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 $x$)

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    $\begingroup$ Could you define the nature of a matrix? $\endgroup$ – Matthew Gunn 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? $\endgroup$ – kjetil b halvorsen 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

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?


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