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Math Help - Linear algebra

  1. #1
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    Linear algebra

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

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

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

    thanks very much
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  2. #2
    MHF Contributor

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    Quote Originally Posted by omert View Post
    1. let T:V \to V and S:V \to V and that <br />
B = (u_1 ,u_2 , \cdots ,u_n )<br />
is a Basis for V.

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

    2. prove that if: M,N are two nxn matrix so: <br />
\begin{array}{l}<br />
N^n = 0,N^{n - 1} \ne 0 \\ <br />
M^n = 0,M^{n - 1} \ne 0 \\ <br />
\end{array}<br />
    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!
    Last edited by NonCommAlg; January 19th 2009 at 11:37 PM. Reason: typo again!
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  3. #3
    Junior Member
    Joined
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    if you can please clarify this stage to me:
    then we'll have hence:
    and I didn't understood how to solve 2
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  4. #4
    Super Member PaulRS's Avatar
    Joined
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    By the linearity ( it's a Linear Transformation): <br />
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)} <br />

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

    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 <br />
M_{n \times 1} \left( \mathbb{R} \right)<br />
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 <br />
M^{n - 1}  \ne \bold {0}_{n \times n} <br />
and <br />
N^{n - 1}  \ne \bold {0}_{n \times n} <br />

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

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

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

    Similarly with Q=<br />
\left\{ {\bold u,S(\bold u),...,S^{n - 1} (\bold u)} \right\}<br />

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

    Hence: <br />
\left[ S \right]_Q  = \left[T \right]_C <br />

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

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

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