# Is there any way to know the algebraic multiplicity of the $0$ eigenvalue in the minimal polynomial when the rank is $1$? [duplicate]

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Say I have a matrix $A$ of $r=rank(A)=1$

I know that in the characteristic polynomial the algebraic multiplicity of $(\lambda-0)$ is $n-r$ which in my case is $n-1$

Is there a rule about the algebraic multiplicity in the minimal polynomial in this case or in the general case?

## marked as duplicate by Marc van Leeuwen, John B, Claude Leibovici, M. Vinay, user91500May 28 '16 at 7:39

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## 3 Answers

Every matrix of rank 1 is similar to one of the Jordan forms $$\left(\begin{array}{cc|ccc}0&1&\\&0&\\ \hline && 0 \\ &&&\ddots \\ &&&&0 \end{array}\right) \qquad\text{or}\qquad \left(\begin{array}{c|ccc}\lambda&\\\hline &0\\&&\ddots\\&&&0\end{array}\right)$$ for some $\lambda\ne 0$.

The first has characteristic polynomial $t^n$ and minimal polynomial $t^2$. The second has caracteristic polynomial $(t-\lambda)t^{n-1}$ and minimal polynomial $(t-\lambda)t$ (or just $t-\lambda$ if $n=1$).

• How can you prove that the jordan normal form is one of those two? why can't it be that the minimal polynomial is $t^3$ or $(t-\lambda)t^2$ or any other combination of algebraic multiplicities? – PanthersFan92 May 27 '16 at 10:22
• @Panther: The rank of a matrix is the sum of the ranks of its Jordan blocks -- and the only possible Jordan blocks of rank $1$ are $({}^0_{\strut}\,{}^1_0)$ and $(\lambda)$. So there is exactly one of those, and the other Jordan blocks are $(0)$. – Henning Makholm May 27 '16 at 10:37
• I'm always a nitpick: your final conclusion requires $n>1$. – Marc van Leeuwen May 28 '16 at 7:35

The algebraic multiplicity of $0$, being at least the geometric one, is at least $n-1$ in your case, more generally $n - r$ with $r$ the rank. Contrary to what you say it is not necessarily equal to this value.

Now, it is always at most $n$, the dimension. Thus, in the characteristic polynomial it is either $n-1$ or $n$. Both can happen. For the $n-1$ consider a matrix $0$ except for $a_{11} = 1$. For the latter consider a matrix $0$ except for $a_{12} = 1$. For a rule, you could say it is $n$ if the matrix is nilpotent, and $n-1$ otherwise. Note that the nilpotent matrices are precisely those with $0$ having algebraic multiplicity $n$.

For the minimal polynomial we get that in the former case the algebraic multiplicity of $0$ is $1$. This is because there is another eigenvalue besides $0$ so the dimension of the characteristic space associated to $0$ cannot be the full space and thus has dimension at most $n-1$, that is it coincides with the kernel of $A=A^1$.

In the latter case, one gets that the algebraic multiplicity $m$ in the minimal polynomial is $2$. It cannot be more than two as $\ker A^1 \subset \ker A^2 \dots \subset \ker A^m$ needs to have strict inclusions and since the dimension of $\ker A$ is already $n-1$ there can only be one additional entry in this chain. It also cannot be $1$, as this would mean $\ker A$ is the full space, which is impossible for a rank $1$ matrix.

• But i'm asking specifically for the minimal polynomial. Could it be that the Characteristic Polynomial is $\lambda^{n-1}(tr(A)-\lambda)$ and the Minimal Polynomial is $\lambda(tr(A)-\lambda)$? – PanthersFan92 May 27 '16 at 10:06
• Sorry, I will fix that. – quid May 27 '16 at 11:05
• I explanded the anwer to adress the actual question. – quid May 27 '16 at 11:25
• Sorry for yet another comment. Yes it is exactly as you say (except for the tangential fact that usually the min pol. is normed.) The multiplicty then depends on the trace being 0 or not. – quid May 27 '16 at 11:40

Matrices of rank$~1$ are easy to understand due to their large eigenspace for $\lambda=0$. Let $A$ be square of size$~n$ and have trace$~c$. Then its characteristic polynomial is $X^{n-1}(X-c)$, so the algebraic multiplicity of the eigenvalue$~0$ is either $n-1$ when $c\neq0$, or $n$ when $c=0$. The minimal polynomial is $X(X-c)$ provided $n>1$ (when $n=1$ the minimal polynomial is of course $X-c$). I explained these things in this answer.

The image of$~A$ is always en $A$-stable subspace, and for a rank$~1$ matrix it has dimension$~1$, so any spanning vector of the image is an eigenvector. The eigenvalue of such a vector equals the trace$~c$ of$~A$, and therefore $X(X-c)$ annihilates $A$. In particular the image of$~A$ is contained in its kernel if and only if $c=0$, and (only) in this case $A$ is not diagonalisable.