6
$\begingroup$

The question is: The square matrix $A$ satisfies $p(A) = 0$, where $p(x)$ is a polynomial such that $p(0) \ne 0$. Show that $A$ is invertible.

I'm lost, I don't know if there's something more I have to learn to do this. I've gotten this far (I'm most likely not on the right track):

$$ p(A) = a_0I+a_1A+a_2A^2+ ...+a_nA^n $$ $$ p(0) = a_0I+(a_1\cdot 0)+(a_2\cdot 0^2)+\ldots +(a_n\cdot 0^n) $$ $$ p(0) = a_oI$$ $$ p(A) = p(0)+a_1A+a_2A^2 +\ldots +a_nA^n $$

I don't quite know what to do further. I know that if $AX=B$, where $A$ is the square matrix, $B$ is a matrix vector, if there's only one solution $X$ for all $B$, then $A$ is invertible.

$\endgroup$
  • $\begingroup$ @A.G. No, but I'm going to google it. $\endgroup$ – E.Bob Nov 9 '16 at 20:14
  • $\begingroup$ Forget it, the answer below is much better. $\endgroup$ – A.Γ. Nov 9 '16 at 20:15
27
$\begingroup$

If $p(0)$ is nonzero then $a_0$ is nonzero. Therefore, one has: $$I=-\sum_{i=1}^n\frac{a_i}{a_0}A^i=-A\sum_{i=0}^{n-1}\frac{a_{i+1}}{a_0}A^i.$$

$\endgroup$
  • $\begingroup$ Shouldn't that be $a_{i+1}$ in your last fraction? $\endgroup$ – Malloc Nov 10 '16 at 5:46
  • $\begingroup$ It should indeed! Thank you! $\endgroup$ – C. Falcon Nov 10 '16 at 6:36
5
$\begingroup$

There's a simple way to do this without manipulating an expansion of the polynomial, or knowing anything about determinants or characteristic polynomials.

First, recall that $A$ is non-invertible if and only if there exists some non-zero vector $\bf{x}$ such that $A{\bf x} = {\bf 0}$. Suppose that there exists such a vector.

Since $p(A)$ is a sum of matrices ($a_n A^n + \dots + a_1 A + a_0 I$) we may compute ${\bf x}^T p(A) \bf{x}$. The fact that $A {\bf x} = 0$ implies that ${\bf x}^T A^n {\bf x} = {\bf 0}$ for any $n>0$, so we have that ${\bf x}^T p(A) {\bf x} = {\bf x}^T a_0 {\bf x} = p(0) \left| {\bf x} \right|^2$. Since we are told that $p(A) = 0$, we have that $p(0) \left| {\bf x} \right|^2 = 0$. Since ${\bf x}$ is non-zero, we must have $p(0) = 0$, contradicting the assumption that $p(0) \ne 0$.

To summarise: we are given that $p(A) = 0$ and that $p(0) \ne 0$. If $A$ is not an invertible matrix, then the argument of the previous two paragraphs shows that the conditions $p(A) = 0$ and $p(0) \ne 0$ cannot both hold. So $A$ must be invertible.

$\endgroup$
  • $\begingroup$ I really like this answer. It's always cool to come up with a solution to a problem that requires a minimum amount of vocabulary (or in this case, "math tricks"). This is much more akin to the first solution you would come up with as you're learning this material. $\endgroup$ – NoseKnowsAll Nov 10 '16 at 9:46
3
$\begingroup$

$p(0)\neq 0$ implies that $0$ is not a root of characteristic polynomial $p(x)$ which is turn says that $0$ is not an eigenvalue of $A$. As $det(A)=$ product of eigenvalues gives $det(A)\neq 0$ which suggests that $A$ is invertible.

$\endgroup$
2
$\begingroup$

Well,

if $p(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ is the characteristic polynomial then it is known from theory that $a_0$ represents the determinant of the matrix. Since $p(0)=a_0 \neq 0$ this means that $\det A \neq 0$ . Thus $A$ is invertible.

$\endgroup$
  • $\begingroup$ I haven't learned determinants yet, but it looks easier to understand by doing it that way. $\endgroup$ – E.Bob Nov 9 '16 at 20:51
  • 6
    $\begingroup$ $p(x)$ is not necessarily the characteristic polynomial, in this exercise. $\endgroup$ – Federico Poloni Nov 9 '16 at 23:16
1
$\begingroup$

All eigenvalues of A must be roots of P.

(See the eigenvector X and P(A)X)

=> all eigennumbers of A is not zero => |A| is not zero => A is invertible

$\endgroup$
  • $\begingroup$ I haven't learned eigennumbers and eigenvectors yet, but thank's for future answer. $\endgroup$ – E.Bob Nov 9 '16 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.