Sum of Symmetric Positive Definite Matrix and Scalar of Identity












1












$begingroup$


If $A$ is an $ntimes n$ symmetric positive definite matrix with the smallest eigenvalue $lambda$, then for any $mu>-lambda$, $A+mu I$ is positive definite.



I am trying to show this, but I am stuck on one part. Here is what I have so far:
$$
begin{align*}
langle x,left(A+mu Iright)xrangle&=langle x,Ax+mu xrangle\
&=langle x,Axrangle+langle x,mu xrangle\
&>0+mulangle x,xrangle\
&>-lambdalangle x,xrangle.
end{align*}
$$



I'm stuck on showing that $mulangle x,xrangle$ is positive because I only know that $mu>-lambda$. Any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors positive-definite







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  • 2




    $begingroup$
    $langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
    $endgroup$
    – DisintegratingByParts
    Dec 10 '18 at 18:22
















1












$begingroup$


If $A$ is an $ntimes n$ symmetric positive definite matrix with the smallest eigenvalue $lambda$, then for any $mu>-lambda$, $A+mu I$ is positive definite.



I am trying to show this, but I am stuck on one part. Here is what I have so far:
$$
begin{align*}
langle x,left(A+mu Iright)xrangle&=langle x,Ax+mu xrangle\
&=langle x,Axrangle+langle x,mu xrangle\
&>0+mulangle x,xrangle\
&>-lambdalangle x,xrangle.
end{align*}
$$



I'm stuck on showing that $mulangle x,xrangle$ is positive because I only know that $mu>-lambda$. Any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors positive-definite







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
    $endgroup$
    – DisintegratingByParts
    Dec 10 '18 at 18:22














1












1








1


1



$begingroup$


If $A$ is an $ntimes n$ symmetric positive definite matrix with the smallest eigenvalue $lambda$, then for any $mu>-lambda$, $A+mu I$ is positive definite.



I am trying to show this, but I am stuck on one part. Here is what I have so far:
$$
begin{align*}
langle x,left(A+mu Iright)xrangle&=langle x,Ax+mu xrangle\
&=langle x,Axrangle+langle x,mu xrangle\
&>0+mulangle x,xrangle\
&>-lambdalangle x,xrangle.
end{align*}
$$



I'm stuck on showing that $mulangle x,xrangle$ is positive because I only know that $mu>-lambda$. Any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors positive-definite







share|cite|improve this question









$endgroup$




If $A$ is an $ntimes n$ symmetric positive definite matrix with the smallest eigenvalue $lambda$, then for any $mu>-lambda$, $A+mu I$ is positive definite.



I am trying to show this, but I am stuck on one part. Here is what I have so far:
$$
begin{align*}
langle x,left(A+mu Iright)xrangle&=langle x,Ax+mu xrangle\
&=langle x,Axrangle+langle x,mu xrangle\
&>0+mulangle x,xrangle\
&>-lambdalangle x,xrangle.
end{align*}
$$



I'm stuck on showing that $mulangle x,xrangle$ is positive because I only know that $mu>-lambda$. Any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors positive-definite




linear-algebra matrices eigenvalues-eigenvectors positive-definite






share|cite|improve this question













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asked Dec 10 '18 at 18:18









JakeJake

415312




415312








  • 2




    $begingroup$
    $langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
    $endgroup$
    – DisintegratingByParts
    Dec 10 '18 at 18:22














  • 2




    $begingroup$
    $langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
    $endgroup$
    – DisintegratingByParts
    Dec 10 '18 at 18:22








2




2




$begingroup$
$langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
$endgroup$
– DisintegratingByParts
Dec 10 '18 at 18:22




$begingroup$
$langle Ax,xrangle ge lambdalangle x,xrangle$ by the assumptions on $A$.
$endgroup$
– DisintegratingByParts
Dec 10 '18 at 18:22










1 Answer
1






active

oldest

votes


















1












$begingroup$

If $A$ is a positive definite symmetric matrix with smallest eigenvalue $lambda$, then for all vectors $x$ we have



$langle x, Ax rangle ge lambda langle x, x rangle; tag 1$



the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal matrix $O$, which will preserve the inner product:



$langle Oy, Ox rangle = langle y, O^TOx rangle = langle y, Ix rangle = langle y, x rangle, tag 2$



where we have used the fact that



$O^TO = OO^T = I tag 3$



in (2); then we have



$OAO^T = Lambda = text{diag}(lambda_1, lambda_2, ldots, lambda_n), tag 3$



where $lambda_1, lambda_2, ldots, lambda_n$ are the eigenvalues of $A$. It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that



$A e_i = lambda_i e_i, ; 1 le i le n; tag 4$



we may then write



$x = displaystyle sum_1^n x_i e_i, tag 5$



and



$langle x, Ax rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i Ae_i right rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i lambda_i e_i right rangle = displaystyle sum_{i, j = 1}^n x_ix_j langle e_i, lambda_j e_j rangle$
$= displaystyle sum_{i, j = 1}^n x_ix_j lambda_j langle e_i,e_j rangle = sum_{i, j = 1}^n x_ix_j lambda_j delta_{ij} = sum_1^n lambda_i x_i^2; tag 6$



now if



$lambda = min {lambda_1, lambda_2, ldots, lambda_n } > 0 tag 7$



is the least eigenvalue, then (6) yields



$langle x, Ax rangle = displaystyle sum_1^n lambda_i x_i^2 ge sum_1^n lambda x_i^2 = lambda sum_1^n x_i^2 = lambda langle x, x rangle; tag 8$



therefore,



$langle x, (A + mu I)x rangle = langle x, Ax rangle + mu langle x, x rangle ge lambda langle x, x rangle + mu langle x, x rangle = (lambda + mu) langle x, x rangle; tag 9$



since



$mu > - lambda Longleftrightarrow mu + lambda > 0, tag{10}$



(9) becomes



$langle x, (A + mu I)x rangle ge (mu + lambda ) langle x, x rangle > 0, tag{11}$



which shows that $A + mu I$ is positive definite, the desired result. $OEDelta$.






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    1












    $begingroup$

    If $A$ is a positive definite symmetric matrix with smallest eigenvalue $lambda$, then for all vectors $x$ we have



    $langle x, Ax rangle ge lambda langle x, x rangle; tag 1$



    the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal matrix $O$, which will preserve the inner product:



    $langle Oy, Ox rangle = langle y, O^TOx rangle = langle y, Ix rangle = langle y, x rangle, tag 2$



    where we have used the fact that



    $O^TO = OO^T = I tag 3$



    in (2); then we have



    $OAO^T = Lambda = text{diag}(lambda_1, lambda_2, ldots, lambda_n), tag 3$



    where $lambda_1, lambda_2, ldots, lambda_n$ are the eigenvalues of $A$. It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that



    $A e_i = lambda_i e_i, ; 1 le i le n; tag 4$



    we may then write



    $x = displaystyle sum_1^n x_i e_i, tag 5$



    and



    $langle x, Ax rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i Ae_i right rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i lambda_i e_i right rangle = displaystyle sum_{i, j = 1}^n x_ix_j langle e_i, lambda_j e_j rangle$
    $= displaystyle sum_{i, j = 1}^n x_ix_j lambda_j langle e_i,e_j rangle = sum_{i, j = 1}^n x_ix_j lambda_j delta_{ij} = sum_1^n lambda_i x_i^2; tag 6$



    now if



    $lambda = min {lambda_1, lambda_2, ldots, lambda_n } > 0 tag 7$



    is the least eigenvalue, then (6) yields



    $langle x, Ax rangle = displaystyle sum_1^n lambda_i x_i^2 ge sum_1^n lambda x_i^2 = lambda sum_1^n x_i^2 = lambda langle x, x rangle; tag 8$



    therefore,



    $langle x, (A + mu I)x rangle = langle x, Ax rangle + mu langle x, x rangle ge lambda langle x, x rangle + mu langle x, x rangle = (lambda + mu) langle x, x rangle; tag 9$



    since



    $mu > - lambda Longleftrightarrow mu + lambda > 0, tag{10}$



    (9) becomes



    $langle x, (A + mu I)x rangle ge (mu + lambda ) langle x, x rangle > 0, tag{11}$



    which shows that $A + mu I$ is positive definite, the desired result. $OEDelta$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If $A$ is a positive definite symmetric matrix with smallest eigenvalue $lambda$, then for all vectors $x$ we have



      $langle x, Ax rangle ge lambda langle x, x rangle; tag 1$



      the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal matrix $O$, which will preserve the inner product:



      $langle Oy, Ox rangle = langle y, O^TOx rangle = langle y, Ix rangle = langle y, x rangle, tag 2$



      where we have used the fact that



      $O^TO = OO^T = I tag 3$



      in (2); then we have



      $OAO^T = Lambda = text{diag}(lambda_1, lambda_2, ldots, lambda_n), tag 3$



      where $lambda_1, lambda_2, ldots, lambda_n$ are the eigenvalues of $A$. It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that



      $A e_i = lambda_i e_i, ; 1 le i le n; tag 4$



      we may then write



      $x = displaystyle sum_1^n x_i e_i, tag 5$



      and



      $langle x, Ax rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i Ae_i right rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i lambda_i e_i right rangle = displaystyle sum_{i, j = 1}^n x_ix_j langle e_i, lambda_j e_j rangle$
      $= displaystyle sum_{i, j = 1}^n x_ix_j lambda_j langle e_i,e_j rangle = sum_{i, j = 1}^n x_ix_j lambda_j delta_{ij} = sum_1^n lambda_i x_i^2; tag 6$



      now if



      $lambda = min {lambda_1, lambda_2, ldots, lambda_n } > 0 tag 7$



      is the least eigenvalue, then (6) yields



      $langle x, Ax rangle = displaystyle sum_1^n lambda_i x_i^2 ge sum_1^n lambda x_i^2 = lambda sum_1^n x_i^2 = lambda langle x, x rangle; tag 8$



      therefore,



      $langle x, (A + mu I)x rangle = langle x, Ax rangle + mu langle x, x rangle ge lambda langle x, x rangle + mu langle x, x rangle = (lambda + mu) langle x, x rangle; tag 9$



      since



      $mu > - lambda Longleftrightarrow mu + lambda > 0, tag{10}$



      (9) becomes



      $langle x, (A + mu I)x rangle ge (mu + lambda ) langle x, x rangle > 0, tag{11}$



      which shows that $A + mu I$ is positive definite, the desired result. $OEDelta$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If $A$ is a positive definite symmetric matrix with smallest eigenvalue $lambda$, then for all vectors $x$ we have



        $langle x, Ax rangle ge lambda langle x, x rangle; tag 1$



        the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal matrix $O$, which will preserve the inner product:



        $langle Oy, Ox rangle = langle y, O^TOx rangle = langle y, Ix rangle = langle y, x rangle, tag 2$



        where we have used the fact that



        $O^TO = OO^T = I tag 3$



        in (2); then we have



        $OAO^T = Lambda = text{diag}(lambda_1, lambda_2, ldots, lambda_n), tag 3$



        where $lambda_1, lambda_2, ldots, lambda_n$ are the eigenvalues of $A$. It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that



        $A e_i = lambda_i e_i, ; 1 le i le n; tag 4$



        we may then write



        $x = displaystyle sum_1^n x_i e_i, tag 5$



        and



        $langle x, Ax rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i Ae_i right rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i lambda_i e_i right rangle = displaystyle sum_{i, j = 1}^n x_ix_j langle e_i, lambda_j e_j rangle$
        $= displaystyle sum_{i, j = 1}^n x_ix_j lambda_j langle e_i,e_j rangle = sum_{i, j = 1}^n x_ix_j lambda_j delta_{ij} = sum_1^n lambda_i x_i^2; tag 6$



        now if



        $lambda = min {lambda_1, lambda_2, ldots, lambda_n } > 0 tag 7$



        is the least eigenvalue, then (6) yields



        $langle x, Ax rangle = displaystyle sum_1^n lambda_i x_i^2 ge sum_1^n lambda x_i^2 = lambda sum_1^n x_i^2 = lambda langle x, x rangle; tag 8$



        therefore,



        $langle x, (A + mu I)x rangle = langle x, Ax rangle + mu langle x, x rangle ge lambda langle x, x rangle + mu langle x, x rangle = (lambda + mu) langle x, x rangle; tag 9$



        since



        $mu > - lambda Longleftrightarrow mu + lambda > 0, tag{10}$



        (9) becomes



        $langle x, (A + mu I)x rangle ge (mu + lambda ) langle x, x rangle > 0, tag{11}$



        which shows that $A + mu I$ is positive definite, the desired result. $OEDelta$.






        share|cite|improve this answer











        $endgroup$



        If $A$ is a positive definite symmetric matrix with smallest eigenvalue $lambda$, then for all vectors $x$ we have



        $langle x, Ax rangle ge lambda langle x, x rangle; tag 1$



        the easiest way I know to see this is to diagonalize $A$ by a suitable orthogonal matrix $O$, which will preserve the inner product:



        $langle Oy, Ox rangle = langle y, O^TOx rangle = langle y, Ix rangle = langle y, x rangle, tag 2$



        where we have used the fact that



        $O^TO = OO^T = I tag 3$



        in (2); then we have



        $OAO^T = Lambda = text{diag}(lambda_1, lambda_2, ldots, lambda_n), tag 3$



        where $lambda_1, lambda_2, ldots, lambda_n$ are the eigenvalues of $A$. It is well-known that $A$ is also possessed of an orthonormal eigenbasis of vectors $e_i$ such that



        $A e_i = lambda_i e_i, ; 1 le i le n; tag 4$



        we may then write



        $x = displaystyle sum_1^n x_i e_i, tag 5$



        and



        $langle x, Ax rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i Ae_i right rangle = left langle displaystyle sum_1^n x_ie_i, sum_1^n x_i lambda_i e_i right rangle = displaystyle sum_{i, j = 1}^n x_ix_j langle e_i, lambda_j e_j rangle$
        $= displaystyle sum_{i, j = 1}^n x_ix_j lambda_j langle e_i,e_j rangle = sum_{i, j = 1}^n x_ix_j lambda_j delta_{ij} = sum_1^n lambda_i x_i^2; tag 6$



        now if



        $lambda = min {lambda_1, lambda_2, ldots, lambda_n } > 0 tag 7$



        is the least eigenvalue, then (6) yields



        $langle x, Ax rangle = displaystyle sum_1^n lambda_i x_i^2 ge sum_1^n lambda x_i^2 = lambda sum_1^n x_i^2 = lambda langle x, x rangle; tag 8$



        therefore,



        $langle x, (A + mu I)x rangle = langle x, Ax rangle + mu langle x, x rangle ge lambda langle x, x rangle + mu langle x, x rangle = (lambda + mu) langle x, x rangle; tag 9$



        since



        $mu > - lambda Longleftrightarrow mu + lambda > 0, tag{10}$



        (9) becomes



        $langle x, (A + mu I)x rangle ge (mu + lambda ) langle x, x rangle > 0, tag{11}$



        which shows that $A + mu I$ is positive definite, the desired result. $OEDelta$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 19:37

























        answered Dec 10 '18 at 19:16









        Robert LewisRobert Lewis

        45.6k23065




        45.6k23065






























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