Question

Question: 3. Let A be an \(n \times n\) symmetric matrix of rank r. Explain why the spectral decomposition of A represents A as the sum of r rank 1 matrices.

Short answer

The matrix A can be represented as a sum of r terms each of rank 1 in the spectral decomposition of A.

Step-by-step solution

Apply spectral decomposition theorem

From the spectral decomposition theorem, we can write the matrix A in the form \(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + \cdots + {\lambda _n}{{\bf{u}}_n}{\bf{u}}_n^T\).

Apply rank and nullity theorem

Apply this theorem on matrix A as:

\(\begin{array}{c}{\rm{Rank}}A + {\rm{Nullity}}A = n\\{\rm{Nullity}}A = n - {\rm{Rank}}A\\{\rm{Nullity}}A = n - r\end{array}\)

It follows that 0 is an eigenvalue of A with the multiplicity of\(n - r\). So, the remaining r terms are matrices of size\(n \times n\)with rank\(1\).

So, the matrix A can be represented as a sum of r terms each of rank 1 in the spectral decomposition of A.