# If the multiplicity of $\lambda_1$ is $r_1$ in characteristic polynomial of $A$, then $\operatorname{null}(A-\lambda_1 I)^{r_1} = r_1$

Let $$A: V \to V$$ be a linear transform over a field $$F$$, and the characteristic polynomial is $$f(\lambda) = (\lambda-\lambda_1)^{r_1}\ldots(\lambda-\lambda_s)^{r_s}$$. Prove that: $$\operatorname{rank}(A - \lambda_1I)^{r_1} = n - r_1$$

I think this essentially requires me to show that $$r_1 = \operatorname{null}(A-\lambda_1I)^{r_1}$$. But I am not sure how to use the knowledge of minimal polynomial to prove that. Could someone help me with it?

• Is $f$ the characteristic polynomial or the minimal polynomial? Your post/tags mention both. – angryavian Jun 5 at 4:41
• You could look at the Jordan normal form for $A$. $r_1$ is the number of times $\lambda_1$ appears on the diagonal. If you subtract $r_1I$, each Jordan block with a $\lambda_1$ on the diagonal has a zero, and each Jordan block corresponding to $\lambda_i$, for $i\not=1$ has a nonzero entry on the diagonal $\lambda_i-\lambda_1$. The size of largest Jordan block of A corresponding to $\lambda_1$ is at most $r_1$, and each Jordan block corresponding to $\lambda_1$ is now nilpotent (after subtracting $\lambda_1I$ – user124910 Jun 5 at 4:59
• @user124910 you don't need the sledgehammer of Jordan normal form. You have the decomposition $V=\bigoplus_i\ker(A-\lambda_i)^{r_i}$ from looking at the characteristic polynomial and Bezout in $F[X]$ (or explicit computation). On each $\ker(A-\lambda_i)^{r_i}$ you can triangularise $A$ by induction. So the result follows. – user10354138 Jun 5 at 5:13
• @angryavian The problem mentions characteristic polynomial; but it belongs to the minimal polynomial section in the book, so I put them both as tags. – mathdoge Jun 5 at 7:45
• @user10354138 I am more interested in the method you mentioned. Yes I think we could get that decomposition for $V$. So taking a basis for each $\text{ker}(A-\lambda_i I)^{r_i}$, and put the bases together, we obtain a basis for $V$, under which the matrix representation of $A$ should be in the form of $\text{diag}\{A_1, \cdots, A_s \}$. Is it what you meant? – mathdoge Jun 5 at 7:48

## 2 Answers

I won't use any other eigenvalues explicitly, so set $$\lambda=\lambda_1$$ and $$r=r_1$$ for brevity.

Let $$P=(X-\lambda)^r$$ and let $$Q=f/P$$ be the remaining factor of the characteristic polynomial $$f$$. By assumption (certainly you meant the $$\lambda_i$$ to be distinct, although this is not said explicitly), $$\lambda$$ is not a root of$$~Q$$, and so $$P$$ and $$Q$$ are relatively prime in $$F[X]$$. Therefore there exist Bézout coefficients $$S,T\in F[X]$$ with $$SP+TQ=1$$. Since $$(PQ)[A]=0$$ (by Cayley-Hamilton) it is easy to see that $$(SP)[A]$$ and $$(TQ)[A]$$ are projectors on complementary subspaces $$U,W$$ of $$V$$. Then $$P[A]$$ acts invertibly on $$U$$ and vanishes on $$W$$, so $$W=\ker(P[A])=\ker((A-\lambda I)^r)$$, and $$U$$ is the image of $$P[A]$$, in particular $$\operatorname{rank}(P[A])=\dim(U)$$.

Also the characteristic polynomial $$f$$ of $$A$$ is the product of the characteristic polynomials of the restrictions of$$~A$$ to $$U$$ and $$W$$, the first of which does not have $$\lambda$$ as a root (as $$P[A]$$ acts invertibly on $$U$$) while the latter is a power of $$X-\lambda$$ (since $$(X-\lambda)^r$$ is an annihilating polynomial). But then the characteristic polynomial of the restriction of$$~A$$ to$$~W$$ is $$(X-\lambda)^r$$, and $$\dim(W)$$ is its degree$$~r$$, and $$\dim(U)=\dim(V)-r$$, completing the proof.

Note that my second paragraph only depends of $$PQ$$ being an annihilating polynomial with $$P$$ grouping all its factors $$X-\lambda$$; in particular it could have been the minimal polynomial, and the exponent could have been different from $$r$$. This is in fact a bit confusing in a proof of something that eventually says something about$$~r$$. Eventually $$r$$ is gotten as $$\dim(W)$$, not as $$\deg(P)$$.

• Thanks! I will be asking a rather stupid question: how to see that the char poly of $A$ restricted on $W$ is $(x-\lambda)^r$? – mathdoge Jun 7 at 14:42
• @mathdoge Because as I say, there cannot be any factors $X-\lambda$ coming from the summand $U$, and we know there are $r$ such factors in all. – Marc van Leeuwen Jun 7 at 15:16

We know that matrix A will satisfy its characteristic polynomial, such that

$$(A-\lambda_1I)^{r_1}\ldots(A-\lambda_sI)^{r_s} = 0$$, Now every matrix is similar to a upper triangular matrix. Let B is the upper triangular matrix similar to A. Where the diagonal elements are the eigen values of A and arranged similarly from left to right as in the equation above (by grouping together the same eigenvalues in Schur decomposition). Then the above can be written as

$$(B-\lambda_1I)^{r_1}\ldots(B-\lambda_sI)^{r_s} = 0$$. Now note that B is an upper triangular matrix and $$(B-\lambda_1I)$$ is an upper triangular matrix, with diagonal elements 0 in first $$r_1$$ column only (no other diagonal elements are 0) also with the first column vector 0 (all elements in column one are 0).

Now we see that in $$(B-\lambda_1I)^{2}$$ first two columns (vectors) are surely zero, and this continues, for ex. in $$(B-\lambda_1I)^{x}$$ first x columns (vectors) ($$x \leq r_1$$) are surely zero.

We also note that in any such multiplication process described above the diagonal elements in rest of the columns can not be zero (they actually powered up). So if we perform $$(B-\lambda_1I)^{r_1}$$, only first $$r_1$$ columns (vectors) will be zero (and no other column vector). Now looking at the rest of the $$n-r_1$$ columns in $$(B-\lambda_1I)^{r_1}$$, we see that there diagonal elements are not zero but all elements below diagonal elements are zero. That makes these column vectors linearly independent.

And proof is completed as $$(B-\lambda_1I)^{r_1}$$ is similar to $$(A-\lambda_1I)^{r_1}$$.

• One cannot easily group equal eigenvalues together in a triangular matrix, as "shuffling columns" (using conjugation by a permutation matrix, as we need to remain similar) destroys the triangular form. One can get the right form basically by separating out eigenvalues before seeking a triangular form, but seeking a triangular form initially does not really help. – Marc van Leeuwen Jun 5 at 11:22
• @MarcvanLeeuwen, my bad, you are right. The proper statement should be something like "grouping together the same eigenvalues in Schur decomposition". I am editing it now. – amitava Jun 5 at 12:10