Nilpotent matrix generated by a polynomial formula

Question

Having non-singular matrix $A$ (real, dim. $n \times n$) we can generate singular matrix $B$ using polynomial formula $B=A-\lambda{I}$, where $\lambda$ is an eigenvalue of matrix $A$. If $A$ has $n$ distinct eigenvalues then there are at least $n$ different ways of generating from non-singular matrix a matrix which is singular.
If we additionally use polynomial formula based fully on the characteristic equation we can obtain, according to Cayley-Hamilton theorem, simply the zero matrix.

Nilpotency is the concept somewhere between general singularity and zero matrix.

• I'm interested whether it is possible to generate non-zero nilpotent matrix $B$ from non-singular matrix $A$ with the use of polynomial formula...

It seems that the problem is tractable when we assume that $A$ is diagonalizable.

Indeed if $B$ is generated by the polynomial formula we have
$B= A^n+a_{n-1}A^{n-1}+ \dots+a_1A+a_0I$.

But if $A$ is diagonalizable then $A=SDS^{- 1}$ and with polynomial formula we obtain expression $B=SD_2S^{-1}$ where $D_2$ is some diagonal matrix. On the other hand we require $B^k=(SD_2S^{-1})^k=0$ i.e. $(D_2)^k=0$ what is possible only when diagonal matrix $D_2 =0$ so in this case it would be $B=SD_2S^{-1} =0$ what is in contradiction with $B$ assumed being non-zero matrix. So it seems there is no way to generate non-zero nilpotent matrix in this case.
I hope that my reasoning is correct..

However I don't know how to solve the problem if $A$ is non-singular non-diagonalizable matrix?

• Is it possible in this case to generate with the use of polynomial formula non-zero nilpotent matrix? What is the necessary and sufficient condition for that?

Answer 1

Any matrix $A$ has a minimal polynomial $q(x)$ defined as the monic polynomial with minimum degree such that $q(A)=0$. An easy corollary is that

If $p(x)$ is a polynomial, then $$p(A) =0 \iff q(x)|p(x) $$

Let now $B=p(A)$ be a non-zaero nilpotent matrix. This means that $B^k=p(A)^k=0$ and by the corollary, $q(x)|p(x)^k$ and $q(x)\not| p(x)$. This can happen if and only if $q(x)$ has irreducible factors with multiplicity at least 2. If we're working on the field of complex numbers, then we have also an useful lemma:

The matrix $A$ is diagonalizable if and only if its minimal polynomial is square-free

This result gives us the answer to your question:

There exists a polynomial $p(x)$ such that $B=p(A)$ is a nilpotent non-zero matrix if and only if $A$ isn't diagonalizable. For example, take $p(x)$ as the product of $x-\lambda$, with $\lambda$ eigenvalues without multiplicity

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Some examples:

Take the matrix $$ A=\begin{pmatrix} 1 & 1 & \\ & 1 & \\ & & 2 \end{pmatrix} $$ this is not diagonalizable, so we take $p(x)=(x-1)(x-2)$ and obtain $$ p(A)=\begin{pmatrix} 0 & -1 & \\ & 0 & \\ & & 0 \end{pmatrix} $$ that is nilpotent and non-zero.

The matrix must be non-diagonalizable in $\mathbb C$: for example take $$ A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$ this is not diagonalizable in $\mathbb R$, but in $\mathbb C$ its diagonal form is $$ A=\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} $$ and you can use your own reasoning to prove that there doesn't exist $B=p(A)$ non-zero nilpotent.

To find a real matrix non-diagonalizable in $\mathbb R$ that is also non-diagonalizable in $\mathbb C$, you have to reach dimension 4. $$ A=\begin{pmatrix} 0 & -1 & 1 & \\ 1 & 0 & & 1 \\ & & 0 & -1 \\ & & 1 & 0 \end{pmatrix} $$ and in this case, the eigenvalues are $i,i,-i,-i$, so $p(x) = (x-i)(x+i) = x^2+1$. In fact, $$ p(A)=\begin{pmatrix} 0 & & & -2 \\ & 0 & 2 & \\ & & 0 & \\ & & & 0 \end{pmatrix} $$ that is non-zero nilpotent.