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Thread: Linear algebra

  1. #1
    Junior Member
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    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
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  2. #2
    MHF Contributor

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    Quote Originally Posted by omert View Post
    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!
    Last edited by NonCommAlg; Jan 19th 2009 at 11:37 PM. Reason: typo again!
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  3. #3
    Junior Member
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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): $\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}
    $
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