Raising the bar

Diagonalization of symmetric matrices

We are looking for conditions under which a matrix in some orthonormal basis takes a diagonal form.

Definition 1. We say that $latex A$ is diagonalizable if there exists an orthonormal basis $latex U=(u_1,...,u_n)$ such that $latex A_U=U^{-1}AU$ is diagonal: $latex A_U=diag[\lambda_1,...,\lambda_n]$ (all matrices are real).

The usual way of solving a mathematical problem is this. Look for necessary conditions. Then try to find weak sufficient conditions. If they coincide with the necessary ones, you have an exact solution to the problem.

Exercise 1 (necessary condition) Suppose $latex A$ is diagonalizable by an orthogonal matrix. Then it is symmetric.

Proof. A diagonal matrix is obviously symmetric. Hence, from $latex A=UA_UU^{-1}$ and using Exercise 1 on the orthogonal matrix inverse $latex A^T=(U^{-1})^TA_U^TU^T$ $latex =(U^T)^TA_UU^{-1}=A.$

Exercise 2. If $latex A$ is symmetric and a subspace $latex L\subset R^n$ is invariant with respect to $latex A$, then $latex L^\perp$ is also an invariant subspace of $latex A$.

Proof. Let $latex y\in L^\perp.$ We need to show that $latex Ay\in L^\perp.$ Take any $latex x\in L.$ Since $latex L$ is invariant, we have $latex Ax\in L.$ Hence, by Exercise 1 $latex 0=(Ax)\cdot y=x\cdot(Ay).$ Since $latex x$ is arbitrary, it follows that $latex Ay\in L^\perp.$

Exercise 3 (sufficient condition) If $latex A$ is symmetric, then it is diagonalizable by an orthogonal matrix.

Proof. Let $latex A$ be symmetric. By Exercise 3 $latex A$ has at least one eigenvector $latex u_1\in R^n.$ Denote $latex L_1$ the set of vectors orthogonal to $latex u_1.$ It has dimension $latex n-1$ and it is invariant by Exercise 2. By Exercise 4 $latex A$ has an eigenvector $latex u_2$ in $latex L_1.$

Let $latex L_2$ be the set of vectors from $latex L_1$ which are orthogonal to $latex u_2.$ As above, $latex \dim L_2=n-2$, $latex L_2$ is invariant and in $latex L_2$ there is an eigenvector $latex u_3$ of $latex A$.

Continuing in this way, eventually we obtain a system $latex u_1,...,u_n.$ Since $latex u_1$ is orthogonal to $latex L_1$ and all of $latex u_2,u_3,...$ belong to $latex L_1$, $latex u_1$ is orthogonal to all of $latex u_2,u_3,...$. Similarly, at each step $latex u_i$ is orthogonal to all of $latex u_{i+1},u_{i+2},...$ Thus, the system $latex u_1,...,u_n$ is orthogonal. Its elements can be scaled to satisfy $latex \|u_i\|=1$ for all $latex i.$

The resulting system $latex U=(u_1,...,u_n)$ will be an orthonormal basis. Its completeness follows from the fact that there are $latex n$ elements in this system and they are linearly independent.

Denoting $latex \lambda_i$ the eigenvalue corresponding to $latex u_i,$ we have $latex Au_i=\lambda_iu_i.$ This means that $latex A$ is of diagonal form in the basis $latex U.$ The transition matrix from the basis consisting of unit vectors $latex e_i$ to $latex u_i$ is orthogonal. If $latex x\in R^n,$ then the same $latex x$ is the vector of coefficients of $latex x$ in the basis $latex E=(e_1,...,e_n).$ By Exercise 3, $latex A_U=U^{-1}AU.$

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