# Show that a symmetric and idempotent matrix $P$ is the projection matrix onto some subspace.

I am reading "Seminar of Linear Algebra" by Kenichi Kanatani.
In this book, there is the following problem:

Show that a symmetric and idempotent matrix $$P$$ is the orthogonal projection matrix onto some subspace.

As is well known, an $$n \times n$$ symmetric matrix $$P$$ has $$n$$ real eigenvalues $$\lambda_1, \dots, \lambda_n$$, the corresponding eigenvectors $$u_1, \dots, u_n$$ being an orthonormal system. If we multiply $$P u_i = \lambda_i u_i$$ by $$P$$ from left on both sides, we have
$$P^2 u_i = \lambda_i P u_i = \lambda_i^2 u_i.$$ If $$P$$ is idempotent, the left side is $$P u_i = \lambda_i u_i$$. Hence, $$\lambda_i = \lambda_i^2$$, i.e., $$\lambda_i = 0, 1$$. Let $$\lambda_1 = \cdots = \lambda_r = 1$$, $$\lambda_{r+1} = \cdots = \lambda_n = 0$$. Then, $$P u_i=u_i, i = 1, \dots, r,$$ $$P u_i = 0, i = r+1, \dots, n.$$ We see that $$P$$ is the orthogonal projection matrix onto the subspace spanned by $$u_1, \dots, u_r$$.

I cannot understand this statement:

As is well known, an $$n \times n$$ symmetric matrix $$P$$ has $$n$$ real eigenvalues $$\lambda_1, \dots, \lambda_n$$, the corresponding eigenvectors $$u_1, \dots, u_n$$ being an orthonormal system.

• I know that an $$n \times n$$ symmetric matrix $$P$$ has $$n$$ real eigenvalues $$\lambda_1, \dots, \lambda_n$$.
• And I know if $$\lambda_i \ne \lambda_j$$, then $$u_i$$ is orthogonal to $$u_j$$.
• And I know that every $$n \times n$$ symmetric matrix $$P$$ doesn't have distinct $$n$$ real eigenvalues.

Let $$P$$ be an $$n \times n$$ symmetric matrix $$P$$.
Are there eigenvecotrs of $$P$$ that are mutually orthogonal?

Yes, for any symmetric matrix $$P$$ of size $$n$$, it is possible to find a set of eigenvectors $$(u_i)_{i=1}^n$$ such that if $$i\neq j$$, then $$u_i . u_j = 0$$ (scalar product), and $$||u_i||^2 = 1$$ for all $$i$$.

Basically, a symmetric matrix always admits a set of eigenvectors which is an orthonormal basis of the vector space.

In matrix notations, it is possible to find an orthogonal matrix $$O$$ and a diagonal matrix $$D$$ (whose diagonal are the eigenvalues of $$P$$) such that $$P = O D O^T = O D O^{-1}$$.

• Thank you very much, seamp. Oct 24, 2018 at 10:17