Linear algebra: proving equality equations











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If A and B are square, nonsingular matrices and X is a square matrix, then
$ (A + XBX^T)^{−1} = A^{−1} − A^{−1}X(B^{−1} + X^TA^{−1}X)^{−1}X^TA^{−1} $



This equation so complex, I can't see the key variable to transform and solve it. Plz help me!










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    If A and B are square, nonsingular matrices and X is a square matrix, then
    $ (A + XBX^T)^{−1} = A^{−1} − A^{−1}X(B^{−1} + X^TA^{−1}X)^{−1}X^TA^{−1} $



    This equation so complex, I can't see the key variable to transform and solve it. Plz help me!










    share|cite|improve this question









    New contributor




    AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















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      If A and B are square, nonsingular matrices and X is a square matrix, then
      $ (A + XBX^T)^{−1} = A^{−1} − A^{−1}X(B^{−1} + X^TA^{−1}X)^{−1}X^TA^{−1} $



      This equation so complex, I can't see the key variable to transform and solve it. Plz help me!










      share|cite|improve this question









      New contributor




      AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      If A and B are square, nonsingular matrices and X is a square matrix, then
      $ (A + XBX^T)^{−1} = A^{−1} − A^{−1}X(B^{−1} + X^TA^{−1}X)^{−1}X^TA^{−1} $



      This equation so complex, I can't see the key variable to transform and solve it. Plz help me!







      linear-algebra linear-transformations






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      AnNg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question









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      edited yesterday









      Crazy for maths

      4737




      4737






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          First premultiply RHS by $A+XBX^T$, you will get identity matrix.



          Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.



          Therefore, it will be the inverse of this matrix by definition.



          I am showing the procedure for postmultiplication for instance,



          $$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
          $Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
          $Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$



          Now, let $B^{-1}+X^TA^{-1}X = M$,



          In order to prove that this expression is I, we have to show that,
          $$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
          $$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
          $$Longleftrightarrow MB = I+X^TA^{-1}XB$$
          $$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
          $$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$



          Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
          Hope it is helpful.






          share|cite|improve this answer























          • You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
            – AnNg
            22 hours ago












          • I have edited the question to show how it will occur.
            – Crazy for maths
            21 hours ago










          • many thanks to you
            – AnNg
            20 hours ago










          • For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
            – AnNg
            20 hours ago












          • That is what I am doing in the last step.
            – Crazy for maths
            20 hours ago











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          up vote
          0
          down vote



          accepted










          First premultiply RHS by $A+XBX^T$, you will get identity matrix.



          Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.



          Therefore, it will be the inverse of this matrix by definition.



          I am showing the procedure for postmultiplication for instance,



          $$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
          $Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
          $Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$



          Now, let $B^{-1}+X^TA^{-1}X = M$,



          In order to prove that this expression is I, we have to show that,
          $$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
          $$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
          $$Longleftrightarrow MB = I+X^TA^{-1}XB$$
          $$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
          $$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$



          Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
          Hope it is helpful.






          share|cite|improve this answer























          • You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
            – AnNg
            22 hours ago












          • I have edited the question to show how it will occur.
            – Crazy for maths
            21 hours ago










          • many thanks to you
            – AnNg
            20 hours ago










          • For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
            – AnNg
            20 hours ago












          • That is what I am doing in the last step.
            – Crazy for maths
            20 hours ago















          up vote
          0
          down vote



          accepted










          First premultiply RHS by $A+XBX^T$, you will get identity matrix.



          Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.



          Therefore, it will be the inverse of this matrix by definition.



          I am showing the procedure for postmultiplication for instance,



          $$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
          $Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
          $Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$



          Now, let $B^{-1}+X^TA^{-1}X = M$,



          In order to prove that this expression is I, we have to show that,
          $$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
          $$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
          $$Longleftrightarrow MB = I+X^TA^{-1}XB$$
          $$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
          $$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$



          Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
          Hope it is helpful.






          share|cite|improve this answer























          • You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
            – AnNg
            22 hours ago












          • I have edited the question to show how it will occur.
            – Crazy for maths
            21 hours ago










          • many thanks to you
            – AnNg
            20 hours ago










          • For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
            – AnNg
            20 hours ago












          • That is what I am doing in the last step.
            – Crazy for maths
            20 hours ago













          up vote
          0
          down vote



          accepted







          up vote
          0
          down vote



          accepted






          First premultiply RHS by $A+XBX^T$, you will get identity matrix.



          Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.



          Therefore, it will be the inverse of this matrix by definition.



          I am showing the procedure for postmultiplication for instance,



          $$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
          $Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
          $Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$



          Now, let $B^{-1}+X^TA^{-1}X = M$,



          In order to prove that this expression is I, we have to show that,
          $$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
          $$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
          $$Longleftrightarrow MB = I+X^TA^{-1}XB$$
          $$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
          $$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$



          Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
          Hope it is helpful.






          share|cite|improve this answer














          First premultiply RHS by $A+XBX^T$, you will get identity matrix.



          Then postmultiply RHS by $A+XBX^T$, you will get identity matrix again.



          Therefore, it will be the inverse of this matrix by definition.



          I am showing the procedure for postmultiplication for instance,



          $$(A^{-1}-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1})(A+XBX^T)$$
          $Longrightarrow I+A^{-1}XBX^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^T-A^{-1}X(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XBX^T$
          $Longrightarrow I+A^{-1}X[B-(B^{-1}+X^TA^{-1}X)^{-1}-(B^{-1}+X^TA^{-1}X)^{-1}X^TA^{-1}XB]X^T$



          Now, let $B^{-1}+X^TA^{-1}X = M$,



          In order to prove that this expression is I, we have to show that,
          $$B-M^{-1}-M^{-1}X^TA^{-1}XB = 0$$
          $$Longleftrightarrow B = M^{-1}(I+X^TA^{-1}XB)$$
          $$Longleftrightarrow MB = I+X^TA^{-1}XB$$
          $$Longleftrightarrow M = (I+X^TA^{-1}XB)B^{-1}$$
          $$Longleftrightarrow M = B^{-1}+X^TA^{-1}X$$



          Hence, this turns out to be I and similarly the other one will be I(After quite lengthy calculation).
          Hope it is helpful.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 21 hours ago

























          answered yesterday









          Crazy for maths

          4737




          4737












          • You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
            – AnNg
            22 hours ago












          • I have edited the question to show how it will occur.
            – Crazy for maths
            21 hours ago










          • many thanks to you
            – AnNg
            20 hours ago










          • For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
            – AnNg
            20 hours ago












          • That is what I am doing in the last step.
            – Crazy for maths
            20 hours ago


















          • You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
            – AnNg
            22 hours ago












          • I have edited the question to show how it will occur.
            – Crazy for maths
            21 hours ago










          • many thanks to you
            – AnNg
            20 hours ago










          • For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
            – AnNg
            20 hours ago












          • That is what I am doing in the last step.
            – Crazy for maths
            20 hours ago
















          You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
          – AnNg
          22 hours ago






          You mean that I premultiply firstly RHS by $A+XBX^T$, you will get identity matrix. And then, postmultiply LHS by $B^{−1}+X^TA^{−1}X$, you will get identity matrix again. After that, I solve the adjusted equation and get the result which equals iif X is orthogonal matrix.
          – AnNg
          22 hours ago














          I have edited the question to show how it will occur.
          – Crazy for maths
          21 hours ago




          I have edited the question to show how it will occur.
          – Crazy for maths
          21 hours ago












          many thanks to you
          – AnNg
          20 hours ago




          many thanks to you
          – AnNg
          20 hours ago












          For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
          – AnNg
          20 hours ago






          For the expression of M, can I expand in this way? $B=M^{−1}(I+X^TA^{−1}XB) Leftrightarrow MB = I + IX^TA^{-1}XB Leftrightarrow MI = IB^{-1}+IX^TA^{-1}XI$, then I get the result
          – AnNg
          20 hours ago














          That is what I am doing in the last step.
          – Crazy for maths
          20 hours ago




          That is what I am doing in the last step.
          – Crazy for maths
          20 hours ago










          AnNg is a new contributor. Be nice, and check out our Code of Conduct.










           

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