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

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?

• You mention $T$, but what is it? – enedil Feb 10 at 16:10
• i meant S. sorry... – Johnathan1994 Feb 10 at 16:25

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.
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 $$x^2(x-2)=x^3-2x^2$$.
• My mistake. @enedil it's $\frac{n(n+1)}2$. – Chris Custer Feb 10 at 17:41