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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 $A^3=0$?

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  • $\begingroup$ Hint: Eigenvalues are roots of the characteristic polynomial $\endgroup$ – ab123 Feb 4 at 20:05
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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 $0$.

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  • $\begingroup$ So does it mean that a nilpotent matrix has all eigen values equal to 0? $\endgroup$ – Jor_El Feb 4 at 20:11
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    $\begingroup$ @Jor_El Yes - every eigenvalue of a nilpotent matrix is zero. $\endgroup$ – Jan Feb 4 at 20:13
  • $\begingroup$ 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. $\endgroup$ – Michael Anderson Feb 5 at 1:53
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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.

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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 $A$.

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    $\begingroup$ 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. $\endgroup$ – Mike Earnest Feb 4 at 23:39
  • $\begingroup$ 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 $P$. $\endgroup$ – Paul Sinclair Feb 5 at 0:57
  • $\begingroup$ You're right. I'll fix it right away. $\endgroup$ – Bruno Tassone Feb 5 at 8:56

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