91影视

BecauseA is symmetric, there is an orthonormal eigenvector basis\(\left\{ {{u_1},{u_2},....{u_n}} \right\}\)for\({R^n}\).

Let r be rank of matrix A, if r becomes 0, then rank of A is 0.

The decomposition as given in exercise 4(b) is\(y = \hat y + z\).

When \(r = n\), then the decomposition becomes \(y = y + 0\).

Let\({u_1},{u_2},....,{u_r}\)are the eigenvector corresponding to the r nonzero eigenvalues.

Then\({u_1},{u_2},....,{u_r}\)are in Col A and\({u_{r + 1}},{u_{r + 2}},...{u_n}\)are vector in Nul A.

The vector y in\({R^n}\)can be written as:

\(\begin{array}{l}y = {c_1}{u_1} + {c_2}{u_2} + ... + {c_r}{u_r} + {c_{r + 1}}{u_{r + 1}} + {c_{r + 2}}{u_{r + 2}} + .... + {c_n}{u_n}\\y = \hat y + z\end{array}\)

Hence,\(y = \hat y + z\).

Here, \(\hat y \in {\rm{Col}}A\), and \(z \in {\rm{Nul}}A\).