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

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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

01

Apply spectral decomposition theorem

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

02

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.

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Most popular questions from this chapter

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

Question: Let \({\bf{X}}\) denote a vector that varies over the columns of a \(p \times N\) matrix of observations, and let \(P\) be a \(p \times p\) orthogonal matrix. Show that the change of variable \({\bf{X}} = P{\bf{Y}}\) does not change the total variance of the data. (Hint: By Exercise 11, it suffices to show that \(tr\left( {{P^T}SP} \right) = tr\left( S \right)\). Use a property of the trace mentioned in Exercise 25 in Section 5.4.)

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

19. Show that the columns of\(V\)are eigenvectors of\({A^T}A\), the columns of\(U\)are eigenvectors of\(A{A^T}\), and the diagonal entries of\({\bf{\Sigma }}\)are the singular values of \(A\). (Hint: Use the SVD to compute \({A^T}A\) and \(A{A^T}\).)

In Exercises 17–24, \(A\) is an \(m \times n\) matrix with a singular value decomposition \(A = U\Sigma {V^T}\) , where \(U\) is an \(m \times m\) orthogonal matrix, \({\bf{\Sigma }}\) is an \(m \times n\) “diagonal” matrix with \(r\) positive entries and no negative entries, and \(V\) is an \(n \times n\) orthogonal matrix. Justify each answer.

23. Let \(U = \left( {{u_1}...{u_m}} \right)\) and \(V = \left( {{v_1}...{v_n}} \right)\) where the \({{\bf{u}}_i}\) and \({{\bf{v}}_i}\) are in Theorem 10. Show that \(A = {\sigma _1}{u_1}v_1^T + {\sigma _2}{u_2}v_2^T + ... + {\sigma _r}{u_r}v_r^T\).

In Exercises 3-6, find (a) the maximum value of \(Q\left( {\rm{x}} \right)\) subject to the constraint \({{\rm{x}}^T}{\rm{x}} = 1\), (b) a unit vector \({\rm{u}}\) where this maximum is attained, and (c) the maximum of \(Q\left( {\rm{x}} \right)\) subject to the constraints \({{\rm{x}}^T}{\rm{x}} = 1{\rm{ and }}{{\rm{x}}^T}{\rm{u}} = 0\).

5. \(Q\left( x \right) = x_1^2 + x_2^2 - 10x_1^{}x_2^{}\).

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