Question: 2. Let \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\) be an orthonormal basis for \({\mathbb{R}_n}\) , and let \({\lambda _1},....{\lambda _n}\) be any real scalars. Define

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + ..... + {\lambda _n}{\bf{u}}_n^T\)

a. Show that A is symmetric.

b. Show that \({\lambda _1},....{\lambda _n}\) are the eigenvalues of A

The matrix A is shown below:

\(A = {\lambda _1}{{\bf{u}}_1}{\bf{u}}_1^T + {\lambda _2}{{\bf{u}}_2}{\bf{u}}_2^T + \cdots + {\lambda _n}{{\bf{u}}_n}{\bf{u}}_n^T\)

The transpose of matrix A is shown below:

\(\begin{array}{c}{A^T} = \left( {{\lambda _1}{u_1}u_1^T + {\lambda _2}{u_2}u_2^T + \cdots + {\lambda _j}{u_j}u_j^T + {\lambda _n}{u_n}u_n^T} \right){u_j}\\ = \left( {{\lambda _1}{u_1}u_1^T} \right){u_j} + \left( {{\lambda _2}{u_2}u_2^T} \right){u_j} + ..... + \left( {{\lambda _j}{u_j}u_j^T} \right){u_j} + .... + \left( {{\lambda _n}{u_n}u_n^T} \right){u_j}\\ = {\lambda _1}\left( {{u_j}u_1^T} \right){u_j} + {\lambda _2}\left( {{u_j}u_2^T} \right){u_j} + ..... + {\lambda _j}\left( {{u_j}u_j^T} \right){u_j} + ... + {\lambda _n}\left( {{u_n}u_n^T} \right){u_j}\\ = {\lambda _1}\left( 0 \right)\left( {{u_1}} \right) + {\lambda _2}\left( 0 \right)\left( {{u_2}} \right) + .... + {\lambda _j}\left( 1 \right)\left( {{u_j}} \right) + .... + {\lambda _n}\left( 0 \right)\left( {u_n^T} \right)\\ = {\lambda _j}{u_j}\end{array}\)

Next step;

\(\)

\(\begin{array}{l}{A^T} = {\lambda _1}{u_1}u_1^T + {\lambda _2}{u_2}u_2^T + \cdots + {\lambda _n}{u_n}u_n^T\\{A^T} = A\end{array}\)

That’s why the matrix is symmetric

Since the set\(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},....,{{\bf{u}}_n}} \right\}\)is an orthogonal basis, so\({\bf{u}}_i^T{{\bf{u}}_j} = \left\{ {\begin{array}{*{20}{c}}{1{\rm{ for }}i = j}\\{0{\rm{ for }}i \ne j}\end{array}} \right.\).

For the product\(A{u_j}\):

\(\begin{array}{c}{A^T} = {\left( {{\lambda _1}{u_1}u_1^T + {\lambda _2}{u_2}u_2^T + \cdots + {\lambda _n}{u_n}u_n^T} \right)^T}\\ = {\left( {{\lambda _1}{u_1}u_1^T} \right)^T} + {\left( {{\lambda _2}{u_2}u_2^T} \right)^T} + ..... + {\left( {{\lambda _n}{u_n}u_n^T} \right)^T}\\ = {\lambda _1}{\left( {{u_1}u_1^T} \right)^T} + {\lambda _2}{\left( {{u_2}u_2^T} \right)^T} + ..... + {\lambda _n}{\left( {{u_n}u_n^T} \right)^T}\\ = {\lambda _1}{\left( {u_1^T} \right)^T}{\left( {{u_1}} \right)^T} + {\lambda _2}{\left( {u_2^T} \right)^T}{\left( {{u_2}} \right)^T} + .... + {\lambda _n}\left( {{u_n}} \right){\left( {u_n^T} \right)^T}\end{array}\)

Hence,\(A{u_j} = {\lambda _j}{u_j}\).

Therefore, \({\lambda _1},{\lambda _2}, \cdots ,{\lambda _n}\) are the eigenvalues of \(A\).