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I have the following matrix equation

$$(A x - y)^T \cdot A = 0_n^T$$

Whereby $x \in \mathbb{R}^n$, $y \in \mathbb{R}^d$, $A \in \mathbb{R}^{d \times n}$. A is not invertible, but $A^TA$ is invertible and the task is to solve for $x$.

$0_n$ is a $n$-dim vector of 0s, $(0,0,\dots,0)^T$ and $^T$ is the transpose of a matrix.

In order to solve it I transformed the equations in the following way:

\begin{align*} (Ax - y)^T \cdot A &= 0_n^T\\ (x^TA^T-y^T) \cdot A &= 0_n^T\\ x^TA^T \cdot A - y^T A &= 0_n^T\\ x^TA^T \cdot A &= y^T A\\ x^T &= (y^T \cdot A) \cdot (A^TA)^{-1}\\ x &= ((y^T \cdot A) \cdot (A^TA)^{-1})^T\\ x &= A^T \cdot y \cdot ((A^TA)^{-1})^T \end{align*}

Is this legal or have I missed something?

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what do you mean by $A^{T}$ ? –  Abdelmajid Khadari Apr 15 '12 at 13:14
    
Isn't this the hard way? What's wrong with $Ax-y=0,Ax=y,A^tAx=A^ty,x=(A^tA)^{-1}A^ty$? –  Gerry Myerson Apr 15 '12 at 13:14
    
@Matrix, presumably the transpose. –  Gerry Myerson Apr 15 '12 at 13:15
    
@GerryMyerson Sorry I forgot the additional $\cdot A$ on the left side of the equation –  Mahoni Apr 15 '12 at 13:18
    
if $A^{T}$ is the transpose, so i think that equation is equivalent to $Ax - y= 0_n$, because ${(A^T)}^{T}=A$. –  Abdelmajid Khadari Apr 15 '12 at 13:20
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up vote 3 down vote accepted

Up to the sixth line of displayed equations, you are fine, thus $x=(y^TAB^{-1})^T$ with $B=A^TA$ (and $B$ is assumed to be invertible hence $B^{-1}$ exists), that is, $x=(B^{-1})^TA^Ty$ (do not forget that $(MN)^T=N^TM^T$, if only for dimension reasons).

Now, $(B^{-1})^T=(B^T)^{-1}$ (this always holds) and $B^T=(A^TA)^T=A^T(A^T)^T=A^TA=B$ hence $$ x=B^{-1}A^Ty,\qquad B=A^TA. $$ Dimensional analysis: The matrix $A$ has dimension $d\times n$ hence $A^T$ has dimension $n\times d$ and $B=A^TA$ has dimension $n\times$ d $\times$ d $ \times n=n\times n$, as well as $B^{-1}$. As a column vector, $y$ has dimension $d\times1$ hence $B^{-1}A^Ty$ has dimension $n\times$ n $\times$ n $\times$ d $\times$ d $\times 1=n\times 1$. All is well.

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