# A symmetric matrix $A$ has eigenvalues 1 and 2. Find A if $(1, 1, 1)^T$ spans $E_2$

Let $A$ be the symmetric $3 \times 3$ matrix which has eigenvalues $1$ and $2$ and $E_2 = \text{span}\left(\begin{bmatrix} 1\\1\\1\end{bmatrix}\right)$. Find $A$.

What I have so far:

Let $\lambda_1=1$ and $\lambda_2=2$. Since any symmetric matrix is diagonalizable, the algebraic and geometric multiplicites of all eigenvalues must be equal. So if $d_i := \text{dim}(E_i)$ is the geometric multiplicity of $\lambda_i$ and $m_i$ is the algebraic multiplicity of $\lambda_i$, we get $1 = d_1 = m_1$ and $d_2 = m_2 = 3 - 1 = 2$ since $\sum d_i = \sum m_i = n$ and there are only two eigenvalues. So $E_1$ is a two-dimensional plane in $\Bbb R^3$. Since any symmetric matrix has also orthogonal eigenspaces, we know this plane $E_1$ must be orthogonal to the line $E_2$.

How do I proceed from here to find $E_1$?

Having found $E_1$ I would have three linearly independent eigenvectors which make a basis of $\Bbb R^3$. Then I would orthonormalize them and construct the orthogonal $Q$ such that $A = Q \Lambda Q^T$.

• Any basis for $E_1$ will do. Orthogonalizing it lets you use a transpose instead of an inverse when “dediagonalizing,” which might make for a simpler computation.
– amd
Commented Apr 20, 2018 at 20:02

The subspace orthogonal to $[1,1,1]^T$ is given by $x+y+z=0$. Find two vectors satisfying that relationship, and you have your basis for $E_1$.