Linear Algebra/Any Matrix Represents a Linear Map/Solutions

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Solutions[edit]

This exercise is recommended for all readers.
Problem 1

Decide if the vector is in the column space of the matrix.

  1.  \begin{pmatrix}
2  &1  \\
2  &5
\end{pmatrix} ,  \begin{pmatrix} 1 \\ -3 \end{pmatrix}
  2.  \begin{pmatrix}
4  &-8 \\
2  &-4
\end{pmatrix} ,  \begin{pmatrix} 0 \\ 1 \end{pmatrix}
  3.  \begin{pmatrix}
1  &-1  &1  \\
1  &1   &-1 \\
-1  &-1  &1
\end{pmatrix} ,  \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}
Answer
  1. Yes; we are asking if there are scalars  c_1 and  c_2 such that
    
c_1\begin{pmatrix} 2 \\ 2 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ 5 \end{pmatrix}=\begin{pmatrix} 1 \\ -3 \end{pmatrix}
    which gives rise to a linear system
    
\begin{array}{*{2}{rc}r}
2c_1  &+  &c_2  &=  &1  \\
2c_1  &+  &5c_2 &=  &-3
\end{array}
\;\xrightarrow[]{-\rho_1+\rho_2}\;
\begin{array}{*{2}{rc}r}
2c_1  &+  &c_2  &=  &1  \\
&   &4c_2 &=  &-4
\end{array}
    and Gauss' method produces  c_2=-1 and  c_1=1 . That is, there is indeed such a pair of scalars and so the vector is indeed in the column space of the matrix.
  2. No; we are asking if there are scalars c_1 and c_2 such that
    
c_1\begin{pmatrix} 4 \\ 2 \end{pmatrix}+c_2\begin{pmatrix} -8 \\ -4 \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}
    and one way to proceed is to consider the resulting linear system
    
\begin{array}{*{2}{rc}r}
4c_1  &-  &8c_2  &=  &0  \\
2c_1  &-  &4c_2  &=  &1
\end{array}
    that is easily seen to have no solution. Another way to proceed is to note that any linear combination of the columns on the left has a second component half as big as its first component, but the vector on the right does not meet that criterion.
  3. Yes; we can simply observe that the vector is the first column minus the second. Or, failing that, setting up the relationship among the columns
    
c_1\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}
+c_2\begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}
+c_3\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}
=\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}
    and considering the resulting linear system
    
\begin{array}{*{3}{rc}r}
c_1  &-  &c_2  &+  &c_3  &=  &2  \\
c_1  &+  &c_2  &-  &c_3  &=  &0  \\
-c_1  &-  &c_2  &+  &c_3  &=  &0    
\end{array}
\;\xrightarrow[\rho_1+\rho_3]{-\rho_1+\rho_2}\;
\begin{array}{*{3}{rc}r}
c_1  &-  &c_2  &+  &c_3  &=  &2  \\
&   &2c_2 &-  &2c_3 &=  &-2  \\
&   &-2c_2&+  &2c_3 &=  &2    
\end{array}
\;\xrightarrow[]{\rho_2+\rho_3}\;
\begin{array}{*{3}{rc}r}
c_1  &-  &c_2  &+  &c_3  &=  &2  \\
&   &2c_2 &-  &2c_3 &=  &-2  \\
&   &     &   &0    &=  &0    
\end{array}
    gives the additional information (beyond that there is at least one solution) that there are infinitely many solutions. Parametizing gives c_2=-1+c_3 and c_1=1, and so taking c_3 to be zero gives a particular solution of c_1=1, c_2=-1, and c_3=0 (which is, of course, the observation made at the start).
This exercise is recommended for all readers.
Problem 2

Decide if each vector lies in the range of the map from  \mathbb{R}^3 to  \mathbb{R}^2 represented with respect to the standard bases by the matrix.

  1.  \begin{pmatrix}
1  &1  &3  \\
0  &1  &4
\end{pmatrix}  ,  \begin{pmatrix} 1 \\ 3 \end{pmatrix}
  2.  \begin{pmatrix}
2  &0  &3  \\
4  &0  &6
\end{pmatrix}  ,  \begin{pmatrix} 1 \\ 1 \end{pmatrix}
Answer

As described in the subsection, with respect to the standard bases, representations are transparent, and so, for instance, the first matrix describes this map.


\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_{\mathcal{E}_3}
\!\mapsto\begin{pmatrix} 1 \\ 0 \end{pmatrix}_{\mathcal{E}_2}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}
\qquad


\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
\!\mapsto\begin{pmatrix} 1 \\ 1 \end{pmatrix}
\qquad


\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\!\mapsto\begin{pmatrix} 3 \\ 4 \end{pmatrix}

So, for this first one, we are asking whether thare are scalars such that


c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix}+c_2\begin{pmatrix} 1 \\ 1 \end{pmatrix}+c_3\begin{pmatrix} 3 \\ 4 \end{pmatrix}=\begin{pmatrix} 1 \\ 3 \end{pmatrix}

that is, whether the vector is in the column space of the matrix.

  1. Yes. We can get this conclusion by setting up the resulting linear system and applying Gauss' method, as usual. Another way to get it is to note by inspection of the equation of columns that taking c_3=3/4, and c_1=-5/4, and c_2=0 will do. Still a third way to get this conclusion is to note that the rank of the matrix is two, which equals the dimension of the codomain, and so the map is onto— the range is all of \mathbb{R}^2 and in particular includes the given vector.
  2. No; note that all of the columns in the matrix have a second component that is twice the first, while the vector does not. Alternatively, the column space of the matrix is
    
\{c_1\begin{pmatrix} 2 \\ 4 \end{pmatrix}
+c_2\begin{pmatrix} 0 \\ 0 \end{pmatrix}
+c_3\begin{pmatrix} 3 \\ 6 \end{pmatrix} \,\big|\, c_1,c_2,c_3\in\mathbb{R}\}
=\{c\begin{pmatrix} 1 \\ 2 \end{pmatrix}\,\big|\, c\in\mathbb{R}\}
    (which is the fact already noted, but was arrived at by calculation rather than inspiration), and the given vector is not in this set.
This exercise is recommended for all readers.
Problem 3

Consider this matrix, representing a transformation of \mathbb{R}^2, and these bases for that space.


\frac{1}{2}\cdot
\begin{pmatrix}
1  &1  \\
-1 &1
\end{pmatrix}
\qquad
B=\langle \begin{pmatrix} 0 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ 0 \end{pmatrix} \rangle 
\quad 
D=\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ -1 \end{pmatrix} \rangle
  1. To what vector in the codomain is the first member of B mapped?
  2. The second member?
  3. Where is a general vector from the domain (a vector with components x and y) mapped? That is, what transformation of  \mathbb{R}^2 is represented with respect to  B,D by this matrix?
Answer
  1. The first member of the basis
    
\begin{pmatrix} 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \end{pmatrix}_B
    is mapped to
    
\begin{pmatrix} 1/2 \\ -1/2 \end{pmatrix}_D
    which is this member of the codomain.
    
\frac{1}{2}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}
-\frac{1}{2}\cdot\begin{pmatrix} 1 \\ -1 \end{pmatrix}
=\begin{pmatrix} 0 \\ 1 \end{pmatrix}
  2. The second member of the basis is mapped
    
\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \end{pmatrix}_B
\mapsto
\begin{pmatrix} (1/2 \\ 1/2 \end{pmatrix}_D
    to this member of the codomain.
    
\frac{1}{2}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}
+\frac{1}{2}\cdot\begin{pmatrix} 1 \\ -1 \end{pmatrix}
=\begin{pmatrix} 1 \\ 0 \end{pmatrix}
  3. Because the map that the matrix represents is the identity map on the basis, it must be the identity on all members of the domain. We can come to the same conclusion in another way by considering
    
\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} y \\ x \end{pmatrix}_B
    which is mapped to
    
\begin{pmatrix} (x+y)/2 \\ (x-y)/2 \end{pmatrix}_D
    which represents this member of \mathbb{R}^2.
    
\frac{x+y}{2}\cdot\begin{pmatrix} 1 \\ 1 \end{pmatrix}
+\frac{x-y}{2}\cdot\begin{pmatrix} 1 \\ -1 \end{pmatrix}
=\begin{pmatrix} x \\ y \end{pmatrix}
Problem 4

What transformation of  F=\{a\cos\theta+b\sin\theta\,\big|\, a,b\in\mathbb{R}\} is represented with respect to  B=\langle \cos\theta-\sin\theta,\sin\theta \rangle  and  D=\langle \cos\theta+\sin\theta,\cos\theta \rangle  by this matrix?


\begin{pmatrix}
0  &0  \\
1  &0
\end{pmatrix}
Answer

A general member of the domain, represented with respect to the domain's basis as


a\cos\theta+b\sin\theta=\begin{pmatrix} a \\ a+b \end{pmatrix}_B

is mapped to


\begin{pmatrix} 0 \\ a \end{pmatrix}_D
\quad\text{representing}\quad
0\cdot(\cos\theta+\sin\theta)+a\cdot(\cos\theta)

and so the linear map represented by the matrix with respect to these bases


a\cos\theta+b\sin\theta  
\mapsto
a\cos\theta

is projection onto the first component.

This exercise is recommended for all readers.
Problem 5

Decide if 1+2x is in the range of the map from \mathbb{R}^3 to \mathcal{P}_2 represented with respect to \mathcal{E}_3 and \langle 1,1+x^2,x \rangle by this matrix.


\begin{pmatrix}
1  &3  &0  \\
0  &1  &0  \\
1  &0  &1
\end{pmatrix}
Answer

Denote the given basis of \mathcal{P}_2 by B. Then application of the linear map is represented by matrix-vector addition. Thus, the first vector in \mathcal{E}_3 is mapped to the element of \mathcal{P}_2 represented with respect to B by


\begin{pmatrix}
1  &3  &0  \\
0  &1  &0  \\
1  &0  &1
\end{pmatrix}
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
=
\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}

and that element is 1+x. The other two images of basis vectors are calculated similarly.


\begin{pmatrix}
1  &3  &0  \\
0  &1  &0  \\
1  &0  &1
\end{pmatrix}
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
=
\begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}={\rm Rep}_{B}(4+x^2)
\quad
\begin{pmatrix}
1  &3  &0  \\
0  &1  &0  \\
1  &0  &1
\end{pmatrix}
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}={\rm Rep}_{B}(x)

We can thus decide if 1+2x is in the range of the map by looking for scalars c_1, c_2, and c_3 such that


c_1\cdot(1)+c_2\cdot(1+x^2)+c_3\cdot(x)=1+2x

and obviously c_1=1, c_2=0, and c_3=1 suffice. Thus it is in the range, and in fact it is the image of this vector.


1\cdot\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+0\cdot\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
+1\cdot\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
Problem 6

Example 2.8 gives a matrix that is nonsingular, and is therefore associated with maps that are nonsingular.

  1. Find the set of column vectors representing the members of the nullspace of any map represented by this matrix.
  2. Find the nullity of any such map.
  3. Find the set of column vectors representing the members of the rangespace of any map represented by this matrix.
  4. Find the rank of any such map.
  5. Check that rank plus nullity equals the dimension of the domain.
Answer

Let the matrix be G, and suppose that it rperesents g:V\to W with respect to bases B and D. Because G has two columns, V is two-dimensional. Because G has two rows, W is two-dimensional. The action of g on a general member of the domain is this.


\begin{pmatrix} x \\ y \end{pmatrix}_B 
\;\mapsto\; 
\begin{pmatrix} x+2y \\ 3x+6y \end{pmatrix}_D
  1. The only representation of the zero vector in the codomain is
    
{\rm Rep}_{D}(\vec{0})=\begin{pmatrix} 0 \\ 0 \end{pmatrix}_D
    and so the set of representations of members of the nullspace is this.
    
\{\begin{pmatrix} x \\ y \end{pmatrix}_B\,\big|\, x+2y=0\text{ and }3x+6y=0\}
=\{y\cdot\begin{pmatrix} -1/2 \\ 1 \end{pmatrix}_D\,\big|\, y\in\mathbb{R}\}
  2. The representation map \mbox{Rep}_D:W\to \mathbb{R}^2 and its inverse are isomorphisms, and so preserve the dimension of subspaces. The subspace of \mathbb{R}^2 that is in the prior item is one-dimensional. Therefore, the image of that subspace under the inverse of the representation map— the nullspace of G, is also one-dimensional.
  3. The set of representations of members of the rangespace is this.
    
\{\begin{pmatrix} x+2y \\ 3x+6y \end{pmatrix}_D\,\big|\, x,y\in\mathbb{R}\}
=\{k\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix}_D\,\big|\, k\in\mathbb{R}\}
  4. Of course, Theorem 2.3 gives that the rank of the map equals the rank of the matrix, which is one. Alternatively, the same argument that was used above for the nullspace gives here that the dimension of the rangespace is one.
  5. One plus one equals two.
This exercise is recommended for all readers.
Problem 7

Because the rank of a matrix equals the rank of any map it represents, if one matrix represents two different maps  H={\rm Rep}_{B,D}(h)={\rm Rep}_{\hat{B},\hat{D}}(\hat{h}) (where  h,\hat{h}:V\to W ) then the dimension of the rangespace of  h equals the dimension of the rangespace of  \hat{h} . Must these equal-dimensioned rangespaces actually be the same?

Answer

No, the rangespaces may differ. Example 2.2 shows this.

This exercise is recommended for all readers.
Problem 8

Let  V be an  n -dimensional space with bases  B and  D . Consider a map that sends, for  \vec{v}\in V, the column vector representing  \vec{v} with respect to  B to the column vector representing  \vec{v} with respect to  D . Show that is a linear transformation of  \mathbb{R}^n .

Answer

Recall that the represention map


V\stackrel{\text{Rep}_{B}}{\longmapsto}\mathbb{R}^n

is an isomorphism. Thus, its inverse map \mbox{Rep}_B^{-1}:\mathbb{R}^n\to V is also an isomorphism. The desired transformation of \mathbb{R}^n is then this composition.


\mathbb{R}^n\stackrel{\text{Rep}_{B}^{-1}}{\longmapsto}
V\stackrel{\text{Rep}_{D}}{\longmapsto}\mathbb{R}^n

Because a composition of isomorphisms is also an isomorphism, this map \mbox{Rep}_{D}\circ \mbox{Rep}_{B}^{-1} is an isomorphism.

Problem 9

Example 2.2 shows that changing the pair of bases can change the map that a matrix represents, even though the domain and codomain remain the same. Could the map ever not change? Is there a matrix  H , vector spaces  V and  W , and associated pairs of bases  B_1,D_1 and  B_2,D_2 (with  B_1\neq B_2 or  D_1\neq D_2 or both) such that the map represented by  H with respect to  B_1,D_1 equals the map represented by  H with respect to  B_2,D_2 ?

Answer

Yes. Consider


H=\begin{pmatrix}
1  &0  \\
0  &1
\end{pmatrix}

representing a map from  \mathbb{R}^2 to  \mathbb{R}^2 . With respect to the standard bases  B_1=\mathcal{E}_2, D_1=\mathcal{E}_2 this matrix represents the identity map. With respect to


B_2=D_2=\langle \begin{pmatrix} 1 \\ 1 \end{pmatrix},\begin{pmatrix} 1 \\ -1 \end{pmatrix} \rangle

this matrix again represents the identity. In fact, as long as the starting and ending bases are equal— as long as  B_i=D_i — then the map represented by H is the identity.

This exercise is recommended for all readers.
Problem 10

A square matrix is a diagonal matrix if it is all zeroes except possibly for the entries on its upper-left to lower-right diagonal— its  1,1 entry, its  2,2 entry, etc. Show that a linear map is an isomorphism if there are bases such that, with respect to those bases, the map is represented by a diagonal matrix with no zeroes on the diagonal.

Answer

This is immediate from Corollary 2.6.

Problem 11

Describe geometrically the action on  \mathbb{R}^2 of the map represented with respect to the standard bases \mathcal{E}_2,\mathcal{E}_2 by this matrix.


\begin{pmatrix}
3  &0  \\
0  &2
\end{pmatrix}

Do the same for these.


\begin{pmatrix}
1  &0  \\
0  &0
\end{pmatrix}
\quad
\begin{pmatrix}
0  &1  \\
1  &0
\end{pmatrix}
\quad
\begin{pmatrix}
1  &3  \\
0  &1
\end{pmatrix}
Answer

The first map


\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}_{\mathcal{E}_2}
\mapsto
\begin{pmatrix} 3x \\ 2y \end{pmatrix}_{\mathcal{E}_2}=\begin{pmatrix} 3x \\ 2y \end{pmatrix}

stretches vectors by a factor of three in the  x direction and by a factor of two in the  y direction. The second map


\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}_{\mathcal{E}_2}
\mapsto
\begin{pmatrix} x \\ 0 \end{pmatrix}_{\mathcal{E}_2}=\begin{pmatrix} x \\ 0 \end{pmatrix}

projects vectors onto the  x axis. The third


\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}_{\mathcal{E}_2}
\mapsto
\begin{pmatrix} y \\ x \end{pmatrix}_{\mathcal{E}_2}=\begin{pmatrix} y \\ x \end{pmatrix}

interchanges first and second components (that is, it is a reflection about the line  y=x ). The last


\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x \\ y \end{pmatrix}_{\mathcal{E}_2}
\mapsto
\begin{pmatrix} x+3y \\ y \end{pmatrix}_{\mathcal{E}_2}=\begin{pmatrix} x+3y \\ y \end{pmatrix}

stretches vectors parallel to the  y axis, by an amount equal to three times their distance from that axis (this is a skew.)

Problem 12

The fact that for any linear map the rank plus the nullity equals the dimension of the domain shows that a necessary condition for the existence of a homomorphism between two spaces, onto the second space, is that there be no gain in dimension. That is, where h:V\to W is onto, the dimension of W must be less than or equal to the dimension of V.

  1. Show that this (strong) converse holds: no gain in dimension implies that there is a homomorphism and, further, any matrix with the correct size and correct rank represents such a map.
  2. Are there bases for \mathbb{R}^3 such that this matrix
    
H=\begin{pmatrix}
1  &0  &0 \\
2  &0  &0 \\
0  &1  &0 
\end{pmatrix}
    represents a map from \mathbb{R}^3 to \mathbb{R}^3 whose range is the xy plane subspace of \mathbb{R}^3?
Answer
  1. This is immediate from Theorem 2.3.
  2. Yes. This is immediate from the prior item. To give a specific example, we can start with \mathcal{E}_3 as the basis for the domain, and then we require a basis D for the codomain \mathbb{R}^3. The matrix H gives the action of the map as this
    
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}_{\mathcal{E}_3}
\mapsto\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}_D
\quad       
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}_{\mathcal{E}_3}
\mapsto\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}_D
\quad       
\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}_{\mathcal{E}_3}
\mapsto\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}_D
    and there is no harm in finding a basis D so that
    
{\rm Rep}_{D}(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix})=\begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}_D
\quad\mbox{and}\quad       
{\rm Rep}_{D}(\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix})=\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}_D
    that is, so that the map represented by H with respect to \mathcal{E}_3,D is projection down onto the xy plane. The second condition gives that the third member of D is \vec{e}_2. The first condition gives that the first member of D plus twice the second equals \vec{e}_1, and so this basis will do.
    
D=\langle \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1/2 \\ 1/2 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \rangle
Problem 13

Let  V be an  n -dimensional space and suppose that  \vec{x}\in\mathbb{R}^n . Fix a basis  B for  V and consider the map  h_{\vec{x}}:V\to \mathbb{R} given \vec{v}\mapsto\vec{x}\cdot{\rm Rep}_{B}(\vec{v}) by the dot product.

  1. Show that this map is linear.
  2. Show that for any linear map  g:V\to \mathbb{R} there is an  \vec{x}\in\mathbb{R}^n such that  g=h_{\vec{x}} .
  3. In the prior item we fixed the basis and varied the  \vec{x} to get all possible linear maps. Can we get all possible linear maps by fixing an  \vec{x} and varying the basis?
Answer
  1. Recall that the representation map \mbox{Rep}_{B}:V\to \mathbb{R}^n is linear (it is actually an isomorphism, but we do not need that it is one-to-one or onto here). Considering the column vector x to be a n \! \times \! 1 matrix gives that the map from \mathbb{R}^n to \mathbb{R} that takes a column vector to its dot product with \vec{x} is linear (this is a matrix-vector product and so Theorem 2.1 applies). Thus the map under consideration h_{\vec{x}} is linear because it is the composistion of two linear maps.
    
\vec{v}\mapsto {\rm Rep}_{B}(\vec{v})
\mapsto \vec{x}\cdot{\rm Rep}_{B}(\vec{v})
  2. Any linear map g:V\to \mathbb{R} is represented by some matrix
    
\begin{pmatrix}
g_1  &g_2 &\cdots &g_n
\end{pmatrix}
    (the matrix has n columns because V is n-dimensional and it has only one row because \mathbb{R} is one-dimensional). Then taking \vec{x} to be the column vector that is the transpose of this matrix
    
\vec{x}=\begin{pmatrix} g_1 \\ \vdots \\ g_n \end{pmatrix}
    has the desired action.
    
\vec{v}=\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}
\mapsto
\begin{pmatrix} g_1 \\ \vdots \\ g_n \end{pmatrix}\cdot\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}
=g_1v_1+\dots+g_nv_n
  3. No. If  \vec{x} has any nonzero entries then  h_{\vec{x}} cannot be the zero map (and if  \vec{x} is the zero vector then  h_{\vec{x}} can only be the zero map).
Problem 14

Let  V,W,X be vector spaces with bases  B,C,D .

  1. Suppose that  h:V\to W is represented with respect to  B,C by the matrix  H . Give the matrix representing the scalar multiple  rh (where  r\in\mathbb{R} ) with respect to  B,C by expressing it in terms of  H .
  2. Suppose that  h,g:V\to W are represented with respect to  B,C by  H and  G . Give the matrix representing  h+g with respect to  B,C by expressing it in terms of  H and  G .
  3. Suppose that  h:V\to W is represented with respect to  B,C by  H and  g:W\to X is represented with respect to  C,D by  G . Give the matrix representing  g\circ h with respect to  B,D by expressing it in terms of  H and  G .
Answer

See the following section.