Can't understand this question from Matrices

Question

Let $P=\frac{\underline{x}\underline{x}^{T}}{\underline{x}^{T}\underline{x}}$ be an $n×n (n>1)$ matrix, where $\underline{x}$ is a nonzero column vector. Then which one of the following statement is False?

• P is idempotent
• P is Orthogonal
• P is symmetric
• Rank of P is one

As $\underline{x}$ is a column vector should I assume it to be an $n×1$ column vector in the given context then (from here I am not using underline to represent vector I used it because it was given in the question) $xx^T$ is $n×n$ matrix and $x^Tx$ is $1×1$ scalar while individually trying to see which option is False I did For P is idempotent $P^2=P$ $$P^2=\left(\frac{\underline{x}\underline{x}^{T}}{\underline{x}^{T}\underline{x}}\right)^2=\frac{\underline{x}\left(\underline{x}^{T}\underline{x}\right)\underline{x}^{T}}{\underline{x}^{T}\underline{x}\underline{x}^{T}\underline{x}}$$ now should I consider $x^Tx$ as a scalar and take it out and prove that the given matrix is idempotent? If it's true after this matrix can be shown symmetric easily implies $PP^T=PP$ which is equal to $P$ since idempotent so the matrix is not Orthogonal hence 2nd option is FALSE but what about the fourth option [Rank of P is one] how that can be true I don't know how to prove it. Any suggestions

Answer 1

The option: "$P$ is orthogonal" is false.

As you have observed $P$ is symmetric and $PP^T = P^2$. So proving P is idempotent (i.e. $P^2 = P$) will eliminate the choice P is orthogonal ($P^2 = I$).

In your equation $P^2 = \frac{x(x^Tx)x^T}{(x^Tx)(x^Tx)}$ take out the scalar $x^Tx$ and cancel with the denominator, you get $P^2=P$.

Rank of $P$ is number of linearly independent rows (or columns) of $P$, let us denote $x^T = r$ ($r$ is a row vector) and $r = x^T = \{ a_1, a_2, a_3, \cdots a_n \}$. Then $xx^T$ is a matrix with rows $a_1r,a_2r , \cdots$.

So all other rows depend on only one row. So the rank of $P$ is one.

Answer 2

Hint for the last part: $Pv=\frac{x x^T}{x^T x}v=\frac{x^T v}{x^T x}x$, a scalar multiple of $x$.

Answer 3

Consider $$ y=\frac{1}{\sqrt{x^Tx}}x $$ so that your matrix $P$ is, more simply, $$ P=yy^T $$ Since $y\ne0$, the matrix $P$ is nonzero (just consider its diagonal coefficients, one of them is nonzero). The rank of a product of matrices cannot be greater than the rank of each factor, so the rank of $P$ is at most $1$. From this and $P\ne0$, we have that the rank of $P$ is $1$.

Now $P$ is symmetric, because $P^T=(yy^T)^T=yy^T$; next $$ P^2=yy^Tyy^T=yy^T=P $$ because $y^Ty=1$ by construction. So $P$ is idempotent and not orthogonal, because the rank is $1<n$ and so $P^TP=P\ne I$.