Question

Show that if p(x) is a polynomial with a nonzero constant term, and if A is a square matrix for which p(A) = 0, then A is invertible.

Answer 1

The polynomial is
\[p(x)= a _{n}x ^{n}+a _{n-1}x ^{n-1}+.....+a_{1}x+a _{0}\]

Then,

\[p(A)= a _{n}A ^{n}+a _{n-1}A ^{n-1}+......a _{1}A+a _{0}I=0\]

so,

\[a _{n}A ^{n}+a _{n-1}A ^{n-1}+........a _{1}A=-a _{0}I\]

\[A(a _{n}A ^{n-1}+ a _{n-1}A ^{n-2}+....a _{1}I)=-a _{0}I\]

If A is invertible

\[AA ^{-1}=I\]

Hence, you can calculate \[A ^{-1}\]

Answer 2

@Narad
okay that makes a it easier to see. Thanks

Answer 3

You are welcome

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Narad
The polynomial is
\[p(x)= a _{n}x ^{n}+a _{n-1}x ^{n-1}+.....+a_{1}x+a _{0} = 0\]

Then,

\[p(A)= a _{n}A ^{n}+a _{n-1}A ^{n-1}+......a _{1}A+a _{0}I=0\]

\[a _{n}A ^{n}+a _{n-1}A ^{n-1}+........a _{1}A=-a _{0}I\]

\[\frac{a _{n}}{-a_{0}}A ^{n}+ \frac{a _{n-1}}{-a_{0}}A ^{n-1}+....\frac{a _{1}}{-a_{0}}A=I\]

\[A(\frac{a _{n}}{-a_{0}}A ^{n-1}+ \frac{a _{n-2}}{-a_{0}}A ^{n-1}+....\frac{a _{1}}{-a_{0}}I)=I\]

If A is invertible

\[AA ^{-1}=I\]

Hence, you can calculate \[A ^{-1}\]
\(\color{#0cbb34}{\text{End of Quote}}\)