I came across the following exercise from my professor and had no idea how to start. I tried it with the standard basis in $\mathbb{R}^{2}$, which results in a rank of $1$ and eigenvalue $0$, but this must be shown without an example using an explanation of different properties. The literal question can be found below:
Given two orthonormal vectors x, y in $\mathbb{R}^{2}$.
Consider the matrix $A = xy^T$.
- What is the rank of A, and what does it tell you about the eigenvalues?
- Calculate $A^2$ and deduce what the eigenvalues of $A$ are.
Note: I do not want to see an example, not even with vectors and matrices with letters!
Write $x_i=\langle x,e_i\rangle$ for $e_1,e_2$ the standard basis, and the same for $y$. As $x,y$ are orthogonal, we know that $x_1y_1 +x_2y_2 = 0$. Note that $M=(x_iy_j)_{i,j}$ in the standard basis. Computing the characteristic polynomial, we see that $$ \chi(\lambda) = \lambda^2-\lambda(x_1y_1+x_2y_2)+x_1x_2y_1y_2-x_1x_2y_1y_2 = \lambda^2$$ using the equality above. We see that the only eigenvalue is $0$, thus the matrix must have rank $1$ or $0$. It is easy to check that $M$ being the zero matrix implies that either $x$ or $y$ is the zero vector, which is not possible. Thus it has rank $1$.
Edit: Following @Stiftungwarentest 's hint provides a much shorter and cleaner proof.