# Eigenvalues of a matrix whose square is zero

Let $$A$$ be a nonzero $$3 \times 3$$ matrix such that $$A^2=0$$. Then what is the number of non-zero eigenvalues of the matrix? I am unable to figure out the eigenvalues of the above matrix.

P.S.: how would the answer change if it were given that $$A^3=0$$?

• Hint: Eigenvalues are roots of the characteristic polynomial – ab123 Feb 4 at 20:05

A square matrix $$A$$ is called nilpotent if there is a $$p \in \mathbb{N}$$ such that $$A^p=0$$. So let $$A$$ be a nilpotent matrix. Then we have by definition of an eigenvalue

$$Av=\lambda v,$$

where $$\lambda$$ is an eigenvalue of $$A$$ and $$v\neq 0$$ is an eigenvector of $$A$$ to the corresponding eigenvalue. Because $$A$$ is nilpotent we also have

$$0=A^p v=\lambda^p v$$

and because $$v \neq 0$$ it follows $$\lambda^p=0$$, i. e. $$\lambda=0$$. So to your question: The number of non zero eigenvalues is in this case $$0$$.

• So does it mean that a nilpotent matrix has all eigen values equal to 0? – Jor_El Feb 4 at 20:11
• @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero. – Jan Feb 4 at 20:13
• As an aside, the same argument used when $B^p=I$ shows that the eigen-vectors of $B$ will be $p$-th roots of unity. – Michael Anderson Feb 5 at 1:53

Clearly, as the characteristic polynomial of the matrix $$A$$ is $$x^n = 0$$, and the eigenvalues are roots of the characteristic polynomial, there can not be any non-zero eigenvalue.

Another approach is this one:

Since $$A^2 = 0$$, the polynomial $$g(x) = x^2$$ annihilates A (and this means that the Linear operator defined by $$g(T)$$ is the null operator). However, the minimal polynomial of A must divide every polynomial that annihilates $$A$$, so if $$m(x)$$ is such polynomial, $$m$$ must divide $$g$$.

Hence, $$m(x) = x^2$$, because $$A ≠ 0$$.

Thus, the characteristical polynomial of $$A$$ is $$p(x) = x^3$$, because $$p(x)$$ has the same roots of $$m(x)$$ (why?), and $$p(x)$$ annihilates $$A$$ (by Cayley-Hamilton theorem).

In conclusion, the characteristical polynomial of $$A$$ has only a single root, $$0$$, and since every eigenvalue of $$A$$ is a root of it's characteristical polynomial, we have that 0 is the only eigenvalue of $$A$$.

• One small note, the characteristic polynomial of $A$ would be $x^3$. The degree of the char poly of a matrix is equal to its dimension. – Mike Earnest Feb 4 at 23:39
• More generally, if some linear operator $B$ has $Bv = \lambda v$ with $v\ne 0$, and $P$ is a polynomial, then it is an easy calculation that $P(B)v = P(\lambda)v$. Therefore for any polynomial with $P(B) = 0$, every eigenvalue $\lambda$ of $B$ must be a root of $P$. – Paul Sinclair Feb 5 at 0:57
• You're right. I'll fix it right away. – Bruno Tassone Feb 5 at 8:56