= Solution
An <eigenvalue> of a <matrix> $A$ is a <scalar> $\lambda$ for which $Av=\lambda v$ for some nonzero <eigenvector> $v$. Its corresponding <eigenspace> is
$$
E_\lambda=\ker(A-\lambda I).
$$
Write $v=(a,b,c,d)^T$. The displayed matrix is the <outer product> $A=vv^T$, so
$$
Ax=v(v^Tx)=v(v\mathbin{\cdot}x).
$$
Its <column space> is contained in $\operatorname{span}\{v\}$ and is nonzero because $Av=\lVert v\rVert^2v\ne0$. Hence $A$ is a <rank-one matrix>.
The vector $v$ is an eigenvector with eigenvalue $\lVert v\rVert^2=a^2+b^2+c^2+d^2$, and every vector in the <orthogonal complement> $v^\perp$ has eigenvalue $0$. Thus
$$
E_{\lVert v\rVert^2}=\operatorname{span}\{v\},
\qquad E_0=v^\perp.
$$
Since $\mathbb R^4=\operatorname{span}\{v\}\oplus v^\perp$ is a <direct sum> of these eigenspaces, $A$ has an <eigenbasis> and is therefore a <diagonalizable matrix>[diagonalizable].
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