2(\lambda) \oplus J_1(\lambda) \oplus J_1(\lambda)$? Doesn't this also work? According to the answers, this "breaks the property of the difference in nullities being non-increasing" but I don't see how this is the case?

I would be grateful if someone could explain this.

  • $\begingroup$ Could you say a bit about how you usually calculate JCFs? Do you use a dot diagram? Are you familiar with cycles of generalized eigenvectors and their relation to a Jordan Canonical basis? $\endgroup$ Commented Aug 19, 2019 at 2:53
  • $\begingroup$ @ArturoMagidin there is no assumption that it is the only one, so in the answers they considered both cases. Have edited the post to include this. I use a block/tower diagram so that there would be 5 blocks at the bottom for the nullity 5, then 3 blocks above that to make 8, then 1 block at the top to make 9. Yes, I am familiar with the cycles of generalised eigenvectors. $\endgroup$
    – scott
    Commented Aug 19, 2019 at 3:02

1 Answer 1


The nullities of $(T-\lambda I)^k$ are related to the Jordan Canonical form, the number of blocks, and the size of the blocks.

Each block contains one and only one eigenvector. Thus, $\mathrm{nullity}(T-\lambda I)$ tells you the number of Jordan blocks associated to $\lambda$.

Now, the Jordan canonical basis consists of cycles of generalized eigenvectors; a block of size $k$ corresponds to a sequence of length $k$ of generalized eigenvectors of the form $$(T-\lambda I)^{k-1}(\mathbf{v}), (T-\lambda I)^{k-2}(\mathbf{v}),\ldots, (T-\lambda I)(\mathbf{v}), \mathbf{v},$$ where $\mathbf{v}\in\mathbf{N}((T-\lambda I)^k)$ and $\mathbf{v}\notin \mathbf{N}((T-\lambda I)^{k-1})$.

(I like my Jordan Canonical Forms to have $1$s above the main diagonal; if you like them having $1$s below the main diagonal, your list will start with $\mathbf{v}$ and end with $(T-\lambda I)^{k-1}(\mathbf{v})$ instead)

In particular, the first $r$ vectors of this list (corresponding to the first $r$ rows/columns of the block), and only the first $r$ vectors of this list, lie in $\mathbf{N}((T-\lambda I)^{r})$.

So the number of blocks that have size at least $2$ would be $\mathrm{nullity}((T-\lambda I)^2) - \mathrm{nullity}(T-\lambda I)$: the first gives you all vectors that are either first or second on their corresponding cycle (either first or second in their block), and the second summand is the number that are first. So the difference is exactly the number of vectors in the second position, i.e., number of blocks of length at least $2$.

Similarly, the number of blocks of length at least $3$ is $$\mathrm{nullity}((T-\lambda I)^3) - \mathrm{nullity}((T-\lambda I)^2)$$ and more generally, the number of blocks of length at least $k$ is $$\mathrm{nullity}((T-\lambda I)^k) - \mathrm{nullity}((T-\lambda I)^{k-1}).$$

Looking at your nullities, then, we know that:

  1. There are exactly 5 blocks corresponding to $\lambda$.
  2. Exactly 3 (which is $\mathrm{nullity}((T-\lambda I)^2) - \mathrm{nullity}(T-\lambda I)$) blocks of size at least $2$.
  3. Exactly 1 block of size at least $3$.
  4. Therefore, there are 2 blocks of size $1$, and 2 blocks of size 2.

As you note, the only thing left is to know what happens to the final vector in our basis. If it is an eigenvector corresponding to another eigenvalue, it goes by itself and we are done: we also have exactly one block of size $3$ for $\lambda$.

If the final generalized eigenvector also correspond to $\lambda$, where could it be? We already know we have exactly two blocks of length $1$, and exactly two blocks of length $2$, and one block of length at least $3$; they accounts for $1+1+2+2 = 6$ plus at least $3$ other vectors... and that one has to go into the block of size “at least $3$”, which must therefore have size exactly $4$.

Had the decomposition had two blocks of length $3$, two blocks of length $1$, and one block of length $2$, then the nullity of $(T-\lambda I)^3$ would have been $10$, not $9$, as every vector in the basis would have been in the first three positions of their cycles.

Your mention “block/tower diagram” in your comment; I know it as the “dot diagram”, following Friedberg/Insel/Spence. But as you note: the bottom has $5$ blocks, then $3$, and then $1$. If you had a decomposition with two block of size $3$, you would need to have $2$ blocks in that third level, not just $1$, which would have required the nullity of $(T-\lambda I)^3$ to be two more than that of $(T-\lambda I)^2$, i.e., you would need to have $\mathrm{nullity}((T-\lambda I)^3) = 10$.


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