Solutions To Mathematics Textbooks/Algebra (9780132413770)/Chapter 4

Exercise 1.1
Let $$f(M): F^{m \times n} \rightarrow F^{l \times p}$$ such that $$f(M) = AMB$$ where $$A \in F^{l \times m}, B \in F^{n \times p}$$. Then for $$M, N \in F^{m \times n}$$ and $$\alpha, \beta \in F$$ it holds $$f(\alpha M + \beta N) = A(\alpha M + \beta N)B = \alpha AMB + \beta ANB = \alpha f(M) + \beta f(N) $$, so $$f$$ is a linear transformation.

Exercise 1.3
The matrix $$A \in F^{m \times n}$$ is a linear mapping $$A: F^n \rightarrow F^m$$. Let $$\mathsf{B} = \{v_1,\ldots,v_n\}$$ be a basis for $$F^n$$. Then the space $$\textrm{im}~ A = \textrm{Span }(A(v_1),\ldots,A(v_n))$$ has dimension at most $$\min(n,m)$$. Then, using the dimension formula we have $$\dim (\textrm{im}~ A) + \dim (\ker A) = \dim F^n = n$$, so rearranging we get $$\dim (\ker A) = n - \dim (\textrm{im}~ A) \geq n-\min(m,n) \geq n-m$$.

Exercise 1.4
Let $$A \in F^{m \times n}$$ be a matrix of rank 1. Then, the image of $$A$$ is a space spanned by a single vector, say $$v \in F^n$$, and $$Av = w$$ for some nonzero $$w \in F^m$$. We can assume that $$v = (1,0,\ldots,0)$$, since the vector is unique up to a scaling and change of basis. Then, the kernel of $$A$$ is given by the vectors $$e_i$$ for $$i = 2,\ldots,n$$. Next, consider the matrix $$B = \frac{1}{v^tv}wv^t$$, so that $$Bv = w$$ as well, and $$Be_i = 0, i = 2,\ldots,n$$. It is easy to see that $$A$$ and $$B$$ describe the same linear transformation, so they are equal as matrices. The representation of $$B$$ is not unique, since we could scale one of the vectors $$v$$ and $$w $$ arbitrarily as long as we scale the other accordingly.

Exercise 1.5
a) It is very easy to see that performing the vector space operations coordinate-wise preserves the vector space structure in the product space.

b) Let $$T(u, v) = u + v$$. Then we have $$T(u_1 + u_2, v_1 + v_2) = u_1 + v_1 + u_2 + v_2 = T(u_1, v_1) + T(u_2, v_1) $$ and $$T(au, av) = au + av = aT(u,v)$$, so $$T$$ is a linear operator.

c) We have $$\dim (\textrm{im}~ T) + \dim (\ker T) = \dim (U \times W)$$ where $$\ker T = \{(u, w) \in U \times W : u = -w\} = \{(v, -v) : v \in U \cap W\} $$ so $$\dim (\ker T) = \dim (U \cap W)$$. Furthermore, we have by definition that $$\textrm{im}~ T = U \oplus W $$, and $$\dim (U \times W) = \dim U + \dim W$$. Therefore, the dimension formula has the form $$\dim (U \cap W) + \dim (U \oplus W) = \dim U + \dim W$$.

Exercise 2.1
We can write $$A = a_{11}e_{11} + a_{12}e_{12} + a_{21}e_{21} + a_{22}e_{22}$$, $$B = b_{11}e_{11} + b_{12}e_{12} + b_{21}e_{21} + b_{22}e_{22}$$ and $$M = m_{11}e_{11} + m_{12}e_{12} + m_{21}e_{21} + m_{22}e_{22}$$. We have the following multiplication table where the first element of a column denotes the matrix that is multiplied from the right by the first element of a given row. Then, $$AM = (a_{11}m_{11} + a_{21}m_{12})e_{11} + (a_{12}m_{11} + a_{22}m_{12})e_{12} + (a_{11}m_{21} + a_{21}m_{22})e_{21} + (a_{12}m_{21} + a_{22}m_{22})e_{22}$$. For $$AMB$$ we have then in the given basis the form

$$\begin{pmatrix} a_{11}m_{11}b_{11} + a_{21}m_{12}b_{11} + a_{11}m_{21}b_{12} + a_{21}m_{22}b_{12} \\ a_{12}m_{11}b_{11} + a_{22}m_{12}b_{11} + a_{12}m_{21}b_{12} + a_{22}m_{22}b_{12} \\ a_{11}m_{11}b_{21} + a_{21}m_{12}b_{21} + a_{11}m_{21}b_{22} + a_{21}m_{22}b_{22} \\ a_{12}m_{11}b_{21} + a_{22}m_{12}b_{21} + a_{12}m_{21}b_{22} + a_{22}m_{22}b_{22} \end{pmatrix}$$.

Exercise 2.3
The matrix with the given property satisfies the equation $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ x \end{pmatrix} = \begin{pmatrix} x \\ 3x \end{pmatrix}$$. Solving this yields that the matrix has to have the form $$\begin{pmatrix} a & 1-a \\ c & 3-c \end{pmatrix}$$ for any $$a,c \in \mathbb{R}.$$