Question

Consider a nilpotent n × n matrix A. Use the result demonstrated in exercise 78 to show that${\mathbf{A}}^{\mathbf{n}}\mathbf{=}\mathbf{0}$.

Short answer

If we take a nilpotent n × n matrix A, then by definition of a nilpotent matrix we get

_{${\mathrm{A}}^{\mathrm{n}}=0$}

Step-by-step solution

 Definition of linearly independent vectors

Let V be a vector space over a field K. Let $v_1,v_2,v_3,...,v_n\in V$if there exist scalars $a_1,a_2,a_3,...,a_n$K, not all of them 0, such that

$a_1{v}_1+a_2{v}_{+}...a_n{v}_n=0$

Then the vectors$v_1,v_2,v_3,...,v_n$ are called linearly dependent vectors.

Otherwise, linearly independent vectors.

Definition of a nilpotent matrix

A matrix A of order n x n is said to be nilpotent if there exists a smallest positive integer ‘m’ for which

$A^m=0but{A}^{m-1}\ne 0$

Then the smallest positive integer ‘m’ is called an index of the nilpotent matrix.

 Considering a nilpotent matrix

Let us consider a nilpotent matrix A of order 3 x 3 such that

$A^2=0butA\ne 0$

Then we have a nilpotent matrix A of index 2 and so ‘m’ = 2.

 To show the vectors v→,Av→,A2v→,...,Am-1v→ are linearly independent.

Considering the vectors $\overrightarrow{v},A\overrightarrow{v},A^2\overrightarrow{v},...,A^{m-1}\overrightarrow{v}$are linearly dependent then, there exist scalars $a_1,a_2,a_3,\dots ,a_m$, not all of them 0, such that

$a_1\overrightarrow{v}+a_2\overrightarrow{v}⃗+a_3{A}^2\overrightarrow{v}+...+a_m{A}^{m-1}\overrightarrow{v}=0$

Since we have considered a nilpotent matrix A of order 3 x 3 such that

$A^2=0butA\ne 0$

where ‘m’ = 2, then we have to show that the vectors are linearly independent.

Let us suppose the vectors $\overrightarrow{v}$and $A\overrightarrow{v}$are linearly dependent, then there exist scalars a and b, not all of them 0, such that

$a\overrightarrow{v}+{}_{b}A\overrightarrow{v}=0$ (1)

Multiply equation (1) by A we get,

$aA\overrightarrow{v}+b{A}^2\overrightarrow{v}=0$ (2)

$$\begin{gathered} \because A^2=0 \\ \Rightarrow b{A}^2\overrightarrow{v}=0(\because A\overrightarrow{v}\ne 0) \end{gathered}$$

Thus from equation (2), we have

$$\begin{gathered} aA\overrightarrow{v}=0 \\ \Rightarrow a=0 \end{gathered}$$

Therefore from equation (1), we have

$bA\overrightarrow{v}=0$

$\Rightarrow b=0(\because A\overrightarrow{v}\ne 0)$

Thus, for $a\overrightarrow{v}+{}_{b}A\overrightarrow{v}=0$we get $a=0andb=0$.

Hence, our supposition that the vectors$\overrightarrow{v}$and$A\overrightarrow{v}$ are linearly dependent is wrong and so the vectors$\overrightarrow{\mathrm{v}}\mathrm{and}\mathrm{A}\overrightarrow{\mathrm{v}}$ are linearly independent.

Since it is true for a 3 x 3 matrix; hence it is true for n x n matrix A such that

$A^m=0but{A}^{m-1}\ne 0$

Thus the vectors $\overrightarrow{v},A\overrightarrow{v},A^2\overrightarrow{v},...,A^{m-1}\overrightarrow{v}$are linearly independent.

 To show An=0

Let $m\le n$. Then

$$\begin{gathered} A^n=A^n{A}^m{A}^{-m}=A^{n-m}{A}^m=0(\because A^m=0) \\ \Rightarrow A^n=0 \end{gathered}$$

Now, let $m>n$.

This implies the number of linearly independent vectors is greater than ‘m’ which is a contradiction as the number of linearly independent vectors in an n x n matrix is m.

Thus,$m\le n$

 Final Answer

If we take a nilpotent 3 × 3 matrix A and choose the smallest number ‘m’ = 2 such that $A^2=0butA\ne 0$and pick a vector $\overrightarrow{v}$in ${\mathrm{ℝ}}^n$such that $A\overrightarrow{v}\ne 0$then the vectors$\overrightarrow{v}$are$A\overrightarrow{v}$ linearly independent.

Since it is true for the 3 x 3 matrix; hence it is true for n x n matrix A such that

$A^m=0but{A}^{m-1}\ne 0$

Thus the vectors$\overrightarrow{v},A\overrightarrow{v},A^2\overrightarrow{v},...,A^{m-1}\overrightarrow{v}$ are linearly independent.

If$m\le n\Rightarrow A^n=0$