Question

Question: 4. Let A be an \(n \times n\) symmetric matrix.

a. Show that \({({\rm{Col}}A)^ \bot } = {\rm{Nul}}A\). (Hint: See Section 6.1.)

b. Show that each y in \({\mathbb{R}^n}\) can be written in the form \(y = \hat y + z\), with \(\hat y\) in \({\rm{Col}}A\) and z in \({\rm{Nul}}A\).

Short answer

(a) It is proved that \({({\rm{Col}}A)^ \bot } = {\rm{Nul}}A\).

(b) It is proved that \(\hat y\) in \({\rm{Col}}A\)and \(z\)is in \({\rm{Nul}}A\).

Step-by-step solution

(a) Find the Null-space of matrix

In 6.1, from Theorem-3, The null-space of is the orthogonal counterpart of column space A, and also null-space of\(A\)and \({A^T}\) are equal.

So, \({\left( {{\rm{Col}}A} \right)^ \bot } = {\rm{Nul}}{A^ \bot } = {\rm{Nul}}A\).

Hence, \({(ColA)^ \bot } = NulA\).

(b) The condition for solving matrix

From orthogonal decomposition Theorem, it can be said that yin\({\mathbb{R}^n}\)has\(y = \hat y + z\).

Here, \(\hat y\) in \({\rm{Col}}A\) and z is in \({\left( {{\rm{Col}}A} \right)^ \bot }\) from part (a).

\({\left( {{\rm{Col}}A} \right)^ \bot } = {\rm{Nul}}A\)

And, Z is in \({\left( {ColA} \right)^ \bot }\).

Hence, zis in \({\rm{Nul}}A\).