Define the kernel and the image of a linear map α \alpha α from R m \mathbb{R}^{m} R m to R n \mathbb{R}^{n} R n .
Let { e 1 , e 2 , … , e m } \left\{\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{m}\right\} { e 1 , e 2 , … , e m } be a basis of R m \mathbb{R}^{m} R m and { f 1 , f 2 , … , f n } \left\{\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\right\} { f 1 , f 2 , … , f n } a basis of R n \mathbb{R}^{n} R n . Explain how to represent α \alpha α by a matrix A A A relative to the given bases.
A second set of bases { e 1 ′ , e 2 ′ , … , e m ′ } \left\{\mathbf{e}_{1}^{\prime}, \mathbf{e}_{2}^{\prime}, \ldots, \mathbf{e}_{m}^{\prime}\right\} { e 1 ′ , e 2 ′ , … , e m ′ } and { f 1 ′ , f 2 ′ , … , f n ′ } \left\{\mathbf{f}_{1}^{\prime}, \mathbf{f}_{2}^{\prime}, \ldots, \mathbf{f}_{n}^{\prime}\right\} { f 1 ′ , f 2 ′ , … , f n ′ } is now used to represent α \alpha α by a matrix A ′ A^{\prime} A ′ . Relate the elements of A ′ A^{\prime} A ′ to the elements of A A A .
Let β \beta β be a linear map from R 2 \mathbb{R}^{2} R 2 to R 3 \mathbb{R}^{3} R 3 defined by
β ( 1 1 ) = ( 1 2 3 ) , β ( 1 − 1 ) = ( 6 4 2 ) \beta\left(\begin{array}{l} 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right), \quad \beta\left(\begin{array}{c} 1 \\ -1 \end{array}\right)=\left(\begin{array}{l} 6 \\ 4 \\ 2 \end{array}\right) β ( 1 1 ) = ⎝ ⎛ 1 2 3 ⎠ ⎞ , β ( 1 − 1 ) = ⎝ ⎛ 6 4 2 ⎠ ⎞
Either find one or more x \mathbf{x} x in R 2 \mathbb{R}^{2} R 2 such that
β x = ( 1 − 2 1 ) \beta \mathbf{x}=\left(\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right) β x = ⎝ ⎛ 1 − 2 1 ⎠ ⎞
or explain why one cannot be found.
Let γ \gamma γ be a linear map from R 3 \mathbb{R}^{3} R 3 to R 2 \mathbb{R}^{2} R 2 defined by
γ ( 1 2 0 ) = ( 1 3 ) , γ ( 0 1 1 ) = ( − 2 1 ) , γ ( 0 1 0 ) = ( 0 1 ) \gamma\left(\begin{array}{l} 1 \\ 2 \\ 0 \end{array}\right)=\left(\begin{array}{l} 1 \\ 3 \end{array}\right), \quad \gamma\left(\begin{array}{l} 0 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{c} -2 \\ 1 \end{array}\right), \quad \gamma\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)=\left(\begin{array}{l} 0 \\ 1 \end{array}\right) γ ⎝ ⎛ 1 2 0 ⎠ ⎞ = ( 1 3 ) , γ ⎝ ⎛ 0 1 1 ⎠ ⎞ = ( − 2 1 ) , γ ⎝ ⎛ 0 1 0 ⎠ ⎞ = ( 0 1 )
Find the kernel of γ \gamma γ .