Solve $A = C B C^t$ for $B$Find $C$, if $A=CBC$, where $A$,$B$,$C$ are symmetric matrices.Is symmetry a necessary condition for positive (or negative) definiteness?Variant of Cholesky Decomposition: solve $B^TB=A$ for general square matrix $A$Finding Rank of the following MatrixProve that a square matrix $A$ is positive definite if and only if $A$+ $A^T$ is positive definiteFinding an orthogonal matrix for a 3x2How to solve for square matrix $y$ given $x = A'yA$ for $A'A = I$Find Condition on the matrix $X in mathbbR^n times m$ where $mleq n$Necessary condition for a matrix equation to be zeroPrincipal-Minors-Test for Positivity of Complex Hermitian MatricesHow to solve for unknown matrix?

Can my sorcerer use a spellbook only to collect spells and scribe scrolls, not cast?

Why is this clock signal connected to a capacitor to gnd?

Why didn't Miles's spider sense work before?

Why no variance term in Bayesian logistic regression?

What mechanic is there to disable a threat instead of killing it?

Should I tell management that I intend to leave due to bad software development practices?

What does the expression "A Mann!" means

How can I determine if the org that I'm currently connected to is a scratch org?

What killed these X2 caps?

Is it acceptable for a professor to tell male students to not think that they are smarter than female students?

Can I run a new neutral wire to repair a broken circuit?

Bullying boss launched a smear campaign and made me unemployable

Detention in 1997

Assassin's bullet with mercury

A category-like structure without composition?

Determining Impedance With An Antenna Analyzer

One verb to replace 'be a member of' a club

Do scales need to be in alphabetical order?

How did the Super Star Destroyer Executor get destroyed exactly?

What does “the session was packed” mean in this context?

I would say: "You are another teacher", but she is a woman and I am a man

How can I deal with my CEO asking me to hire someone with a higher salary than me, a co-founder?

Is there a hemisphere-neutral way of specifying a season?

How do I gain back my faith in my PhD degree?



Solve $A = C B C^t$ for $B$


Find $C$, if $A=CBC$, where $A$,$B$,$C$ are symmetric matrices.Is symmetry a necessary condition for positive (or negative) definiteness?Variant of Cholesky Decomposition: solve $B^TB=A$ for general square matrix $A$Finding Rank of the following MatrixProve that a square matrix $A$ is positive definite if and only if $A$+ $A^T$ is positive definiteFinding an orthogonal matrix for a 3x2How to solve for square matrix $y$ given $x = A'yA$ for $A'A = I$Find Condition on the matrix $X in mathbbR^n times m$ where $mleq n$Necessary condition for a matrix equation to be zeroPrincipal-Minors-Test for Positivity of Complex Hermitian MatricesHow to solve for unknown matrix?













0












$begingroup$


I know this question, but I would like to know the middle square matrix $B$.




Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$



$$A = C B C^t$$











share|cite|improve this question











$endgroup$











  • $begingroup$
    Are A, B and C real or complex matrices?
    $endgroup$
    – PerelMan
    Mar 21 at 6:37











  • $begingroup$
    yes, real only, just simplest case...
    $endgroup$
    – JimSD
    Mar 21 at 6:37











  • $begingroup$
    $CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
    $endgroup$
    – Saucy O'Path
    Mar 21 at 6:38







  • 1




    $begingroup$
    This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 6:54







  • 1




    $begingroup$
    @JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 7:25















0












$begingroup$


I know this question, but I would like to know the middle square matrix $B$.




Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$



$$A = C B C^t$$











share|cite|improve this question











$endgroup$











  • $begingroup$
    Are A, B and C real or complex matrices?
    $endgroup$
    – PerelMan
    Mar 21 at 6:37











  • $begingroup$
    yes, real only, just simplest case...
    $endgroup$
    – JimSD
    Mar 21 at 6:37











  • $begingroup$
    $CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
    $endgroup$
    – Saucy O'Path
    Mar 21 at 6:38







  • 1




    $begingroup$
    This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 6:54







  • 1




    $begingroup$
    @JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 7:25













0












0








0





$begingroup$


I know this question, but I would like to know the middle square matrix $B$.




Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$



$$A = C B C^t$$











share|cite|improve this question











$endgroup$




I know this question, but I would like to know the middle square matrix $B$.




Given positive definite matrix $A in mathbb R^2 times 2$ and non-zero matrix $C in mathbb R^2 times 3$, find $3 times3$ matrix $B$



$$A = C B C^t$$








matrices systems-of-equations matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 7:28









Rodrigo de Azevedo

13.1k41960




13.1k41960










asked Mar 21 at 6:33









JimSDJimSD

114




114











  • $begingroup$
    Are A, B and C real or complex matrices?
    $endgroup$
    – PerelMan
    Mar 21 at 6:37











  • $begingroup$
    yes, real only, just simplest case...
    $endgroup$
    – JimSD
    Mar 21 at 6:37











  • $begingroup$
    $CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
    $endgroup$
    – Saucy O'Path
    Mar 21 at 6:38







  • 1




    $begingroup$
    This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 6:54







  • 1




    $begingroup$
    @JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 7:25
















  • $begingroup$
    Are A, B and C real or complex matrices?
    $endgroup$
    – PerelMan
    Mar 21 at 6:37











  • $begingroup$
    yes, real only, just simplest case...
    $endgroup$
    – JimSD
    Mar 21 at 6:37











  • $begingroup$
    $CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
    $endgroup$
    – Saucy O'Path
    Mar 21 at 6:38







  • 1




    $begingroup$
    This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 6:54







  • 1




    $begingroup$
    @JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 21 at 7:25















$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37





$begingroup$
Are A, B and C real or complex matrices?
$endgroup$
– PerelMan
Mar 21 at 6:37













$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37





$begingroup$
yes, real only, just simplest case...
$endgroup$
– JimSD
Mar 21 at 6:37













$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38





$begingroup$
$CBC^t$ is not a thing, if $B$ is $3times 2$ and $C$ is $3times 3$.
$endgroup$
– Saucy O'Path
Mar 21 at 6:38





1




1




$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54





$begingroup$
This is extremely easy using vectorization. Solve the linear system $$(C otimes C) , mboxvec (B) = mboxvec (A)$$
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 6:54





1




1




$begingroup$
@JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25




$begingroup$
@JimSD Then un-vectorize!! In MATLAB, use reshape to do both vectorization and un-vectorization.
$endgroup$
– Rodrigo de Azevedo
Mar 21 at 7:25










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).



Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.



(*) is equivalent to $C^+A(C^+)^T=SBS$ or



$beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.



Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied



$a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "69"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156430%2fsolve-a-c-b-ct-for-b%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).



    Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.



    (*) is equivalent to $C^+A(C^+)^T=SBS$ or



    $beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.



    Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied



    $a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).



      Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.



      (*) is equivalent to $C^+A(C^+)^T=SBS$ or



      $beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.



      Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied



      $a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).



        Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.



        (*) is equivalent to $C^+A(C^+)^T=SBS$ or



        $beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.



        Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied



        $a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.






        share|cite|improve this answer









        $endgroup$



        Let (*) $A=CBC^T$. Since $A$ is sym. $>0$, necessarily $rank(C)=2$ ( otherwise, there are no solutions in $B$).



        Then $C^+=C^T(CC^T)^-1$ ( cf. the Moore Penrose inverse of $C$) and $CC^+=I_2$. Moreover, $S=C^+C$ is sym. and similar to $diag(1,1,0)$. Up to an orthonormal change of basis, we may assume that $S=C^+C=diag(1,1,0),C=[D_2,2,0_2,1],C^+=beginpmatrixD^-1\0_1,2endpmatrix$.



        (*) is equivalent to $C^+A(C^+)^T=SBS$ or



        $beginpmatrixD^-1A(D^-1)^T&0\0&0endpmatrix=beginpmatrixa&b&0\d&e&0\0&0&0endpmatrix$ where the unknown $B=beginpmatrixa&b&c\d&e&f\g&h&pendpmatrix$.



        Note that $U=[u_i,j]=D^-1A(D^-1)^T$ is a known sym. $>0$ matrix. Then $B$ is a solution of (*) iff the following conditions are satisfied



        $a=u_1,1,e=u_2,2,b=d=u_1,2$; $g,h,p,c,f$ are $5=9-4$ arbitrary parameters.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 22 at 11:58









        loup blancloup blanc

        24.1k21851




        24.1k21851



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156430%2fsolve-a-c-b-ct-for-b%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye

            random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

            How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer