1. ## Linear algebra

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

it is known that $
\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 $
S^n = 0,S^{n - 1} \ne 0
$

prove that there is a another basis to V, C that sustains $[S]_{C } = [T]_B
$

2. prove that if: M,N are two nxn matrix so: $
\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 $T:V \to V$ and $S:V \to V$ and that $
B = (u_1 ,u_2 , \cdots ,u_n )
$
is a Basis for V.

it is known that $
\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 $
S^n = 0,S^{n - 1} \ne 0
$

prove that there is a another basis to V, C that sustains $[S]_{C } = [T]_B
$

2. prove that if: M,N are two nxn matrix so: $
\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 $\dim V = n$ and $S:V \longrightarrow V$ is a linear transformation such that $S^n=0, \ S^{n-1} \neq 0.$ let $v \in V$ be such that

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

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

$a_kS^{n-1}(v)=S^{n-1-k} \left(\sum_{j=k}^{n-1}a_j S^j(v) \right)=0,$ which gives us $a_k = 0,$ because $S^{n-1}(v) \neq 0.$ 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): $
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 $j=k,k+1,...,n$ : $
\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 $\bold{v}\in \mathbb{R}^n$ (written as a column, or $\bold{v} \in
M_{n \times 1} \left( \mathbb{R} \right)
$
if you prefer) such that $N^{n-1}\cdot \bold v \ne \bold 0$ and $\bold u \in \mathbb{R}^n$ such that $M^{n-1}\cdot \bold u \ne \bold 0$. These exist for $
M^{n - 1} \ne \bold {0}_{n \times n}
$
and $
N^{n - 1} \ne \bold {0}_{n \times n}
$

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

Take the basis of $
\mathbb{R}^n
$
given by $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: $
\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 $Q=
\left\{ {\bold u,S(\bold u),...,S^{n - 1} (\bold u)} \right\}
$

We have: $
\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: $
\left[ S \right]_Q = \left[T \right]_C
$

Now let $D$ be the standard basis for $
\mathbb{R}^n
$
, that is: $
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: $
\left[ S \right]_Q = A \cdot \left[ S \right]_D \cdot A^{ - 1} = A \cdot M \cdot A^{ - 1}
$
and: $
\left[ T \right]_C = B^{ - 1} \cdot \left[T \right]_C \cdot B= B^{ - 1} \cdot N \cdot B
$
simply because: $
\left[ T \right]_D = N;\left[ S \right]_D = M
$

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