Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.