How to find projection matrix of the singular matrix onto fundamental subspaces?












0












$begingroup$


begin{align}
A & =begin{bmatrix}2&3\6&9end{bmatrix}
end{align}
I found four fundamental subspaces of A which are



begin{align}
C(A) & =begin{bmatrix}2\6end{bmatrix} , dimension = 1
end{align}



begin{align}
C(A^T) & =begin{bmatrix}2\3end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A) & =begin{bmatrix}-3\2end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A^T) & =begin{bmatrix}3\-1end{bmatrix} , dimension = 1
end{align}



The first question is that "Find the projection matrices onto each of the fundamental subspaces"
and I used the formula that the projection matrix on to C(A) is
begin{align}
P & = A(A^TA)^{-1}A^T
end{align}



However, how to find the projection matrix if



begin{align}
det(A^TA) = 0 ?
end{align}
I thought I couldn't find the projection matrix onto C(A) and even other subspaces because A^TA is singular matrix.



So, please let me know what is projection matrix onto each of the fundamental subsapces and how to find them.



Next question is that "Write the vector b = (2,2) as a linear combination of a vector in C(A) and N(A^T)



I thought this question was asking to solve



begin{align}
a & begin{bmatrix}2\6end{bmatrix} + bbegin{bmatrix}3\-1end{bmatrix} = begin{bmatrix}2\2end{bmatrix}
end{align}



using vectors from C(A) and N(A^T) we found in previous problem.
so, I got a = 2/5 and b = 2/5 from the above equation I made.



Please let me know how to write vector b as a linear combination of a vector in column space and left null space.






linear-algebra vector-spaces vectors projection-matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
    $endgroup$
    – amd
    Dec 8 '18 at 7:05
















0












$begingroup$


begin{align}
A & =begin{bmatrix}2&3\6&9end{bmatrix}
end{align}
I found four fundamental subspaces of A which are



begin{align}
C(A) & =begin{bmatrix}2\6end{bmatrix} , dimension = 1
end{align}



begin{align}
C(A^T) & =begin{bmatrix}2\3end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A) & =begin{bmatrix}-3\2end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A^T) & =begin{bmatrix}3\-1end{bmatrix} , dimension = 1
end{align}



The first question is that "Find the projection matrices onto each of the fundamental subspaces"
and I used the formula that the projection matrix on to C(A) is
begin{align}
P & = A(A^TA)^{-1}A^T
end{align}



However, how to find the projection matrix if



begin{align}
det(A^TA) = 0 ?
end{align}
I thought I couldn't find the projection matrix onto C(A) and even other subspaces because A^TA is singular matrix.



So, please let me know what is projection matrix onto each of the fundamental subsapces and how to find them.



Next question is that "Write the vector b = (2,2) as a linear combination of a vector in C(A) and N(A^T)



I thought this question was asking to solve



begin{align}
a & begin{bmatrix}2\6end{bmatrix} + bbegin{bmatrix}3\-1end{bmatrix} = begin{bmatrix}2\2end{bmatrix}
end{align}



using vectors from C(A) and N(A^T) we found in previous problem.
so, I got a = 2/5 and b = 2/5 from the above equation I made.



Please let me know how to write vector b as a linear combination of a vector in column space and left null space.






linear-algebra vector-spaces vectors projection-matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
    $endgroup$
    – amd
    Dec 8 '18 at 7:05














0












0








0





$begingroup$


begin{align}
A & =begin{bmatrix}2&3\6&9end{bmatrix}
end{align}
I found four fundamental subspaces of A which are



begin{align}
C(A) & =begin{bmatrix}2\6end{bmatrix} , dimension = 1
end{align}



begin{align}
C(A^T) & =begin{bmatrix}2\3end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A) & =begin{bmatrix}-3\2end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A^T) & =begin{bmatrix}3\-1end{bmatrix} , dimension = 1
end{align}



The first question is that "Find the projection matrices onto each of the fundamental subspaces"
and I used the formula that the projection matrix on to C(A) is
begin{align}
P & = A(A^TA)^{-1}A^T
end{align}



However, how to find the projection matrix if



begin{align}
det(A^TA) = 0 ?
end{align}
I thought I couldn't find the projection matrix onto C(A) and even other subspaces because A^TA is singular matrix.



So, please let me know what is projection matrix onto each of the fundamental subsapces and how to find them.



Next question is that "Write the vector b = (2,2) as a linear combination of a vector in C(A) and N(A^T)



I thought this question was asking to solve



begin{align}
a & begin{bmatrix}2\6end{bmatrix} + bbegin{bmatrix}3\-1end{bmatrix} = begin{bmatrix}2\2end{bmatrix}
end{align}



using vectors from C(A) and N(A^T) we found in previous problem.
so, I got a = 2/5 and b = 2/5 from the above equation I made.



Please let me know how to write vector b as a linear combination of a vector in column space and left null space.






linear-algebra vector-spaces vectors projection-matrices







share|cite|improve this question









$endgroup$




begin{align}
A & =begin{bmatrix}2&3\6&9end{bmatrix}
end{align}
I found four fundamental subspaces of A which are



begin{align}
C(A) & =begin{bmatrix}2\6end{bmatrix} , dimension = 1
end{align}



begin{align}
C(A^T) & =begin{bmatrix}2\3end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A) & =begin{bmatrix}-3\2end{bmatrix} , dimension = 1
end{align}



begin{align}
N(A^T) & =begin{bmatrix}3\-1end{bmatrix} , dimension = 1
end{align}



The first question is that "Find the projection matrices onto each of the fundamental subspaces"
and I used the formula that the projection matrix on to C(A) is
begin{align}
P & = A(A^TA)^{-1}A^T
end{align}



However, how to find the projection matrix if



begin{align}
det(A^TA) = 0 ?
end{align}
I thought I couldn't find the projection matrix onto C(A) and even other subspaces because A^TA is singular matrix.



So, please let me know what is projection matrix onto each of the fundamental subsapces and how to find them.



Next question is that "Write the vector b = (2,2) as a linear combination of a vector in C(A) and N(A^T)



I thought this question was asking to solve



begin{align}
a & begin{bmatrix}2\6end{bmatrix} + bbegin{bmatrix}3\-1end{bmatrix} = begin{bmatrix}2\2end{bmatrix}
end{align}



using vectors from C(A) and N(A^T) we found in previous problem.
so, I got a = 2/5 and b = 2/5 from the above equation I made.



Please let me know how to write vector b as a linear combination of a vector in column space and left null space.






linear-algebra vector-spaces vectors projection-matrices




linear-algebra vector-spaces vectors projection-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 2:01









Rudy1996Rudy1996

32




32












  • $begingroup$
    That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
    $endgroup$
    – amd
    Dec 8 '18 at 7:05


















  • $begingroup$
    That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
    $endgroup$
    – amd
    Dec 8 '18 at 7:05
















$begingroup$
That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
$endgroup$
– amd
Dec 8 '18 at 7:05




$begingroup$
That formula only applies when $A$ has full rank. Otherwise, as you’ve discovered, $A^TA$ is not invertible.
$endgroup$
– amd
Dec 8 '18 at 7:05










1 Answer
1






active

oldest

votes


















0












$begingroup$

Projection of a vector $bf{u}$ along the vector $bf{v}$ is given by $textbf{proj}_{bf{v}}bf{u}=left(frac{bf{u} cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $bf{e}_1$ and $bf{e}_2$, i.e.
$$textbf{proj}_{bf{v}}bf{e_1}=left(frac{bf{e}_1 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(frac{bf{e}_2 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$$
Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $bf{v}=begin{bmatrix}2\3end{bmatrix}$. Then,
$$textbf{proj}_{bf{v}}bf{e_1}=left(mathrm{frac{2}{13}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(mathrm{frac{3}{13}}right)bf{v}.$$
The projection matrix is given by
$$P=begin{bmatrix}uparrow & uparrow\ textbf{proj}_{bf{v}}bf{e_1} & textbf{proj}_{bf{v}}bf{e_2}\ downarrow & downarrow end{bmatrix}=begin{bmatrix}frac{4}{13} & frac{6}{13}\frac{6}{13} & frac{9}{13}end{bmatrix}$$
Now you can compute other projection matrices as well.






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%2f3030619%2fhow-to-find-projection-matrix-of-the-singular-matrix-onto-fundamental-subspaces%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









    0












    $begingroup$

    Projection of a vector $bf{u}$ along the vector $bf{v}$ is given by $textbf{proj}_{bf{v}}bf{u}=left(frac{bf{u} cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $bf{e}_1$ and $bf{e}_2$, i.e.
    $$textbf{proj}_{bf{v}}bf{e_1}=left(frac{bf{e}_1 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(frac{bf{e}_2 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$$
    Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $bf{v}=begin{bmatrix}2\3end{bmatrix}$. Then,
    $$textbf{proj}_{bf{v}}bf{e_1}=left(mathrm{frac{2}{13}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(mathrm{frac{3}{13}}right)bf{v}.$$
    The projection matrix is given by
    $$P=begin{bmatrix}uparrow & uparrow\ textbf{proj}_{bf{v}}bf{e_1} & textbf{proj}_{bf{v}}bf{e_2}\ downarrow & downarrow end{bmatrix}=begin{bmatrix}frac{4}{13} & frac{6}{13}\frac{6}{13} & frac{9}{13}end{bmatrix}$$
    Now you can compute other projection matrices as well.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Projection of a vector $bf{u}$ along the vector $bf{v}$ is given by $textbf{proj}_{bf{v}}bf{u}=left(frac{bf{u} cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $bf{e}_1$ and $bf{e}_2$, i.e.
      $$textbf{proj}_{bf{v}}bf{e_1}=left(frac{bf{e}_1 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(frac{bf{e}_2 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$$
      Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $bf{v}=begin{bmatrix}2\3end{bmatrix}$. Then,
      $$textbf{proj}_{bf{v}}bf{e_1}=left(mathrm{frac{2}{13}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(mathrm{frac{3}{13}}right)bf{v}.$$
      The projection matrix is given by
      $$P=begin{bmatrix}uparrow & uparrow\ textbf{proj}_{bf{v}}bf{e_1} & textbf{proj}_{bf{v}}bf{e_2}\ downarrow & downarrow end{bmatrix}=begin{bmatrix}frac{4}{13} & frac{6}{13}\frac{6}{13} & frac{9}{13}end{bmatrix}$$
      Now you can compute other projection matrices as well.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Projection of a vector $bf{u}$ along the vector $bf{v}$ is given by $textbf{proj}_{bf{v}}bf{u}=left(frac{bf{u} cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $bf{e}_1$ and $bf{e}_2$, i.e.
        $$textbf{proj}_{bf{v}}bf{e_1}=left(frac{bf{e}_1 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(frac{bf{e}_2 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$$
        Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $bf{v}=begin{bmatrix}2\3end{bmatrix}$. Then,
        $$textbf{proj}_{bf{v}}bf{e_1}=left(mathrm{frac{2}{13}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(mathrm{frac{3}{13}}right)bf{v}.$$
        The projection matrix is given by
        $$P=begin{bmatrix}uparrow & uparrow\ textbf{proj}_{bf{v}}bf{e_1} & textbf{proj}_{bf{v}}bf{e_2}\ downarrow & downarrow end{bmatrix}=begin{bmatrix}frac{4}{13} & frac{6}{13}\frac{6}{13} & frac{9}{13}end{bmatrix}$$
        Now you can compute other projection matrices as well.






        share|cite|improve this answer









        $endgroup$



        Projection of a vector $bf{u}$ along the vector $bf{v}$ is given by $textbf{proj}_{bf{v}}bf{u}=left(frac{bf{u} cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$. So to get the projection matrix we can see what happens to the standard basis vectors $bf{e}_1$ and $bf{e}_2$, i.e.
        $$textbf{proj}_{bf{v}}bf{e_1}=left(frac{bf{e}_1 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(frac{bf{e}_2 cdot bf{v}}{bf{v} cdot bf{v}}right)bf{v}$$
        Suppose we want the projection matrix for the fundamental space $C(A^T)$ so $bf{v}=begin{bmatrix}2\3end{bmatrix}$. Then,
        $$textbf{proj}_{bf{v}}bf{e_1}=left(mathrm{frac{2}{13}}right)bf{v} qquad textbf{proj}_{bf{v}}bf{e_2}=left(mathrm{frac{3}{13}}right)bf{v}.$$
        The projection matrix is given by
        $$P=begin{bmatrix}uparrow & uparrow\ textbf{proj}_{bf{v}}bf{e_1} & textbf{proj}_{bf{v}}bf{e_2}\ downarrow & downarrow end{bmatrix}=begin{bmatrix}frac{4}{13} & frac{6}{13}\frac{6}{13} & frac{9}{13}end{bmatrix}$$
        Now you can compute other projection matrices as well.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 2:16









        Anurag AAnurag A

        25.9k12249




        25.9k12249






























            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%2f3030619%2fhow-to-find-projection-matrix-of-the-singular-matrix-onto-fundamental-subspaces%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

            Tonle Sap (See)

            Image
            Tonle Sap Satellitenfoto (April 1994) Geographische Lage Kambodscha Zuflüsse Tonle-Sap-Fluss, Siem Reap River Abfluss Tonle-Sap-Fluss Orte in der Nähe Siem Reap, Battambang Daten Koordinaten 12° 52′ 34″  N , 104° 4′ 23″  O 12.876 104.073 Koordinaten: 12° 52′ 34″  N , 104° 4′ 23″  O Höhe über Meeresspiegel f1 0,5 Meter Fläche 2.700–25.000 km² dep1 f5 Länge 250 km (maximal) dep1 f6 Breite 100 km (maximal) dep1 f7 Volumen 80 km³ (maximal) dep1 f8 Maximale Tiefe 14 m f10 Mittlere Tiefe 2–3 m (minimal) f11 Besonderheiten größter See Südostasiens Der Tonle-Sap-See (Khmer: បឹង​ទន្លេសាប , Aussprache: [ boeung tunleː saːp ]) in Kambodscha ist der größte See Südostasiens und eines der fischreichsten Binnengewässer der Erde. Inhaltsverzeichnis 1 Geographie 2 Wirtschaftliche Bedeutung 2.1 Verkehr 2.2 Reis 2.3 Fisch 3 Ökologie und Gefährdung 3.1 Überfischung...
            Read more

            I get strange results when I access the Sqlitedatabase with Unity C# via XAMPP

            Image
            0 <?php $servername = "localhost"; $username="root"; $password=""; $dbName="escape_room"; $user_username = $_POST['Input_user']; $user_password = $_POST['Input_pass']; $conn = new mysqli($servername, $username, $password, $dbName); if(!$conn){ die("Cound not Connect: " . mysqli_connect_error()); } $sql = "SELECT pass FROM escape_room WHERE user = '".$user_username."' "; $result = mysqli_query($conn, $sql); //Get the result and confirm login if(mysqli_num_rows($result)>0){ while($row = mysqli_fetch_assoc($result)){ if($row['pass'] == $user_password){ echo "login success"; ...
            Read more

            Guatemaltekische Davis-Cup-Mannschaft

            Image
            Guatemala Kapitän Anthony Vásquez Aktuelles ITF-Ranking 73 Statistik Erste Teilnahme 1990 Davis-Cup-Teilnahmen 22 Bestes Ergebnis Amerika-Gruppenzone II HF (1995, 2000, 2001) Ewige Bilanz 42:30 Erfolgreichste Spieler Meiste Siege gesamt Daniel Chávez Morales (29) Meiste Einzelsiege Cristian Paiz (21) Meiste Doppelsiege Daniel Chávez Morales (17) Bestes Doppel Jacobo Chávez / Daniel Chávez Morales (9) Meiste Teilnahmen Cristian Paiz (32) Meiste Jahre Daniel Chávez Morales (10) Letzte Aktualisierung der Infobox: 3. Juni 2012 Die guatemaltekische Davis-Cup-Mannschaft ist die Tennisnationalmannschaft Guatemalas. Geschichte | 1990 nahm Guatemala erstmals am Davis Cup teil. Das beste Resultat erreichte die Mannschaft jeweils 1995, 2000 und 2001 mit dem Halbfinale in der Amerika-Gruppenzone II. Bester Spieler ist Daniel Chávez Morales mit 29 Siegen bei insgesamt 31 Teilnahmen. Rekord...
            Read more