1
$\begingroup$

Let $x$ be a non-zero column vector and $$ P= ( xx^\intercal)/(x^\intercal x) $$ be an $n\times n$ matrix where $n>1$. Then what is the nature of matrix $P$? (here $x^\intercal$ denotes transpose of $x$)

$\endgroup$
7
  • 7
    $\begingroup$ Could you define the nature of a matrix? $\endgroup$ Feb 9 '17 at 8:37
  • $\begingroup$ To find out whether P is symmetric, idempotent or orthogonal $\endgroup$
    – Varsha Rao
    Feb 9 '17 at 10:58
  • $\begingroup$ What did you try? $\endgroup$ Feb 9 '17 at 14:10
  • $\begingroup$ I tried taking transpose on both sides but I got stuck. $\endgroup$
    – Varsha Rao
    Feb 9 '17 at 14:34
  • $\begingroup$ Use $ (xx^\prime)^\prime = (x^\prime)^\prime x^\prime = xx^\prime.$ $\endgroup$
    – whuber
    Feb 9 '17 at 15:00
5
$\begingroup$

Checking the symmetry of this matrix is pretty straightforward. Use the property $(xy^T)^T = yx^T$ and the fact that $x^Tx$ is a scalar.

$P^T = \left(\frac{xx^T}{x^Tx}\right)^T = \frac{(xx^T)^T}{x^Tx} = \frac{(x^T)^T x^T}{x^Tx} = \frac{xx^T}{x^Tx} = P.$

Checking idempotence of the matrix also relies on the fact the $x^Tx$ is a scalar.

$PP = \frac{xx^T}{x^Tx}\frac{xx^T}{x^Tx} = \frac{xx^Txx^T}{(x^Tx)(x^Tx)} = \frac{x(x^Tx)x^T}{(x^Tx)^2} = \frac{x^Tx}{x^Tx}\frac{xx^T}{x^Tx} = \frac{xx^T}{x^Tx} = P.$

Checking orthogonality is just a matter of showing that $PP^T = P^TP = I$. I claim that this follows from the fact that the matrix is both symmetric and idempotent. Can you see why that is true?

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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