Find $2times 2$ symmetric matrix $A$ given two eigenvalues and one eigenvector












1












$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







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  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04


















1












$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04
















1












1








1





$begingroup$


I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.



I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!






linear-algebra eigenvalues-eigenvectors




linear-algebra eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 22:06









user376343

3,4033826




3,4033826










asked Dec 7 '18 at 21:52









a hea he

61




61








  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04
















  • 3




    $begingroup$
    Different eigenspaces of a symmetric matrix are orthogonal to each other.
    $endgroup$
    – Berci
    Dec 7 '18 at 21:54






  • 1




    $begingroup$
    ...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
    $endgroup$
    – Jean Marie
    Dec 7 '18 at 22:04










3




3




$begingroup$
Different eigenspaces of a symmetric matrix are orthogonal to each other.
$endgroup$
– Berci
Dec 7 '18 at 21:54




$begingroup$
Different eigenspaces of a symmetric matrix are orthogonal to each other.
$endgroup$
– Berci
Dec 7 '18 at 21:54




1




1




$begingroup$
...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
$endgroup$
– Jean Marie
Dec 7 '18 at 22:04






$begingroup$
...thus $[1,-1]^T$ could be this vector (attached to eigenvalue 4)...
$endgroup$
– Jean Marie
Dec 7 '18 at 22:04












2 Answers
2






active

oldest

votes


















2












$begingroup$

To offer a slightly different perspective, due to the Spectral Theorem you have
$$
A=P_1+4P_2,$$
where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
$$
P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
$$
And then
$$
A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Let :



    $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



    Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



    $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



    such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



    $$(a-lambda)(d-lambda)-bc=0$$



    Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



    $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



    $$Leftrightarrow$$



    $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



    Can you now find $a,b,c$ and $d$ ?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
      $endgroup$
      – M. A. SARKAR
      Dec 7 '18 at 22:10













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    To offer a slightly different perspective, due to the Spectral Theorem you have
    $$
    A=P_1+4P_2,$$
    where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
    $$
    P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
    $$
    And then
    $$
    A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
    $$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      To offer a slightly different perspective, due to the Spectral Theorem you have
      $$
      A=P_1+4P_2,$$
      where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
      $$
      P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
      $$
      And then
      $$
      A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
      $$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        To offer a slightly different perspective, due to the Spectral Theorem you have
        $$
        A=P_1+4P_2,$$
        where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
        $$
        P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
        $$
        And then
        $$
        A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
        $$






        share|cite|improve this answer









        $endgroup$



        To offer a slightly different perspective, due to the Spectral Theorem you have
        $$
        A=P_1+4P_2,$$
        where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus
        $$
        P_1=tfrac12,begin{bmatrix} 1&1end{bmatrix}begin{bmatrix} 1\1end{bmatrix}=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}.
        $$
        And then
        $$
        A=begin{bmatrix}1/2&1/2\1/2&1/2end{bmatrix}+4begin{bmatrix}1/2&-1/2\-1/2&1/2end{bmatrix}=begin{bmatrix}5/2&-3/2\-3/2&5/2 end{bmatrix}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 7 '18 at 22:12









        Martin ArgeramiMartin Argerami

        126k1182180




        126k1182180























            1












            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10


















            1












            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10
















            1












            1








            1





            $begingroup$

            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?






            share|cite|improve this answer









            $endgroup$



            Let :



            $$A = begin{pmatrix} a & b \ c & d end{pmatrix}$$



            Since the eigenvalues are $lambda =1 $ and $lambda = 4$, it must be



            $$det(A-lambda I) = begin{vmatrix} a - lambda & b \ c & d - lambdaend{vmatrix} = (a-lambda)(d-lambda)-bc $$



            such that $lambda = 1$ and $lambda = 4$ are solutions to the equation :



            $$(a-lambda)(d-lambda)-bc=0$$



            Since $(1,1)^mathbf{T}$ is an eigenvector of $A$ for $lambda = 1$, it is :



            $$(A- I)(1,1)^mathbf{T} = 0 Rightarrow begin{pmatrix} a - 1 & b \ c & d - 1end{pmatrix}begin{pmatrix}1 \ 1 end{pmatrix} =begin{pmatrix}0 \ 0 end{pmatrix} $$



            $$Leftrightarrow$$



            $$begin{cases} a-1 + b = 0 \c + d-1 = 0end{cases}$$



            Can you now find $a,b,c$ and $d$ ?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 7 '18 at 22:07









            RebellosRebellos

            14.6k31247




            14.6k31247












            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10




















            • $begingroup$
              Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
              $endgroup$
              – M. A. SARKAR
              Dec 7 '18 at 22:10


















            $begingroup$
            Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
            $endgroup$
            – M. A. SARKAR
            Dec 7 '18 at 22:10






            $begingroup$
            Also use the other two conditions: $ lambda_1 lambda_2=4=ad-bc$ and $ lambda_1+lambda_2=4+1=trace (A)=a+d$.. i.e., $ a+d=5, ad-bc=4$.
            $endgroup$
            – M. A. SARKAR
            Dec 7 '18 at 22:10




















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