how prove that a linear transformation is diagonalizable, given an eigenvalue and the dimension of its kernel

Question

A question from an exam : (First year mechanical engineering, first course in linear algebra):

Let $V$ be the vector space of $2\times2$ matrices, and let $U$ be the subspace of $V$ containing $2\times2$ symmetric matrices. let $S: U \to U$ a linear transformation.

It is known that $2$ is an eigenvalue of $S$ and that $\dim(\ker S) = 2$. the question is:

A. prove that $S$ is diagonalizable

B. write $S$ characteristic polynomial

It's clear that $\dim(\text{Im} T) = 1$, and since $\text{Im}T$ is spanned by that matrix columns, the matrix has 2 linearly dependent columns meaning its singular and has $0$ as an eigenvalue.

How to continue from here?

Answer 1

Let $\{v_1,v_2\}$ be a basis of $\ker S$ and let $v_3$ be an eigenvector of $S$ with eigenvalue $2$. Then the set $\{v_1,v_2,v_3\}$ is linearly independent and, since $\dim U=3$, it is a basis of $U$. So, $U$ has a basis which consists of eigenvectors of $S$. In other words, $S$ is diagonalisable.

Answer 2

The dimension of $U$ is $3$, since a symmetric matrix is determined by $\frac{n(n+1)}2$ entries.

Since there are $2$ eigenvectors for eigenvalue $0$, and $1$ for e-value $2$, there is a basis of eigenvectors, and $S$ is diagonalizable.

Since $S$ is diagonalizable, the algebraic multiplicity of each eigenvalue is equal to the geometric multiplicity. Thus $2$ has multiplicity $1$ and $0$ algebraic multiplicity $2$. Thus the characteristic polynomial is $x^2(x-2)=x^3-2x^2$.