Find $2\times 2$ symmetric matrix $A$ given two eigenvalues and one eigenvector I am having trouble finding the symmetric matrix $A$ given eigenvalues $1$ and $4$ and eigenvector $(1, 1)$ corresponding to eigenvalue $1$.
I feel like I'd have to use the equation $A=PD(P^{-1})$, but I'm having trouble finding the matrix $P$ if I can't find the second eigenvector.
Any help is appreciated, thanks!
 A: To offer a slightly different perspective, due to the Spectral Theorem you have 
$$
A=P_1+4P_2,$$
where $P_1$ is the orthogonal projection onto the span of $(1,1)^t$, and $P_2$ is orthogonal to $P_1$; that is, $P_2=I-P_1$. Thus 
$$
P_1=\tfrac12\,\begin{bmatrix} 1&1\end{bmatrix}\begin{bmatrix} 1\\1\end{bmatrix}=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}.
$$
And then
$$
A=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}+4\begin{bmatrix}1/2&-1/2\\-1/2&1/2\end{bmatrix}=\begin{bmatrix}5/2&-3/2\\-3/2&5/2 \end{bmatrix}
$$
A: Let :
$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
Since the eigenvalues are $\lambda =1 $ and $\lambda = 4$, it must be 
$$\det(A-\lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda\end{vmatrix} = (a-\lambda)(d-\lambda)-bc $$
such that $\lambda = 1$ and $\lambda = 4$ are solutions to the equation :
$$(a-\lambda)(d-\lambda)-bc=0$$
Since $(1,1)^\mathbf{T}$ is an eigenvector of $A$ for $\lambda = 1$, it is :
$$(A- I)(1,1)^\mathbf{T} = 0 \Rightarrow \begin{pmatrix} a - 1 & b \\ c & d - 1\end{pmatrix}\begin{pmatrix}1 \\ 1 \end{pmatrix} =\begin{pmatrix}0 \\ 0 \end{pmatrix} $$
$$\Leftrightarrow$$
$$\begin{cases} a-1 + b = 0 \\c + d-1 = 0\end{cases}$$
Can you now find $a,b,c$ and $d$ ?
