Show that \(U \subseteq \left (U^{\perp }\right )^\perp \), then use Theorem ?? twice.
Challenge Problems for Chapter 9
If \(U\) is a subspace of \(\RR ^n\), show that \(\left (U^{\perp }\right )^\perp = U\).
If \(W\) is a subspace of \(\RR ^n\), show how to find an \(n \times n\) matrix \(A\) such that \(W = \{\vec {x} \mid A\vec {x} = \vec {0}\}\).
If \(U\) and \(W\) are subspaces, define \(U+W\) to be the set of all possible sums of elements of \(U\) with
elements of \(W\).
- (a)
- Is \(U+W\) a subspace? Show that \((U + W)^\perp = U^\perp \cap W^\perp \).
- (b)
- Illustrate this result with a diagram in \(\RR ^3\) where \(U\) and \(W\) are two distinct lines through the origin. What does \(U+W\) look like? What do \(U^{\perp }\) and \(W^{\perp }\) look like?
A square matrix \(E\) satisfying \(E^2=E=E^T\) is called a projection matrix. If \(E\) is a projection matrix,
show that \(I - E\) is also a projection matrix.
Let \(E\) be an \(n \times n\) matrix, and let \(W = \{\vec {x} E \mid \vec {x} \mbox { in } \RR ^n\}\). Show that the following are equivalent.
- (a)
- \(E^{2} = E = E^{T}\) (\(E\) is a projection matrix).
- (b)
- \((\vec {x} - \vec {x}E) \dotp (\vec {y}E) = 0\) for all \(\vec {x}\) and \(\vec {y}\) in \(\RR ^n\).
- (c)
- \(\mbox {proj}_W(\vec {x}) = \vec {x}E\) for all \(\vec {x}\) in \(\RR ^n\). For (ii) implies (iii): Write \(\vec {x} = \vec {x}E + (\vec {x} - \vec {x}E)\) and use the uniqueness argument preceding the definition of \(\mbox {proj}_W(\vec {x})\). For (iii) implies (ii): \(\vec {x} - \vec {x}E\) is in \(W^\perp \) for all \(\vec {x}\) in \(\RR ^n\).
If \(EF = 0 = FE\) and \(E\) and \(F\) are projection matrices, show that \(E + F\) is also a projection matrix.
If \(A\) is \(m \times n\) and \(AA^{T}\) is invertible, show that \(E = A^{T}(AA^{T})^{-1}A\) is a projection matrix.
Click the arrow to see answer.
\(E^T = A^T[(AA^T)^-1]^T(A^T)^T = A^T[(AA^T)^T]^{-1}A = A^T[AA^T]^{-1}A = E\)
Consider \(A = \begin{bmatrix} 0 & a & 0 \\ a & 0 & c \\ 0 & c & 0 \end{bmatrix}\) where one of \(a, c \neq 0\). Show that the characteristic polynomial (see
Definition ??) is given by \(c_{A}(z) = z(z - k)(z + k)\), where \(k = \sqrt {a^2 + c^2}\), and find an orthogonal matrix \(Q\) such that \(Q^{-1}AQ\) is
diagonal.
Click the arrow to see the answer.
\(Q = \frac {1}{\sqrt {2}k}\begin{bmatrix} c\sqrt {2} & a & a \\ 0 & k & -k \\ -a\sqrt {2} & c & c \end{bmatrix}\)
Given \(A = \begin{bmatrix} b & a \\ a & b \end{bmatrix}\), show that the characteristic polynomial (see Definition ??) is given by \(c_{A}(z) = (z - a - b)(z + a - b)\),
and find an orthogonal matrix \(Q\) such that \(Q^{-1}AQ\) is diagonal.
Bibliography
Some of these problems come from Section 7.4 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)
Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 433–438.
Other problems come from the second part of Section 8.1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)
W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 422–423