# Diagonalizing a matrix M in an invariant subspace to find eigenvectors

Suppose we have a matrix M, and some subspace W which is invariant under M. Suppose W = span(a, b) where a and b are vectors. If we know what the action M on a and b results in (say some linear combinations of a and b), how can we find eigenvectors of M using this information?

-
Also what does it mean to "Diagonalize the action of M on W"? – cdc Mar 18 '13 at 20:42

## 1 Answer

I call $T$ the operator defined by $M$, which I take to be the matrix of $T$ in the canonical basis.

If $a$ and $b$ are linearly dependent, you're done: $W$ is one-dimensional and invariant: it is made of eigenvectors for $T$, hence for $M$, when written in the canonical basis.

If they are linearly independent, express $Aa$ and $Ab$ in terms of $a$ and $b$. You get a $2\times 2$ matrix, which represents the restriction of $T$ to $W$. Find eigenvectors for this $2\times 2$. Their coordinates are with respect to $(a,b)$. Go back to the canonical basis. You get eigenvectors of $M$.

-
Thank you! Just as a follow up to make sure I understand: if instead of just two vectors a and b we have {a} and {b} which are both sets of orthonormal vectors with cardinality a and b respectively. Then W = span({a}, {b}). This will only change the dimension of the restricted T to W. Is this correct? Then the eigenvectors be for this a cross b matrix will be the eigenvectors of M when taken to the canonical basis. Am I right? – cdc Mar 18 '13 at 21:38
@cdc Eigenvectors for $T$ restricted to an invariant subspace $W$ are eigenvectors for $T$ a fortiori. The rest is just a matter of coordinates of these vectors. So yes. – 1015 Mar 18 '13 at 21:43