How do I solve that matrix operation?

Question

Let $x$ be a column vector in $\Bbb R^n$, and $x$ different from matrix zero. Let $A = \frac{ x x^{T}}{x^{T} x}$, T denotes tranpose. Calculate $A²$.

I hope that I am not the only one who was astonished with this exercise. Someone here knows what it means $\frac{ x x^{T}}{x^{T} x}$? I can imagine that a division is simple take the inverse of denominator, but so the exercise become so trivial that I discard it. Anyway, someone here knows what it means?

The numerator is an $n\times n$ matrix and the denominator is just a number. — Michal Adamaszek, Dec 29 '20 at 19:27
The calculation is as trivial as you might have thought, but it's nonetheless significant: It shows that $A$ is an example of a projection operator. — Semiclassical, Dec 29 '20 at 19:28

score 2 · Answer 1 · answered Dec 29 '20 at 19:27

If $x$ is the column vector$$\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix},$$then $x^T=\begin{bmatrix}a_1&a_2&\cdots&a_n\end{bmatrix}$ and so$$x^Tx=\begin{bmatrix}a_1&a_2&\cdots&a_n\end{bmatrix}\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}=a_1^{\,2}+a_2^{\,2}+\cdots+a_n^{\,2}.$$It's just a number.

score 1 · Answer 2 · answered Dec 29 '20 at 19:35

The components of $A$ are

$$ A_{ij}=\frac{x_i x_j}{|x|^2}$$

Where $|x|^2$ is the magnitude squared of $x$. Thus the components of $B=A^2$ are

$$ B_{ik} = A_{ij} A_{jk} =\frac{x_i x_j}{|x|^2} \frac{x_j x_k}{|x|^2}=\frac{x_i x_k}{|x|^2}=A_{ik}$$

Which are identical to the components of $A$. Thus $A^2=A$.

score 1 · Answer 3 · answered Dec 29 '20 at 20:46

If $x\in\mathbb R^n$ is nonzero and $A=\dfrac{xx^T}{x^Tx}$, then by associativity of matrix multiplication and scalar multiplication we obtain

$$\begin{array}{rcl}A^2&=&\left(\dfrac{xx^T}{x^Tx}\right)\left(\dfrac{xx^T}{x^Tx}\right)\\ &=&\dfrac{x(x^Tx)x^T}{(x^Tx)^2}\\ &=&\dfrac{x^Tx}{(x^Tx)^2}\cdot xx^T\\ &=&\dfrac{1}{x^Tx}\cdot xx^T\\ &=&\dfrac{xx^T}{x^Tx}\\ &=&A. \end{array}$$

How do I solve that matrix operation?

3 Answers3