Let $A$ be a square matrix such that $A_10=0$ while $A_9≠0$. Let $V$ be the space of all polynomials of $A$. Find the dimension of $V$.

Question

Let $A$ be a square matrix such that $A^{10}=0$ while $A^9≠0$. Let $V$ be the space of all polynomials of $A$. Find the dimension of $V$.

The part that is really throwing me off is that $A_9≠0$. This is throwing me off to find the dimension of $V$.

Answer 1

If $A^{10} =0$ then any polynomial in A is a linear combination of $I,A,A^{2},... , A^{9}$ (because, $A^{10},A^{11},A^{12},..$ are all 0. Now let us show that $I,A,A^{2},...,A^{9}$ are linearly independent. Suppose $\sum_0 ^{9} a_i A^{i} =0$. Multiply both sides by $A^{9}$ to see that $a_1 =0$. Then multiply by $A^{8}$ to get $a_2 =0$ etc. In 10 steps you get $a_i =0$ for $0 \leq i \leq 9$. Hence $I,A,A^{2},...,A^{9}$ form a basis for V and the dimension is 10.