The existence of a matrix?











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2
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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















up vote
2
down vote

favorite












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













up vote
2
down vote

favorite









up vote
2
down vote

favorite











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

243




243








  • 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

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up vote
6
down vote













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




























    up vote
    6
    down vote













    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




























      up vote
      4
      down vote













      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


















      up vote
      1
      down vote













      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

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        4 Answers
        4






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        active

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        up vote
        6
        down vote













        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

























          up vote
          6
          down vote













          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























            up vote
            6
            down vote










            up vote
            6
            down vote









            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






















                up vote
                6
                down vote













                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

























                  up vote
                  6
                  down vote













                  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























                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    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.

                    21.6k22455




                    21.6k22455






















                        up vote
                        4
                        down vote













                        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















                        up vote
                        4
                        down vote













                        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













                        up vote
                        4
                        down vote










                        up vote
                        4
                        down vote









                        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

                        3724




                        3724












                        • 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










                        up vote
                        1
                        down vote













                        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



























                          up vote
                          1
                          down vote













                          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

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            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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