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Chapter 7: Q4SE (page 395)

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

Expert verified

(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

01

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

02

(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 \({\rm{Nul}}A\).

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Most popular questions from this chapter

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17鈥�22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

18. \(\left( {\begin{aligned}{{}}1&{ - 6}&4\\{ - 6}&2&{ - 2}\\4&{ - 2}&{ - 3}\end{aligned}} \right)\)

Orthogonally diagonalize the matrices in Exercises 13鈥�22, giving an orthogonal matrix \(P\) and a diagonal matrix \(D\). To save you time, the eigenvalues in Exercises 17鈥�22 are: (17) \( - {\bf{4}}\), 4, 7; (18) \( - {\bf{3}}\), \( - {\bf{6}}\), 9; (19) \( - {\bf{2}}\), 7; (20) \( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

15. \(\left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right)\)

Compute the quadratic form \({x^T}Ax\), when \(A = \left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\) and

a. \(x = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

b. \(x = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\)

c. \(x = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\)

Classify the quadratic forms in Exercises 9-18. Then make a change of variable, \({\bf{x}} = P{\bf{y}}\), that transforms the quadratic form into one with no cross-product term. Write the new quadratic form. Construct P using the methods of Section 7.1.

10. \({\bf{2}}x_{\bf{1}}^{\bf{2}} + {\bf{6}}{x_{\bf{1}}}{x_{\bf{2}}} - {\bf{6}}x_{\bf{2}}^{\bf{2}}\)

Question: [M] A Landsat image with three spectral components was made of Homestead Air Force Base in Florida (after the base was hit by Hurricane Andrew in 1992). The covariance matrix of the data is shown below. Find the first principal component of the data, and compute the percentage of the total variance that is contained in this component.

\[S = \left[ {\begin{array}{*{20}{c}}{164.12}&{32.73}&{81.04}\\{32.73}&{539.44}&{249.13}\\{81.04}&{246.13}&{189.11}\end{array}} \right]\]
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