Question

An n × n matrix A is called nilpotent if_{${\mathbf{A}}^{\mathbf{m}}\mathbf{=}\mathbf{0}$}for some positive integer m. Examples are triangular matrices whose entries on the diagonal are all 0. Consider a nilpotent n × n matrix A, and choose the smallest number ‘m’ such that ${\mathbf{A}}^{\mathbf{m}}\mathbf{=}\mathbf{0}$. Pick a vector $\overrightarrow{\mathbf{v}}$in ${\mathbf{ℝ}}^{\mathbf{n}}$such that ${\mathit{A}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\overrightarrow{\mathbf{v}}\mathbf{\ne }\mathbf{0}$. Show that the vectors$\overrightarrow{\mathbf{v}}\mathbf{,}\mathit{A}\overrightarrow{\mathbf{v}}\mathbf{,}{\mathit{A}}^{\mathbf{2}}\overrightarrow{\mathbf{v}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{A}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\overrightarrow{\mathbf{v}}$are linearly independent.

Hint: Consider a relation ${\mathit{c}}_{\mathbf{0}}\overrightarrow{\mathbf{v}}\mathbf{+}{\mathit{c}}_{\mathbf{1}}\mathit{A}\overrightarrow{\mathbf{v}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{A}}^{\mathbf{2}}\overrightarrow{\mathbf{v}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{c}}_{\mathbf{m}\mathbf{-}\mathbf{1}}{\mathit{A}}^{\mathbf{m}\mathbf{-}\mathbf{1}}\overrightarrow{\mathbf{v}}\mathbf{=}\mathbf{0}$. Multiply both sides of the equation with ${\mathbf{A}}^{\mathbf{m}\mathbf{-}\mathbf{1}}$to show ${\mathbf{c}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$. Next, show that${\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$,and so on.

Short answer

If we take a nilpotent n × n matrix A and choose the smallest number ‘m’ such that $A^m=0$and pick a vector $\overrightarrow{v}$in ${\mathrm{ℝ}}^n$such that $A^{m-1}\overrightarrow{V}\ne 0$then the vectors $\overrightarrow{v},A\overrightarrow{v},A^2\overrightarrow{v},...,A^{m-1}\overrightarrow{v}$are linearly independent.

Step-by-step solution

 Definition of linearly independent vectors

Let V be avector space over a field K. Let$v_1,v_2,v_3,...,v_n$ 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

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

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_{$a_1,a_2,a_3,...,a_m$}, not all of them 0, suchthat

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

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$\overrightarrow{v}$are$A\overrightarrow{v}$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,

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

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

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$\overrightarrow{v}$ are$A\overrightarrow{v}$ linearly independent.

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 $\overrightarrow{v}$and $A\overrightarrow{v}$are linearly independent.