Find a characteristic polynomial of endomorphism $varphi$The characteristic and minimal polynomial of a...

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Find a characteristic polynomial of endomorphism $varphi$


The characteristic and minimal polynomial of a companion matrixFind characteristic polynomial of $A+I$ if is knowing characteristic polynomial of $A$Proving minimum polynomial equals characteristic polynomial in a cyclic vector spaceMatrix of orthogonal projection endomorphismthe characteristic polynomial and the minimal polynomial of an operatorCharacteristic Polynomial of EndomorphismMinimal polynomial of an endomorphism.Characteristic Polynomial of Restriction to Invariant Subspace Divides Characteristic PolynomialIrreducible factors of characteristic polynomial and minimal polynomial are same?Linear Transformation EndomorphismCharacteristic polynomial of u is irreducible iff u has not stable subspace not trivial













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Let $ A = (alpha_ {1}, ..., alpha_ {n}) $ be a basis of a vector space $ V $ and let $ varphi: V rightarrow V $ be the endomorphism given by the conditions $varphi(alpha _{i})=alpha _{i+1}$ for $i=1,...,n-1$ and $varphi(alpha _{n})=a_{0} alpha_{1}+a_{1} alpha_{2}+...+a_{n-1} alpha_{n}$. Find a characteristic polynomial of endomorphism $varphi$.




In this task I have a matrix $M^{A}_{A}$ which has $alpha_{i+1}$ in subsequent columns and this $alpha_{i+1}$. However I completely don't know how to do this task because this fact is insufficient for me.

Can you help me?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
    $endgroup$
    – darij grinberg
    yesterday
















2












$begingroup$



Let $ A = (alpha_ {1}, ..., alpha_ {n}) $ be a basis of a vector space $ V $ and let $ varphi: V rightarrow V $ be the endomorphism given by the conditions $varphi(alpha _{i})=alpha _{i+1}$ for $i=1,...,n-1$ and $varphi(alpha _{n})=a_{0} alpha_{1}+a_{1} alpha_{2}+...+a_{n-1} alpha_{n}$. Find a characteristic polynomial of endomorphism $varphi$.




In this task I have a matrix $M^{A}_{A}$ which has $alpha_{i+1}$ in subsequent columns and this $alpha_{i+1}$. However I completely don't know how to do this task because this fact is insufficient for me.

Can you help me?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
    $endgroup$
    – darij grinberg
    yesterday














2












2








2


1



$begingroup$



Let $ A = (alpha_ {1}, ..., alpha_ {n}) $ be a basis of a vector space $ V $ and let $ varphi: V rightarrow V $ be the endomorphism given by the conditions $varphi(alpha _{i})=alpha _{i+1}$ for $i=1,...,n-1$ and $varphi(alpha _{n})=a_{0} alpha_{1}+a_{1} alpha_{2}+...+a_{n-1} alpha_{n}$. Find a characteristic polynomial of endomorphism $varphi$.




In this task I have a matrix $M^{A}_{A}$ which has $alpha_{i+1}$ in subsequent columns and this $alpha_{i+1}$. However I completely don't know how to do this task because this fact is insufficient for me.

Can you help me?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$





Let $ A = (alpha_ {1}, ..., alpha_ {n}) $ be a basis of a vector space $ V $ and let $ varphi: V rightarrow V $ be the endomorphism given by the conditions $varphi(alpha _{i})=alpha _{i+1}$ for $i=1,...,n-1$ and $varphi(alpha _{n})=a_{0} alpha_{1}+a_{1} alpha_{2}+...+a_{n-1} alpha_{n}$. Find a characteristic polynomial of endomorphism $varphi$.




In this task I have a matrix $M^{A}_{A}$ which has $alpha_{i+1}$ in subsequent columns and this $alpha_{i+1}$. However I completely don't know how to do this task because this fact is insufficient for me.

Can you help me?






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Bernard

122k741116




122k741116










asked yesterday









MP3129MP3129

47910




47910












  • $begingroup$
    Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
    $endgroup$
    – darij grinberg
    yesterday


















  • $begingroup$
    Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
    $endgroup$
    – darij grinberg
    yesterday
















$begingroup$
Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
$endgroup$
– darij grinberg
yesterday




$begingroup$
Your matrix $M_A^A$ is the companion matrix of a monic polynomial with non-leading coefficients $a_0, a_1, ldots, a_{n-1}$ (maybe reversed, maybe transposed, depending on how a companion matrix is defined). As for computing the characteristic polynomial of a companion matrix, see math.stackexchange.com/questions/10216 (although the asker of that question has excluded the simplest answer).
$endgroup$
– darij grinberg
yesterday










1 Answer
1






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oldest

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$begingroup$

Hint:
The matrix of this endomorphism is
$$begin{bmatrix}
0&0&0&dots&0&a_0\
1&0&0&dots&0&a_1\
0&1&0&dots &0&a_2 \[-1ex]
vdots&vdots&ddots&&vdots&vdots\
0&0&0&dots&0&a_{n-2}\
0&0&0&dots&1&a_{n-1}
end{bmatrix},$$
so its characteristic polynomial is the determinant
$$begin{vmatrix}
-x&0&0&dots&0&a_0\
1&-x&0&dots&0&a_1\
0&1&-x&dots &0&a_2 \[-1ex]
vdots&vdots&ddots&&vdots&vdots\
0&0&0&dots&-x&a_{n-2}\
0&0&0&dots&1&a_{n-1}-x
end{vmatrix}.$$
Denote it as $D(a_0,a_1,dots,a_{n-1})$ and, expanding along the first row, prove the recurrence relation:
$$D(a_0,a_1,dots,a_{n-1})=-xD(a_1,dots,a_{n-1})+(-1)^{n-1}a_0.$$
Deduce from this relation that
$$D(a_0,a_1,dots,a_{n-1})=(-1)^n(x^n-a_{n-1}x^{n-1}-dots-a_1x-a_0).$$






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    1 Answer
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    1 Answer
    1






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    active

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    2












    $begingroup$

    Hint:
    The matrix of this endomorphism is
    $$begin{bmatrix}
    0&0&0&dots&0&a_0\
    1&0&0&dots&0&a_1\
    0&1&0&dots &0&a_2 \[-1ex]
    vdots&vdots&ddots&&vdots&vdots\
    0&0&0&dots&0&a_{n-2}\
    0&0&0&dots&1&a_{n-1}
    end{bmatrix},$$
    so its characteristic polynomial is the determinant
    $$begin{vmatrix}
    -x&0&0&dots&0&a_0\
    1&-x&0&dots&0&a_1\
    0&1&-x&dots &0&a_2 \[-1ex]
    vdots&vdots&ddots&&vdots&vdots\
    0&0&0&dots&-x&a_{n-2}\
    0&0&0&dots&1&a_{n-1}-x
    end{vmatrix}.$$
    Denote it as $D(a_0,a_1,dots,a_{n-1})$ and, expanding along the first row, prove the recurrence relation:
    $$D(a_0,a_1,dots,a_{n-1})=-xD(a_1,dots,a_{n-1})+(-1)^{n-1}a_0.$$
    Deduce from this relation that
    $$D(a_0,a_1,dots,a_{n-1})=(-1)^n(x^n-a_{n-1}x^{n-1}-dots-a_1x-a_0).$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint:
      The matrix of this endomorphism is
      $$begin{bmatrix}
      0&0&0&dots&0&a_0\
      1&0&0&dots&0&a_1\
      0&1&0&dots &0&a_2 \[-1ex]
      vdots&vdots&ddots&&vdots&vdots\
      0&0&0&dots&0&a_{n-2}\
      0&0&0&dots&1&a_{n-1}
      end{bmatrix},$$
      so its characteristic polynomial is the determinant
      $$begin{vmatrix}
      -x&0&0&dots&0&a_0\
      1&-x&0&dots&0&a_1\
      0&1&-x&dots &0&a_2 \[-1ex]
      vdots&vdots&ddots&&vdots&vdots\
      0&0&0&dots&-x&a_{n-2}\
      0&0&0&dots&1&a_{n-1}-x
      end{vmatrix}.$$
      Denote it as $D(a_0,a_1,dots,a_{n-1})$ and, expanding along the first row, prove the recurrence relation:
      $$D(a_0,a_1,dots,a_{n-1})=-xD(a_1,dots,a_{n-1})+(-1)^{n-1}a_0.$$
      Deduce from this relation that
      $$D(a_0,a_1,dots,a_{n-1})=(-1)^n(x^n-a_{n-1}x^{n-1}-dots-a_1x-a_0).$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint:
        The matrix of this endomorphism is
        $$begin{bmatrix}
        0&0&0&dots&0&a_0\
        1&0&0&dots&0&a_1\
        0&1&0&dots &0&a_2 \[-1ex]
        vdots&vdots&ddots&&vdots&vdots\
        0&0&0&dots&0&a_{n-2}\
        0&0&0&dots&1&a_{n-1}
        end{bmatrix},$$
        so its characteristic polynomial is the determinant
        $$begin{vmatrix}
        -x&0&0&dots&0&a_0\
        1&-x&0&dots&0&a_1\
        0&1&-x&dots &0&a_2 \[-1ex]
        vdots&vdots&ddots&&vdots&vdots\
        0&0&0&dots&-x&a_{n-2}\
        0&0&0&dots&1&a_{n-1}-x
        end{vmatrix}.$$
        Denote it as $D(a_0,a_1,dots,a_{n-1})$ and, expanding along the first row, prove the recurrence relation:
        $$D(a_0,a_1,dots,a_{n-1})=-xD(a_1,dots,a_{n-1})+(-1)^{n-1}a_0.$$
        Deduce from this relation that
        $$D(a_0,a_1,dots,a_{n-1})=(-1)^n(x^n-a_{n-1}x^{n-1}-dots-a_1x-a_0).$$






        share|cite|improve this answer









        $endgroup$



        Hint:
        The matrix of this endomorphism is
        $$begin{bmatrix}
        0&0&0&dots&0&a_0\
        1&0&0&dots&0&a_1\
        0&1&0&dots &0&a_2 \[-1ex]
        vdots&vdots&ddots&&vdots&vdots\
        0&0&0&dots&0&a_{n-2}\
        0&0&0&dots&1&a_{n-1}
        end{bmatrix},$$
        so its characteristic polynomial is the determinant
        $$begin{vmatrix}
        -x&0&0&dots&0&a_0\
        1&-x&0&dots&0&a_1\
        0&1&-x&dots &0&a_2 \[-1ex]
        vdots&vdots&ddots&&vdots&vdots\
        0&0&0&dots&-x&a_{n-2}\
        0&0&0&dots&1&a_{n-1}-x
        end{vmatrix}.$$
        Denote it as $D(a_0,a_1,dots,a_{n-1})$ and, expanding along the first row, prove the recurrence relation:
        $$D(a_0,a_1,dots,a_{n-1})=-xD(a_1,dots,a_{n-1})+(-1)^{n-1}a_0.$$
        Deduce from this relation that
        $$D(a_0,a_1,dots,a_{n-1})=(-1)^n(x^n-a_{n-1}x^{n-1}-dots-a_1x-a_0).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        BernardBernard

        122k741116




        122k741116






























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