Determining eigenvalues, eigenvectors of $A\in \mathbb{R}^{n\times n}(n\geq 2)$.

HINT 1: Your matrix $A$ is $$(a-b)I + b e e^T$$

Can you now compute the eigenvalues?

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HINT 2: Make use of the fact that $\text{eigen}(\lambda A) = \lambda \text{eigen}(A)$

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HINT 3: $\text{eigen}(I + \alpha e e^T)$ are $1 + n\alpha$ and $1$($n-1$ times).

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$$\text{eigen} ((a-b)I + b e e^T) = (a-b) \text{eigen} \left( I + \dfrac{b}{a-b} e e^T\right)$$ Hence, all we need is to find the eigenvalues of $I + \alpha ee^T$, where $\alpha = \dfrac{b}{a-b}$ in our case. Note that $ee^T$ is a rank one matrix and its eigen values are $e^Te = n$ and $n-1$ zeros. If $\lambda$ is an eigen value of $I + \alpha ee^T$, then $$\det (I + \alpha ee^T - \lambda I) = \alpha^n \det \left(ee^T + \dfrac{(1-\lambda)}{\alpha}I \right) = 0$$ This means that $-\dfrac{(1-\lambda)}{\alpha}$ are the eigenvalues of $ee^T$. Hence, we get that $$-\dfrac{(1-\lambda)}{\alpha} = n \text{ or }0 \text{ ($n-1$ times)}.$$ Hence, we get that $$\lambda = 1 + n \alpha, 1 \text{ ($n-1$ times)}$$ Hence, the eigenvalues of the initial matrix are $$\lambda = (a-b) + nb, (a-b) \text{ ($n-1$ times)}$$ The determinant of this matrix is $$((a-b)+nb)(a-b)^{n-1} = (a-b)^n + nb(a-b)^{n-1}$$

To take up the last part of the problem: if $\lambda=a-b$ then $A-\lambda I$ is a matrix of rank one, hence, of nullity $n-1$, so the eigenspace for the eigenvalue $\lambda$ has dimension $n-1$. That means that to diagonalize $A$ it's not enough to find one eigenvector; you must find $n-1$ linearly independent eigenvectors; $n-1$ linearly independent vectors in the nullspace of the all-$b$ matrix. Can you do that?