The Subspace of Matrices that are Diagonalized by a Fixed Matrix

Problem 33

Suppose that $S$ is a fixed invertible $3$ by $3$ matrix. This question is about all the matrices $A$ that are diagonalized by $S$, so that $S^{-1}AS$ is diagonal. Show that these matrices $A$ form a subspace of $3$ by $3$ matrix space.

Check the following criteria for a subset to be a subspace.

Theorem. (Subspace criteria)

A subset $W$ of a vector space $V$ is a subspace if and only if

The zero vector in $V$ is in $W$.

For any vectors $A, B \in W$, the addition $A+B\in W$.

For any vector $A\in W$ and a scalar $c$, the scalar multiplication $cA\in W$.

Proof.

Let $V$ be the vector space of all $3$ by $3$ matrices.
Define the subset of $V$
\[ W=\{ A\in V \mid S^{-1}AS \text{ is diagonal.} \}. \]
We want to show that $W$ is a subspace of $V$.

To do this, it suffices to show the following subspace criteria;

The zero vector in $V$ is in $W$.

For any vectors $A, B \in W$, the addition $A+B\in W$.

For any vector $A\in W$ and a scalar $c$, the scalar multiplication $cA\in W$.

The zero vector in the vector space $V$ is the $3$ by $3$ zero matrix $O$.
Since $S^{-1}OS=O$, we have $O\in W$.

To show the second criterion, take $A, B \in W$.
Then we have $S^{-1}(A+B)S=S^{-1}AS+S^{-1}BS$ and since $A, B \in W$, the matrices $S^{-1}AS, S^{-1}BS$ are diagonal.

The addition of diagonal matrices is still diagonal, thus $S^{-1}(A+B)S$ is diagonal and $A+B \in W$.

We finally check the third criterion.
Take $A \in W$ and let $c$ be a scalar.
Then we have $S^{-1}(cA)S=cS^{-1}AS$. The matrix $S^{-1}AS$ is diagonal since $A \in W$.

Since a scalar multiplication of a diagonal matrix is still diagonal, we conclude that the matrix $S^{-1}(cA)S$ is diagonal and $cA\in W$.

Therefore the subset $W$ of $V$ satisfies the criteria to be a subspace.

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