Find a matrix P such that $P^{-1}AP=J$, where $J$ is in Jordan normal form.












0














$A=begin{pmatrix} -6 & 0 & 1 & -5 \ 5 & - 1 & -1 & 5 \ 0 & 0 & -1 & 0 \ 5 & 0 & -1 & 4
end{pmatrix}$



I found the Jordan normal form to be



$J=begin{pmatrix} -1 & 1 &0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1
end{pmatrix}$.



I know next that I have to solve $(A+1)v_2=v_1$ to find $v_1= begin{pmatrix} 1 \ -1\0\-1 end{pmatrix}$ and $v_2=begin{pmatrix} 0 \ 0 \ 1 \ 0 end{pmatrix}$, where you can then go on to find two more vectors, but I dont understand why this proccess works, can someone explain? Thank you!






linear-algebra matrices linear-transformations







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  • Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
    – qbert
    Dec 1 at 17:37
















0














$A=begin{pmatrix} -6 & 0 & 1 & -5 \ 5 & - 1 & -1 & 5 \ 0 & 0 & -1 & 0 \ 5 & 0 & -1 & 4
end{pmatrix}$



I found the Jordan normal form to be



$J=begin{pmatrix} -1 & 1 &0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1
end{pmatrix}$.



I know next that I have to solve $(A+1)v_2=v_1$ to find $v_1= begin{pmatrix} 1 \ -1\0\-1 end{pmatrix}$ and $v_2=begin{pmatrix} 0 \ 0 \ 1 \ 0 end{pmatrix}$, where you can then go on to find two more vectors, but I dont understand why this proccess works, can someone explain? Thank you!






linear-algebra matrices linear-transformations







share|cite|improve this question
























  • Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
    – qbert
    Dec 1 at 17:37














0












0








0







$A=begin{pmatrix} -6 & 0 & 1 & -5 \ 5 & - 1 & -1 & 5 \ 0 & 0 & -1 & 0 \ 5 & 0 & -1 & 4
end{pmatrix}$



I found the Jordan normal form to be



$J=begin{pmatrix} -1 & 1 &0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1
end{pmatrix}$.



I know next that I have to solve $(A+1)v_2=v_1$ to find $v_1= begin{pmatrix} 1 \ -1\0\-1 end{pmatrix}$ and $v_2=begin{pmatrix} 0 \ 0 \ 1 \ 0 end{pmatrix}$, where you can then go on to find two more vectors, but I dont understand why this proccess works, can someone explain? Thank you!






linear-algebra matrices linear-transformations







share|cite|improve this question















$A=begin{pmatrix} -6 & 0 & 1 & -5 \ 5 & - 1 & -1 & 5 \ 0 & 0 & -1 & 0 \ 5 & 0 & -1 & 4
end{pmatrix}$



I found the Jordan normal form to be



$J=begin{pmatrix} -1 & 1 &0 & 0 \ 0 & -1 & 0 & 0 \ 0 & 0 & -1 & 0 \ 0 & 0 & 0 & -1
end{pmatrix}$.



I know next that I have to solve $(A+1)v_2=v_1$ to find $v_1= begin{pmatrix} 1 \ -1\0\-1 end{pmatrix}$ and $v_2=begin{pmatrix} 0 \ 0 \ 1 \ 0 end{pmatrix}$, where you can then go on to find two more vectors, but I dont understand why this proccess works, can someone explain? Thank you!






linear-algebra matrices linear-transformations




linear-algebra matrices linear-transformations






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edited Dec 1 at 17:28









greedoid

37.6k114794




37.6k114794










asked Dec 1 at 17:25









Brad Scott

1528




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  • Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
    – qbert
    Dec 1 at 17:37


















  • Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
    – qbert
    Dec 1 at 17:37
















Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
– qbert
Dec 1 at 17:37




Depending on the proof of the Jordan canonical form you know, this falls right out. I recommend Artin's algebra
– qbert
Dec 1 at 17:37










1 Answer
1






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oldest

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1














Let's follow the algorithm described here.



It seems that you have found that the characteristic polynomial of $A$ is
$$
chi_A(t)=(t+1)^4
$$
so the only eigenvalue of $A$ is $lambda=-1$ with algebraic multiplicity $m=4$. Note that
begin{align*}
A+I &= left[begin{array}{rrrr}
-5 & 0 & 1 & -5 \
5 & 0 & -1 & 5 \
0 & 0 & 0 & 0 \
5 & 0 & -1 & 5
end{array}right]
&
(A+I)^2
&=
left[begin{array}{rrrr}
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{array}right]
end{align*}
so
begin{align*}
dimoperatorname{Null}(A+I) &= 3 & dimoperatorname{Null}((A+I)^2) &= 4
end{align*}
we then compute the numbers
begin{align*}
d_1 &= dimoperatorname{Null}(A+I) & d_2 &= dimoperatorname{Null}((A+I)^2)-dimoperatorname{Null}(A+I) \
&= 3 & &= 4-3 \
& & &=1
end{align*}
so we must fill the boxes
$$
begin{array}{ccc}
Box & Box & Box \
Box
end{array}
$$
with vectors. Note that $e_1=langle1, 0, 0, 0rangle$ satisfies $e_1inoperatorname{Null}((A+I)^2)$ but $e_1notinoperatorname{Null}(A+I)$. Put
$$
v=(A+I)e_1=leftlangle-5,,5,,0,,5rightrangle
$$
so the diagram takes the form
$$
begin{array}{ccc}
v & Box & Box \
e_1
end{array}
$$
Since $w=leftlangle1,,0,,0,,-1rightrangle$ and $x=leftlangle0,,0,,5,,1rightrangle$ are two linearly independent vectors in $operatorname{Null}(A+I)$ that yield a basis ${v, w, x}$ of $operatorname{Null}(A+I)$ we may complete our diagram with
$$
begin{array}{ccc}
v & w & x \
e_1
end{array}
$$
This gives $A=PJP^{-1}$ where
$$
P = left[begin{array}{rrrr}
-5 & 1 & 1 & 0 \
5 & 0 & 0 & 0 \
0 & 0 & 0 & 5 \
5 & 0 & -1 & 1
end{array}right]
$$






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    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    1














    Let's follow the algorithm described here.



    It seems that you have found that the characteristic polynomial of $A$ is
    $$
    chi_A(t)=(t+1)^4
    $$
    so the only eigenvalue of $A$ is $lambda=-1$ with algebraic multiplicity $m=4$. Note that
    begin{align*}
    A+I &= left[begin{array}{rrrr}
    -5 & 0 & 1 & -5 \
    5 & 0 & -1 & 5 \
    0 & 0 & 0 & 0 \
    5 & 0 & -1 & 5
    end{array}right]
    &
    (A+I)^2
    &=
    left[begin{array}{rrrr}
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0
    end{array}right]
    end{align*}
    so
    begin{align*}
    dimoperatorname{Null}(A+I) &= 3 & dimoperatorname{Null}((A+I)^2) &= 4
    end{align*}
    we then compute the numbers
    begin{align*}
    d_1 &= dimoperatorname{Null}(A+I) & d_2 &= dimoperatorname{Null}((A+I)^2)-dimoperatorname{Null}(A+I) \
    &= 3 & &= 4-3 \
    & & &=1
    end{align*}
    so we must fill the boxes
    $$
    begin{array}{ccc}
    Box & Box & Box \
    Box
    end{array}
    $$
    with vectors. Note that $e_1=langle1, 0, 0, 0rangle$ satisfies $e_1inoperatorname{Null}((A+I)^2)$ but $e_1notinoperatorname{Null}(A+I)$. Put
    $$
    v=(A+I)e_1=leftlangle-5,,5,,0,,5rightrangle
    $$
    so the diagram takes the form
    $$
    begin{array}{ccc}
    v & Box & Box \
    e_1
    end{array}
    $$
    Since $w=leftlangle1,,0,,0,,-1rightrangle$ and $x=leftlangle0,,0,,5,,1rightrangle$ are two linearly independent vectors in $operatorname{Null}(A+I)$ that yield a basis ${v, w, x}$ of $operatorname{Null}(A+I)$ we may complete our diagram with
    $$
    begin{array}{ccc}
    v & w & x \
    e_1
    end{array}
    $$
    This gives $A=PJP^{-1}$ where
    $$
    P = left[begin{array}{rrrr}
    -5 & 1 & 1 & 0 \
    5 & 0 & 0 & 0 \
    0 & 0 & 0 & 5 \
    5 & 0 & -1 & 1
    end{array}right]
    $$






    share|cite|improve this answer




























      1














      Let's follow the algorithm described here.



      It seems that you have found that the characteristic polynomial of $A$ is
      $$
      chi_A(t)=(t+1)^4
      $$
      so the only eigenvalue of $A$ is $lambda=-1$ with algebraic multiplicity $m=4$. Note that
      begin{align*}
      A+I &= left[begin{array}{rrrr}
      -5 & 0 & 1 & -5 \
      5 & 0 & -1 & 5 \
      0 & 0 & 0 & 0 \
      5 & 0 & -1 & 5
      end{array}right]
      &
      (A+I)^2
      &=
      left[begin{array}{rrrr}
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0 \
      0 & 0 & 0 & 0
      end{array}right]
      end{align*}
      so
      begin{align*}
      dimoperatorname{Null}(A+I) &= 3 & dimoperatorname{Null}((A+I)^2) &= 4
      end{align*}
      we then compute the numbers
      begin{align*}
      d_1 &= dimoperatorname{Null}(A+I) & d_2 &= dimoperatorname{Null}((A+I)^2)-dimoperatorname{Null}(A+I) \
      &= 3 & &= 4-3 \
      & & &=1
      end{align*}
      so we must fill the boxes
      $$
      begin{array}{ccc}
      Box & Box & Box \
      Box
      end{array}
      $$
      with vectors. Note that $e_1=langle1, 0, 0, 0rangle$ satisfies $e_1inoperatorname{Null}((A+I)^2)$ but $e_1notinoperatorname{Null}(A+I)$. Put
      $$
      v=(A+I)e_1=leftlangle-5,,5,,0,,5rightrangle
      $$
      so the diagram takes the form
      $$
      begin{array}{ccc}
      v & Box & Box \
      e_1
      end{array}
      $$
      Since $w=leftlangle1,,0,,0,,-1rightrangle$ and $x=leftlangle0,,0,,5,,1rightrangle$ are two linearly independent vectors in $operatorname{Null}(A+I)$ that yield a basis ${v, w, x}$ of $operatorname{Null}(A+I)$ we may complete our diagram with
      $$
      begin{array}{ccc}
      v & w & x \
      e_1
      end{array}
      $$
      This gives $A=PJP^{-1}$ where
      $$
      P = left[begin{array}{rrrr}
      -5 & 1 & 1 & 0 \
      5 & 0 & 0 & 0 \
      0 & 0 & 0 & 5 \
      5 & 0 & -1 & 1
      end{array}right]
      $$






      share|cite|improve this answer


























        1












        1








        1






        Let's follow the algorithm described here.



        It seems that you have found that the characteristic polynomial of $A$ is
        $$
        chi_A(t)=(t+1)^4
        $$
        so the only eigenvalue of $A$ is $lambda=-1$ with algebraic multiplicity $m=4$. Note that
        begin{align*}
        A+I &= left[begin{array}{rrrr}
        -5 & 0 & 1 & -5 \
        5 & 0 & -1 & 5 \
        0 & 0 & 0 & 0 \
        5 & 0 & -1 & 5
        end{array}right]
        &
        (A+I)^2
        &=
        left[begin{array}{rrrr}
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0
        end{array}right]
        end{align*}
        so
        begin{align*}
        dimoperatorname{Null}(A+I) &= 3 & dimoperatorname{Null}((A+I)^2) &= 4
        end{align*}
        we then compute the numbers
        begin{align*}
        d_1 &= dimoperatorname{Null}(A+I) & d_2 &= dimoperatorname{Null}((A+I)^2)-dimoperatorname{Null}(A+I) \
        &= 3 & &= 4-3 \
        & & &=1
        end{align*}
        so we must fill the boxes
        $$
        begin{array}{ccc}
        Box & Box & Box \
        Box
        end{array}
        $$
        with vectors. Note that $e_1=langle1, 0, 0, 0rangle$ satisfies $e_1inoperatorname{Null}((A+I)^2)$ but $e_1notinoperatorname{Null}(A+I)$. Put
        $$
        v=(A+I)e_1=leftlangle-5,,5,,0,,5rightrangle
        $$
        so the diagram takes the form
        $$
        begin{array}{ccc}
        v & Box & Box \
        e_1
        end{array}
        $$
        Since $w=leftlangle1,,0,,0,,-1rightrangle$ and $x=leftlangle0,,0,,5,,1rightrangle$ are two linearly independent vectors in $operatorname{Null}(A+I)$ that yield a basis ${v, w, x}$ of $operatorname{Null}(A+I)$ we may complete our diagram with
        $$
        begin{array}{ccc}
        v & w & x \
        e_1
        end{array}
        $$
        This gives $A=PJP^{-1}$ where
        $$
        P = left[begin{array}{rrrr}
        -5 & 1 & 1 & 0 \
        5 & 0 & 0 & 0 \
        0 & 0 & 0 & 5 \
        5 & 0 & -1 & 1
        end{array}right]
        $$






        share|cite|improve this answer














        Let's follow the algorithm described here.



        It seems that you have found that the characteristic polynomial of $A$ is
        $$
        chi_A(t)=(t+1)^4
        $$
        so the only eigenvalue of $A$ is $lambda=-1$ with algebraic multiplicity $m=4$. Note that
        begin{align*}
        A+I &= left[begin{array}{rrrr}
        -5 & 0 & 1 & -5 \
        5 & 0 & -1 & 5 \
        0 & 0 & 0 & 0 \
        5 & 0 & -1 & 5
        end{array}right]
        &
        (A+I)^2
        &=
        left[begin{array}{rrrr}
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0 \
        0 & 0 & 0 & 0
        end{array}right]
        end{align*}
        so
        begin{align*}
        dimoperatorname{Null}(A+I) &= 3 & dimoperatorname{Null}((A+I)^2) &= 4
        end{align*}
        we then compute the numbers
        begin{align*}
        d_1 &= dimoperatorname{Null}(A+I) & d_2 &= dimoperatorname{Null}((A+I)^2)-dimoperatorname{Null}(A+I) \
        &= 3 & &= 4-3 \
        & & &=1
        end{align*}
        so we must fill the boxes
        $$
        begin{array}{ccc}
        Box & Box & Box \
        Box
        end{array}
        $$
        with vectors. Note that $e_1=langle1, 0, 0, 0rangle$ satisfies $e_1inoperatorname{Null}((A+I)^2)$ but $e_1notinoperatorname{Null}(A+I)$. Put
        $$
        v=(A+I)e_1=leftlangle-5,,5,,0,,5rightrangle
        $$
        so the diagram takes the form
        $$
        begin{array}{ccc}
        v & Box & Box \
        e_1
        end{array}
        $$
        Since $w=leftlangle1,,0,,0,,-1rightrangle$ and $x=leftlangle0,,0,,5,,1rightrangle$ are two linearly independent vectors in $operatorname{Null}(A+I)$ that yield a basis ${v, w, x}$ of $operatorname{Null}(A+I)$ we may complete our diagram with
        $$
        begin{array}{ccc}
        v & w & x \
        e_1
        end{array}
        $$
        This gives $A=PJP^{-1}$ where
        $$
        P = left[begin{array}{rrrr}
        -5 & 1 & 1 & 0 \
        5 & 0 & 0 & 0 \
        0 & 0 & 0 & 5 \
        5 & 0 & -1 & 1
        end{array}right]
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 at 17:57

























        answered Dec 1 at 17:37









        Brian Fitzpatrick

        21k42958




        21k42958






























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