Let $x$ and $y$ be $(n \times 1)$ column vectors.
Then $xy^T$ is an $(n \times n)$ matrix.
And $y^Tx$ is a $(1 \times 1)$ scalar.
The characteristic polynomial of the matrix $A = xy^T$ has two distinct eigenvalues:
(1) One eigenvalue is $\lambda_1 = y^Tx$, which is a non-zero scalar.
(2) The remaining $(n-1)$ eigenvalues are all zero.
Proof:
Consider the eigenvalue equation $Av = \lambda v$.
Substituting $A=xy^T$, we get:
$$(xy^T)v = \lambda v$$
$$x(y^Tv) = \lambda v$$
Since $y^Tv$ is a scalar, we can write it as a constant, say $c$.
$$xc = \lambda v$$
If $v=x$, we have:
$$x(y^Tx) = \lambda x$$
This implies $\lambda = y^Tx$.
So, one eigenvalue is $\lambda_1 = y^Tx$.
Analysis of Rank:
The rank of the matrix $xy^T$ is 1 (since it's an outer product of non-zero vectors).
A matrix with rank 1 has only one non-zero eigenvalue.
The remaining $(n-1)$ eigenvalues must be zero.
Therefore, the eigenvalues are $\lambda_1 = y^Tx$ and $0$.
SUMMARY :
$A = xy^T, \quad \lambda_1 = y^T x, \quad \lambda_2 = 0,
\text{rank} = 1$
Journey through an Example😃 :
Let $x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $y = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$. Then the matrix $A$ is given by:
$$A = xy^T = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$$
The characteristic equation for this $2 \times 2$ matrix is $\det(A - \lambda I) = 0$.
$$\det \begin{bmatrix} 2-\lambda & 1 \\ 4 & 2-\lambda \end{bmatrix} = 0$$
$$(2-\lambda)(2-\lambda) - (1)(4) = 0$$
$$4 - 4\lambda + \lambda^2 - 4 = 0$$
$$\lambda^2 - 4\lambda = 0$$
$$\lambda(\lambda - 4) = 0$$
The eigenvalues are $\lambda_1 = 4$ and $\lambda_2 = 0$.
Let's check our formula for the non-zero eigenvalue, which should be the trace of the matrix, $y^T x$:
$$y^T x = \begin{bmatrix} 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = (2 \times 1) + (1 \times 2) = 2 + 2 = 4$$
In the exam, you'd recognize it's a rank-1 matrix, know that $n-1$ eigenvalues are zero, and the one non-zero eigenvalue must be the trace, which is $y^T x$(scalar value).
