0.1 Solutions for Note 4

1.

For transformation \(T( \langle a, b \rangle ) = a + b\), consider the following equations for some scalars \(\alpha \) and \(\beta \). \begin{align*} T ( \alpha \langle a_1, b_1 \rangle + \beta \langle a_2, b_2 \rangle ) &= T ( \langle \alpha a_1, \alpha b_1 \rangle + \langle \beta a_2, \beta b_2 \rangle ) \\ &= \alpha a_1 + \alpha b_1 + \beta a_2 + \beta b_2 \\ &= \alpha (a_1 + b_1) + \beta (a_2 + b_2) \\ &= \alpha T ( \langle a_1, b_1 \rangle ) + \beta T ( \langle a_2, b_2 \rangle ) \end{align*}

Therefore by the definition of linear transformation, \(T\) is linear.

For \(T( \langle a, b \rangle ) = ab\), consider the following for arbitrary scalars \(\alpha \) and \(\beta \). \begin{align*} T ( \alpha \langle a_1, b_1 \rangle + \beta \langle a_2, b_2 \rangle ) &= T ( \langle \alpha a_1, \alpha b_1 \rangle + \langle \beta a_2, \beta b_2 \rangle ) \\ &= T ( \langle \alpha a_1 + \beta b_2, \alpha b_1 + \beta b_2 \rangle ) \\ &= (\alpha a_1 + \beta b_2)(\alpha b_1 + \beta b_2) \end{align*}

Moreover, \[ \alpha T ( \langle a_1, b_1 \rangle ) + \beta T ( \langle a_2, b_2 \rangle ) = \alpha a_1 b_1 + \beta a_2 b_2 \] holds. Because \((\alpha a_1 + \beta b_2)(\alpha b_1 + \beta b_2) \neq \alpha a_1 b_1 + \beta a_2 b_2\) in general, the transformation is not linear.

2.

By definition, if \(\lambda \) is an eigenvalue of \(AB\), then there exists some nonzero vector \(v\) such that \(AB v = \lambda v\). First, let \(\lambda \neq 0\). Notice that \(u = Bv \neq \mathbf {0}\). Therefore, \(BA u = BA (Bv) = B (ABv) = B (\lambda v) = \lambda (Bv) = \lambda u\) and \(\lambda \) is also an eigenvalue of \(BA\). Continuing if \(\lambda = 0\), then \(\det (AB) = \det (A) \det (B) = 0 = \det (B) \det (A) = \det (BA)\). In other words, if \(AB\) is singular then \(BA\) is also singular and contains the eigenvalue \(\lambda \). Therefore, \(\lambda \) is an eigenvalue of \(AB\) if and only if it is an eigenvalue of \(BA\).

3.

First, to find bases for the kernel of the matrix, the following equation must be solved. \[ \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} 0 \\ 0 \end {bmatrix} \] Solving the system, \(y = 0\) and \(x = -z\). Therefore the basis for the kernel is \(\left \{ \begin {bmatrix} -1 \\ 0 \\ 1 \end {bmatrix} \right \}\). Continuing, the basis for the image can be found. Consider the following equation. \[ \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} x + y + z \\ y \end {bmatrix} = x \begin {bmatrix} 1 \\ 0 \end {bmatrix} + y \begin {bmatrix} 1 \\ 1 \end {bmatrix} + z \begin {bmatrix} 1 \\ 0 \end {bmatrix} \] Therefore, \[ \left \{ \begin {bmatrix} 1 \\ 0 \end {bmatrix}, \begin {bmatrix} 1 \\ 1 \end {bmatrix} \right \} \] is a basis of the image of the given matrix.

4.

By definition, \(A \cong B\) if and only if there exists an invertible matrix \(P\) such that \(A = P^{-1}BP\). Continuing, \(A = P^{-1}BP\) if and only if \(A^\intercal = (P^{-1}BP)^\intercal = P^\intercal B^\intercal (P^{-1})^\intercal = P^\intercal B^\intercal (P^\intercal )^{-1}\). In other words, \(A \cong B\) if and only if \(A^\intercal \cong B^\intercal \).

5.

To find the eigenvalues for the given matrix say \(A\), it suffices to solve \(\det (A - \lambda I) = 0\) where \(\lambda \) is an eigenvalue. \[ \det \left ( \begin {bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 1 - \lambda & 0 \\ 0 & 1 & 2 - \lambda \end {bmatrix} \right ) = 0 \] Continuing with the expansion along the first column, \begin{align*} &\quad \ \det \left ( \begin {bmatrix} 1 - \lambda & 2 & 3 \\ 0 & 1 - \lambda & 0 \\ 0 & 1 & 2 - \lambda \end {bmatrix} \right ) \\ &= (1 - \lambda ) \det \left ( \begin {bmatrix} 1 - \lambda & 0 \\ 1 & 2 - \lambda \end {bmatrix} \right ) \\ &= (1 - \lambda )(1 - \lambda )(2 - \lambda ) \end{align*}

Therefore, the eigenvalues are \(1\) and \(2\). For eigenvalue \(1\), it suffices to solve the equation \(Av = v\) or \((A - I)v = 0\). Consider the following system. \[ \begin {bmatrix} 0 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} 0 \\ 0 \\ 0 \end {bmatrix} \] Therefore, \(2y + 3z = 0\) and \(y + z = 0\). In other words, \(y = z = 0\). Thus, \(S_1 = \Span \left \{ \begin {bmatrix} 1 \\ 0 \\ 0 \end {bmatrix} \right \}\) holds for \(\lambda = 1\). Continuing with \(\lambda = 2\), it suffices to solve \(Av = 2v\) or \((A - 2I) v = 0\). \[ \begin {bmatrix} -1 & 2 & 3 \\ 0 & -1 & 0 \\ 0 & 1 & 0 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} 0 \\ 0 \\ 0 \end {bmatrix} \] Solving the system, \(y = 0\) and \(x = 3z\). Thus, \(S_2 = \Span \left \{ \begin {bmatrix} 3 \\ 0 \\ 1 \end {bmatrix} \right \}\) holds for \(\lambda = 2\).