2

Let $A$ be a $m \times n$ matrix of rank $m$ with $n > m$. If for some non-zero real number $\alpha$, we have $x^t A A^t x = \alpha x^t x$, for all $x \in \Bbb{R}^m$ then $A^tA$ has exactly two distinct eigenvalues where $0$ is an eigenvalue with multiplicity $n-m$ and $\alpha$ is a non-zero eigenvalue.

How do I approach this question?. I could only think that $AA^t$ would be an $m \times m$ matrix.I also cannot take determinant on both the sides of the euation above as all are of different order matrices which i thought would solve the problem, also i think there must be some kind trick associated with the rank of the matrix $A$.

BAYMAX
  • 5,094

1 Answers1

2

If $x^tAA^tx=\alpha x^tx$ than $ x^t(AA^t-\alpha I)x=0$ for all $x$ and so by identity polarization $AA^t-\alpha I=0$ and $AA^t=\alpha I$

In fact you can prove that if $Y$ is a symmetric matrix such that $x^tYx=0$ for each $x$ then $Y$ must be zero:

You can observe that

$(y-x)^tY(y-x)=y^tYy-y^tYx-x^tYy+x^tYx$

but

$x^tYy=(x^tYy)^t=y^tY^tx=y^tYx$

because $Y$ is a symmetric matrix so you have the following equation:

$2y^tYx=y^tYy+x^tYx-(y-x)^tY(y-x)$

that is zero for each $y,x$ so

$y^tYx=0$ for each $y,x$ and in particular if you choose $y=Yx$ you have that

$||Yx||^2=(Yx)^tYx=0$ for each $x$ so $Y=O$

Federico Fallucca
  • 10,123