1. ## Linear algebra

1. let $\displaystyle T:V \to V$ and $\displaystyle S:V \to V$ and that $\displaystyle B = (u_1 ,u_2 , \cdots ,u_n )$ is a Basis for V.

it is known that $\displaystyle \begin{array}{l} T(u_i ) = u_{i + 1} \to 1 \le i \le n - 1 \\ T(u_n ) = 0 \\ \end{array}$ and also that $\displaystyle S^n = 0,S^{n - 1} \ne 0$
prove that there is a another basis to V, C that sustains $\displaystyle [S]_{C } = [T]_B$

2. prove that if: M,N are two nxn matrix so: $\displaystyle \begin{array}{l} N^n = 0,N^{n - 1} \ne 0 \\ M^n = 0,M^{n - 1} \ne 0 \\ \end{array}$
then M is similar to N

thanks very much

2. Originally Posted by omert
1. let $\displaystyle T:V \to V$ and $\displaystyle S:V \to V$ and that $\displaystyle B = (u_1 ,u_2 , \cdots ,u_n )$ is a Basis for V.

it is known that $\displaystyle \begin{array}{l} T(u_i ) = u_{i + 1} \to 1 \le i \le n - 1 \\ T(u_n ) = 0 \\ \end{array}$ and also that $\displaystyle S^n = 0,S^{n - 1} \ne 0$
prove that there is a another basis to V, C that sustains $\displaystyle [S]_{C } = [T]_B$

2. prove that if: M,N are two nxn matrix so: $\displaystyle \begin{array}{l} N^n = 0,N^{n - 1} \ne 0 \\ M^n = 0,M^{n - 1} \ne 0 \\ \end{array}$
then M is similar to N

thanks very much
the point which will easily solve both parts of your problem is this: suppose that $\displaystyle \dim V = n$ and $\displaystyle S:V \longrightarrow V$ is a linear transformation such that $\displaystyle S^n=0, \ S^{n-1} \neq 0.$ let $\displaystyle v \in V$ be such that

$\displaystyle S^{n-1}(v) \neq 0.$ then the set $\displaystyle C=\{v,S(v), S^2(v), \cdots , S^{n-1}(v) \}$ is a basis for $\displaystyle V.$ to prove this, we only need to show that the elements of $\displaystyle C$ are linearly independent. so suppose $\displaystyle \sum_{j=0}^{n-1}a_jS^j(v)=0,$

where $\displaystyle a_j$ are some scalars. if $\displaystyle a_j=0$ for all $\displaystyle j,$ we're done. so suppose that $\displaystyle \exists j: \ a_j \neq 0.$ let $\displaystyle A=\{j: \ a_j \neq 0 \}$ and $\displaystyle k=\min A.$ then we'll have $\displaystyle \sum_{j=k}^{n-1}a_j S^j(v)=\sum_{j=0}^{n-1}a_jS^j(v)=0, \ a_k \neq 0.$ hence:

$\displaystyle a_kS^{n-1}(v)=S^{n-1-k} \left(\sum_{j=k}^{n-1}a_j S^j(v) \right)=0,$ which gives us $\displaystyle a_k = 0,$ because $\displaystyle S^{n-1}(v) \neq 0.$ contradiction!

3. if you can please clarify this stage to me:
then we'll have hence:
and I didn't understood how to solve 2

4. By the linearity ( it's a Linear Transformation): $\displaystyle S^{n - 1 - k} \left( {\sum\limits_{j = k}^{n - 1} {a_j \cdot S^j \left( v \right)} } \right) = \sum\limits_{j = k}^{n - 1} {a_j \cdot S^{n - 1 - k} \left( {S^j \left( v \right)} \right)} = \sum\limits_{j = k}^{n - 1} {a_j \cdot S^{n - 1 + j - k} \left( v \right)}$

And now check for $\displaystyle j=k,k+1,...,n$ : $\displaystyle \begin{gathered} S^{n - 1 + k - k} \left( v \right) = S^{n - 1} \left( v \right) \hfill \\ S^{n - 1 + \left( {k + 1} \right) - k} \left( v \right) = S^n \left( v \right) = \bold 0 \hfill \\ S^{n - 1 + \left( {k + 2} \right) - k} \left( v \right) = S^{n + 1} \left( v \right) = S\left( {S^n \left( v \right)} \right) = S\left( \bold 0 \right) = \bold 0 \hfill \\ \vdots \hfill \\ \end{gathered}$

To prove 2, I will assume you are working on R because you do not mention what field you are working with, it's totally analogous though for the other cases. Take $\displaystyle \bold{v}\in \mathbb{R}^n$ (written as a column, or $\displaystyle \bold{v} \in M_{n \times 1} \left( \mathbb{R} \right)$ if you prefer) such that $\displaystyle N^{n-1}\cdot \bold v \ne \bold 0$ and $\displaystyle \bold u \in \mathbb{R}^n$ such that $\displaystyle M^{n-1}\cdot \bold u \ne \bold 0$. These exist for $\displaystyle M^{n - 1} \ne \bold {0}_{n \times n}$ and $\displaystyle N^{n - 1} \ne \bold {0}_{n \times n}$

Consider the transformations $\displaystyle T:\mathbb{R}^n \to \mathbb{R}^n$ and $\displaystyle S:\mathbb{R}^n \to \mathbb{R}^n$ defined by : $\displaystyle T\left( \bold x \right) = N \cdot \bold x$ and $\displaystyle S\left( \bold x \right) = M \cdot \bold x$

Take the basis of $\displaystyle \mathbb{R}^n$ given by $\displaystyle C= \left\{ {\bold v,T( \bold v),...,T^{n - 1}( \bold v)} \right\}$ (the fact that it's a basis is justified by NonCommAlg's post)

Now note that: $\displaystyle \left[ T \right]_C = \left( {\begin{array}{*{20}c} 0 & 0 & \cdots & \cdots & 0 & 0 \\ 1 & 0 & {} & {} & 0 & \vdots \\ 0 & 1 & {} & {} & \vdots & 0 \\ 0 & 0 & {} & {} & \vdots & 0 \\ \vdots & \vdots & {} & {} & 0 & \vdots \\ 0 & 0 & \cdots & \cdots & 1 & 0 \\ \end{array} } \right)$

Similarly with $\displaystyle Q= \left\{ {\bold u,S(\bold u),...,S^{n - 1} (\bold u)} \right\}$

We have: $\displaystyle \left[ S \right]_Q = \left( {\begin{array}{*{20}c} 0 & 0 & \cdots & \cdots & 0 & 0 \\ 1 & 0 & {} & {} & 0 & \vdots \\ 0 & 1 & {} & {} & \vdots & 0 \\ 0 & 0 & {} & {} & \vdots & 0 \\ \vdots & \vdots & {} & {} & 0 & \vdots \\ 0 & 0 & \cdots & \cdots & 1 & 0 \\ \end{array} } \right)$

Hence: $\displaystyle \left[ S \right]_Q = \left[T \right]_C$

Now let $\displaystyle D$ be the standard basis for $\displaystyle \mathbb{R}^n$, that is: $\displaystyle D = \left\{ {\left( {1,0,0,...,0} \right);\left( {0,1,0,...,0} \right);...;\left( {0,0,...,0,1} \right)} \right\}$

By the change of basis theorem: $\displaystyle \left[ S \right]_Q = A \cdot \left[ S \right]_D \cdot A^{ - 1} = A \cdot M \cdot A^{ - 1}$ and: $\displaystyle \left[ T \right]_C = B^{ - 1} \cdot \left[T \right]_C \cdot B= B^{ - 1} \cdot N \cdot B$ simply because: $\displaystyle \left[ T \right]_D = N;\left[ S \right]_D = M$

$\displaystyle B^{ - 1} \cdot N \cdot B = A \cdot M \cdot A^{ - 1}$ then: $\displaystyle N = BA \cdot M \cdot A^{ - 1} B^{ - 1}$ thus: $\displaystyle N = \left( {BA} \right) \cdot M \cdot \left( {BA} \right)^{ - 1}$