Find a $ 2 \times 2$ symmetric matrix if we know the eigenvalues and one eigenvector. Suppose that $A$ is a symmetric matrix $2 \times 2$ with eigenvalues $\lambda_{1}=1$ y $\lambda_{2}=9$, and $v_{1}=(1,3)^{T}$ is a eigenvector for $\lambda_{1}$.
a) Find a eigenvector for $\lambda_{2}$.
b)Find the matrix $A$.
I thought that $v_{1}$ and $v_{2}$ has to be lineary independent so it doesn't matter what $v_{2}$ I choose since it has only fulfill one condition, for example $v_{2}=(1,4)$ but I didn't obtain $A$ such that have the two eigenvalues.
Can you help me?
 A: In general, two eigenvectors with different eigenvalues must be linearly independent, true. But for a symmetric matrix we know even more: We know that they are orthogonal!
To show this, note first that the symmetry of $A$ means that $v_1^TAv_2=v_2^TAv_1$. Then use the fact that they are both eigenvectors to get $9v_1^Tv_2=1v_2^Tv_1$. This can only happen if $v_1^Tv_2=v_2^Tv_1=0$.
Now that we know they are orthogonal, this means we know we can set $v_2=(3,-1)^T$ (or any nonzero multiple of this; which one we choose ultimately doesn't matter).
Finally, to construct the matrix, I will use diagonalization. If we use $v_1,v_2$ as basis, the linear transformation $A$ is represented by the matrix $$\begin{bmatrix}1&0\\0&9\end{bmatrix}$$Translating this into the standard basis is just a matter of inserting translation matrices: the matrix whose columns are $v_1$ and $v_2$, and its inverse. We get
$$
\begin{bmatrix}v_1&v_2\end{bmatrix}\begin{bmatrix}1&0\\0&9\end{bmatrix}\begin{bmatrix}v_1&v_2\end{bmatrix}^{-1}
$$
Insert here and calculate, and that will give you the matrix representation of the linear transformation $A$ in the standard basis.
A: It is a standard result in linear algebra that if $A$ is symmetric, then eigenvectors from different eigenspace are orthogonal. In your question, since the two eigenvalues are different, the two vectors must be orthogonal. If $v_1=[1,3]^T$, $v_2$ can be taken as $[-3,1]^T $. The matrix $A$ can be computed easily, it is
$$\begin{bmatrix}8.2& -2.4\\ -2.4& 1.8 \end{bmatrix}.$$
