0
$\begingroup$

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 $A = Q \Lambda Q^T$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – amd
    Commented Apr 20, 2018 at 20:02

1 Answer 1

1
$\begingroup$

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 $E_1$.

$\endgroup$
3
  • 1
    $\begingroup$ A plane orthogonal to a line can be anywhere on that line (far away from zero or near). So how do I know whichever plane I find that this is in fact the eigenspace? $\endgroup$
    – mdcq
    Commented Apr 20, 2018 at 15:48
  • 1
    $\begingroup$ Got it. It must be the plane crossing zero, because an eigenspace is a subspace and therefore must contain zero. And this plane is in fact unique. $\endgroup$
    – mdcq
    Commented Apr 20, 2018 at 15:59
  • $\begingroup$ @philmcole That's the one. I could've been clearer. $\endgroup$
    – Arthur
    Commented Apr 20, 2018 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .