The existence of a matrix?












2














Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$



What is the requirement for the matrix $A$? Thank you!






linear-algebra matrices







share|cite|improve this question




















  • 1




    You mean $rank(A)$?
    – Patricio
    Dec 4 at 8:02






  • 2




    @Yasmin, $A$ needs not be square, I think
    – Patricio
    Dec 4 at 8:03








  • 1




    @Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
    – Patricio
    Dec 4 at 8:10






  • 2




    @Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
    – lisyarus
    Dec 4 at 10:41
















2














Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$



What is the requirement for the matrix $A$? Thank you!






linear-algebra matrices







share|cite|improve this question




















  • 1




    You mean $rank(A)$?
    – Patricio
    Dec 4 at 8:02






  • 2




    @Yasmin, $A$ needs not be square, I think
    – Patricio
    Dec 4 at 8:03








  • 1




    @Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
    – Patricio
    Dec 4 at 8:10






  • 2




    @Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
    – lisyarus
    Dec 4 at 10:41














2












2








2







Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$



What is the requirement for the matrix $A$? Thank you!






linear-algebra matrices







share|cite|improve this question















Suppose that the matrix $Ain{mathbb{R}}^{ntimes r}$, $textrm{rank}(A)=r$, and $I_{n}$ is the identity matrix. Is there a matrix $Bin{mathbb{R}}^{rtimes n}$, such that
$$AB=I_{n}?$$



What is the requirement for the matrix $A$? Thank you!






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 at 8:13









Alex Silva

2,70331332




2,70331332










asked Dec 4 at 7:55









Wei Jiang

293




293








  • 1




    You mean $rank(A)$?
    – Patricio
    Dec 4 at 8:02






  • 2




    @Yasmin, $A$ needs not be square, I think
    – Patricio
    Dec 4 at 8:03








  • 1




    @Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
    – Patricio
    Dec 4 at 8:10






  • 2




    @Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
    – lisyarus
    Dec 4 at 10:41














  • 1




    You mean $rank(A)$?
    – Patricio
    Dec 4 at 8:02






  • 2




    @Yasmin, $A$ needs not be square, I think
    – Patricio
    Dec 4 at 8:03








  • 1




    @Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
    – Patricio
    Dec 4 at 8:10






  • 2




    @Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
    – lisyarus
    Dec 4 at 10:41








1




1




You mean $rank(A)$?
– Patricio
Dec 4 at 8:02




You mean $rank(A)$?
– Patricio
Dec 4 at 8:02




2




2




@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03






@Yasmin, $A$ needs not be square, I think
– Patricio
Dec 4 at 8:03






1




1




@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10




@Yasmin, that's my point, $A$ is $n times r$ and $n$ and $r$ need not be equal to each other.
– Patricio
Dec 4 at 8:10




2




2




@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41




@Yasmin In case $n=r$, the question states that $rank(A)=n$, so it is automatically invertible. I guess the meat of the question is what happens if $n neq r$.
– lisyarus
Dec 4 at 10:41










4 Answers
4






active

oldest

votes


















6














From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.






share|cite|improve this answer





























    6














    Consider
    $$
    A=begin{bmatrix}1\0end{bmatrix}
    $$
    then
    $$
    AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
    $$
    Do you see the trouble?






    share|cite|improve this answer





























      4














      One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.

      Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.

      If $n = r$, $B$ is the inverse of $A$.

      If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.

      Look at the corresponding Wikipendia entry for example.






      share|cite|improve this answer























      • The Moore-Penrose inverse is just one of many other possibilities for $B$.
        – A.Γ.
        Dec 4 at 8:46






      • 1




        @A.Γ. Of course. I was editing the answer when you commented it.
        – Damien
        Dec 4 at 8:50










      • OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
        – Federico Poloni
        Dec 4 at 17:49










      • @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
        – Damien
        Dec 4 at 18:17










      • I think OP is asking precisely an exercise about this basic point.
        – Federico Poloni
        Dec 4 at 18:37



















      1














      Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.






      share|cite|improve this answer























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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6














        From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.






        share|cite|improve this answer


























          6














          From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.






          share|cite|improve this answer
























            6












            6








            6






            From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.






            share|cite|improve this answer












            From $n=rank (AB)leqmin{rank (A),rank (B)}$ and also $rank(A)leqmin{n,r}$ you can deduce something about the size of $A$ and its rank.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 at 8:12









            AnyAD

            2,098812




            2,098812























                6














                Consider
                $$
                A=begin{bmatrix}1\0end{bmatrix}
                $$
                then
                $$
                AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
                $$
                Do you see the trouble?






                share|cite|improve this answer


























                  6














                  Consider
                  $$
                  A=begin{bmatrix}1\0end{bmatrix}
                  $$
                  then
                  $$
                  AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
                  $$
                  Do you see the trouble?






                  share|cite|improve this answer
























                    6












                    6








                    6






                    Consider
                    $$
                    A=begin{bmatrix}1\0end{bmatrix}
                    $$
                    then
                    $$
                    AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
                    $$
                    Do you see the trouble?






                    share|cite|improve this answer












                    Consider
                    $$
                    A=begin{bmatrix}1\0end{bmatrix}
                    $$
                    then
                    $$
                    AB=begin{bmatrix}1\0end{bmatrix}begin{bmatrix}b_1 & b_2end{bmatrix}=begin{bmatrix}b_1 & b_2\0 & 0end{bmatrix}.
                    $$
                    Do you see the trouble?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 at 8:23









                    A.Γ.

                    21.8k22455




                    21.8k22455























                        4














                        One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.

                        Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.

                        If $n = r$, $B$ is the inverse of $A$.

                        If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.

                        Look at the corresponding Wikipendia entry for example.






                        share|cite|improve this answer























                        • The Moore-Penrose inverse is just one of many other possibilities for $B$.
                          – A.Γ.
                          Dec 4 at 8:46






                        • 1




                          @A.Γ. Of course. I was editing the answer when you commented it.
                          – Damien
                          Dec 4 at 8:50










                        • OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                          – Federico Poloni
                          Dec 4 at 17:49










                        • @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                          – Damien
                          Dec 4 at 18:17










                        • I think OP is asking precisely an exercise about this basic point.
                          – Federico Poloni
                          Dec 4 at 18:37
















                        4














                        One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.

                        Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.

                        If $n = r$, $B$ is the inverse of $A$.

                        If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.

                        Look at the corresponding Wikipendia entry for example.






                        share|cite|improve this answer























                        • The Moore-Penrose inverse is just one of many other possibilities for $B$.
                          – A.Γ.
                          Dec 4 at 8:46






                        • 1




                          @A.Γ. Of course. I was editing the answer when you commented it.
                          – Damien
                          Dec 4 at 8:50










                        • OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                          – Federico Poloni
                          Dec 4 at 17:49










                        • @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                          – Damien
                          Dec 4 at 18:17










                        • I think OP is asking precisely an exercise about this basic point.
                          – Federico Poloni
                          Dec 4 at 18:37














                        4












                        4








                        4






                        One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.

                        Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.

                        If $n = r$, $B$ is the inverse of $A$.

                        If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.

                        Look at the corresponding Wikipendia entry for example.






                        share|cite|improve this answer














                        One of the solution for $B$ is the Moore-Penrose inverse $A^+$ of $A$.

                        Solution(s) exist as soon as $n le r$ and that $rank(A)=n$.

                        If $n = r$, $B$ is the inverse of $A$.

                        If $n < r$, the set of solutions corresponds to the Moore-Penrose inverse plus an element ($C$ such that $AC = 0$) owing to a set of dimension equal to $r-n$.

                        Look at the corresponding Wikipendia entry for example.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 4 at 9:02

























                        answered Dec 4 at 8:43









                        Damien

                        4474




                        4474












                        • The Moore-Penrose inverse is just one of many other possibilities for $B$.
                          – A.Γ.
                          Dec 4 at 8:46






                        • 1




                          @A.Γ. Of course. I was editing the answer when you commented it.
                          – Damien
                          Dec 4 at 8:50










                        • OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                          – Federico Poloni
                          Dec 4 at 17:49










                        • @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                          – Damien
                          Dec 4 at 18:17










                        • I think OP is asking precisely an exercise about this basic point.
                          – Federico Poloni
                          Dec 4 at 18:37


















                        • The Moore-Penrose inverse is just one of many other possibilities for $B$.
                          – A.Γ.
                          Dec 4 at 8:46






                        • 1




                          @A.Γ. Of course. I was editing the answer when you commented it.
                          – Damien
                          Dec 4 at 8:50










                        • OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                          – Federico Poloni
                          Dec 4 at 17:49










                        • @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                          – Damien
                          Dec 4 at 18:17










                        • I think OP is asking precisely an exercise about this basic point.
                          – Federico Poloni
                          Dec 4 at 18:37
















                        The Moore-Penrose inverse is just one of many other possibilities for $B$.
                        – A.Γ.
                        Dec 4 at 8:46




                        The Moore-Penrose inverse is just one of many other possibilities for $B$.
                        – A.Γ.
                        Dec 4 at 8:46




                        1




                        1




                        @A.Γ. Of course. I was editing the answer when you commented it.
                        – Damien
                        Dec 4 at 8:50




                        @A.Γ. Of course. I was editing the answer when you commented it.
                        – Damien
                        Dec 4 at 8:50












                        OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                        – Federico Poloni
                        Dec 4 at 17:49




                        OP specified that $operatorname{rank}(A)=r$, so this disqualifies cases with $r>n$.
                        – Federico Poloni
                        Dec 4 at 17:49












                        @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                        – Damien
                        Dec 4 at 18:17




                        @FedericoPoloni The problem can only have a solution if $n le r$. I don't think that OP ignores this basic point. Therefore, I assumed it was a typo when OP writes $rank(A)=r$. It is also why in this answer, I insisted on this point in the second line, to avoid confusion.
                        – Damien
                        Dec 4 at 18:17












                        I think OP is asking precisely an exercise about this basic point.
                        – Federico Poloni
                        Dec 4 at 18:37




                        I think OP is asking precisely an exercise about this basic point.
                        – Federico Poloni
                        Dec 4 at 18:37











                        1














                        Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.






                        share|cite|improve this answer




























                          1














                          Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.






                          share|cite|improve this answer


























                            1












                            1








                            1






                            Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.






                            share|cite|improve this answer














                            Yet another sufficient way of doing this. Since $rank(A) = r$, we have $rleq n$. Thinking in terms of columns of $A$ will help you see this. You picked $r$ vectors in a $n$-dimensional space. Since $rank(A)=r$, they are all linearly independent. Since you can have only $n$ independent vectors at most in a $n$-dimensional space, $rleq n$. In case $r=n$, then $B=A^{-1}$. In case $r<n$, $B$ doesn't exist.To see this, observe that determinant of RHS is $1$ ($neq 0$). Now determinant of LHS (i.e. AB) is zero as it is not a full-rank matrix. To see that, you can think of columns of $AB$ as a linear combination of columns $A$. Since you took $r$ independent vectors and constructed $n(>r)$ vectors out of it, they have to be dependent and hence $AB$ should have determinant zero.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 4 at 13:08

























                            answered Dec 4 at 13:00









                            dineshdileep

                            5,85711735




                            5,85711735






























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                                Сан-Квентин

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                                Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
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                                8-я гвардейская общевойсковая армия

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                                8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
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                                Алькесар

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                                Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
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