09/10/2014

Solotion:Vector spaces and linear systems

Solotion:Vector spaces and linear systems:



Problem 1
Transpose each.
  1. \begin{pmatrix} 2 &1 \\ 3 &1 \end{pmatrix}
  2. \begin{pmatrix} 2 &1 \\ 1 &3 \end{pmatrix}
  3. \begin{pmatrix} 1 &4 &3 \\ 6 &7 &8 \end{pmatrix}
  4. \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
  5. \begin{pmatrix} -1 &-2 \end{pmatrix}
Answer
  1. \begin{pmatrix} 2 &3 \\ 1 &1 \end{pmatrix}
  2. \begin{pmatrix} 2 &1 \\ 1 &3 \end{pmatrix}
  3. \begin{pmatrix} 1 &6 \\ 4 &7 \\ 3 &8 \end{pmatrix}
  4. \begin{pmatrix} 0 &0 &0 \end{pmatrix}
  5. \begin{pmatrix} -1 \\ -2 \end{pmatrix}
Problem 2
Decide if the vector is in the row space of the matrix.
  1. \begin{pmatrix} 2 &1 \\ 3 &1 \end{pmatrix} ,  \begin{pmatrix} 1 &0 \end{pmatrix}
  2. \begin{pmatrix} 0 &1 &3 \\ -1 &0 &1 \\ -1 &2 &7 \end{pmatrix} ,  \begin{pmatrix} 1 &1 &1 \end{pmatrix}
Answer
  1. Yes. To see if there are c_1 and c_2 such that  c_1\cdot\begin{pmatrix} 2 &1 \end{pmatrix}+c_2\cdot\begin{pmatrix} 3 &1 \end{pmatrix}=\begin{pmatrix} 1 &0 \end{pmatrix}  we solve
    \begin{array}{*{2}{rc}r} 2c_1 &+ &3c_2 &= &1 \\ c_1 &+ &c_2 &= &0 \end{array}
    and get  c_1=-1  and  c_2=1 . Thus the vector is in the row space.
  2. No. The equation  c_1\begin{pmatrix} 0 &1 &3 \end{pmatrix} +c_2\begin{pmatrix} -1 &0 &1 \end{pmatrix} +c_3\begin{pmatrix} -1 &2 &7 \end{pmatrix} =\begin{pmatrix} 1 &1 &1 \end{pmatrix} has no solution.
    \left(\begin{array}{*{3}{c}|c} 0 &-1 &-1 &1 \\ 1 &0 &2 &1 \\ 3 &1 &7 &1 \end{array}\right) \xrightarrow[]{\rho_1\leftrightarrow\rho_2} \;\xrightarrow[]{-3\rho_1+\rho_2} \;\xrightarrow[]{\rho_2+\rho_3} \left(\begin{array}{*{3}{c}|c} 1 &0 &2 &1 \\ 0 &-1 &-1 &1 \\ 0 &0 &0 &-1 \end{array}\right)
    Thus, the vector is not in the row space.

Problem 3
Decide if the vector is in the column space.
  1. \begin{pmatrix} 1 &1 \\ 1 &1 \end{pmatrix} ,  \begin{pmatrix} 1 \\ 3 \end{pmatrix}
  2. \begin{pmatrix} 1 &3 &1 \\ 2 &0 &4 \\ 1 &-3 &-3 \end{pmatrix} ,  \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
Answer
  1. No. To see if there are  c_1,c_2\in\mathbb{R}  such that
    c_1\begin{pmatrix} 1 \\ 1 \end{pmatrix} +c_2\begin{pmatrix} 1 \\ 1 \end{pmatrix} =\begin{pmatrix} 1 \\ 3 \end{pmatrix}
    we can use Gauss' method on the resulting linear system.
    \begin{array}{*{2}{rc}r} c_1 &+ &c_2 &= &1 \\ c_1 &+ &c_2 &= &3 \end{array} \xrightarrow[]{-\rho_1+\rho_2} \;\begin{array}{*{2}{rc}r} c_1 &+ &c_2 &= &1 \\ & &0 &= &2 \end{array}
    There is no solution and so the vector is not in the column space.
  2. Yes. From this relationship
    c_1\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} +c_2\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} +c_3\begin{pmatrix} 1 \\ 4 \\ 3 \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}
    we get a linear system that, when Gauss' method is applied,
    \left(\begin{array}{*{3}{c}|c} 1 &3 &1 &1 \\ 2 &0 &4 &0 \\ 1 &-3 &-3 &0 \end{array}\right) \xrightarrow[-\rho_1+\rho_3]{-2\rho_1+\rho_2} \;\xrightarrow[]{-\rho_2+\rho_3} \left(\begin{array}{*{3}{c}|c} 1 &3 &1 &1 \\ 0 &-6 &2 &-2 \\ 0 &0 &-6 &1 \end{array}\right)
    yields a solution. Thus, the vector is in the column space.

Problem 4
Find a basis for the row space of this matrix.

\begin{pmatrix} 2 &0 &3 &4 \\ 0 &1 &1 &-1 \\ 3 &1 &0 &2 \\ 1 &0 &-4 &-1 \end{pmatrix}

Answer
A routine Gaussian reduction
\begin{pmatrix} 2 &0 &3 &4 \\ 0 &1 &1 &-1 \\ 3 &1 &0 &2 \\ 1 &0 &-4 &1 \end{pmatrix} \xrightarrow[-(1/2)\rho_1+\rho_4]{-(3/2)\rho_1+\rho_3} \;\xrightarrow[]{-\rho_2+\rho_3} \;\xrightarrow[]{-\rho_3+\rho_4} \begin{pmatrix} 2 &0 &3 &4 \\ 0 &1 &1 &-1 \\ 0 &0 &-11/2 &-3 \\ 0 &0 &0 &0 \end{pmatrix}
suggests this basis \langle \begin{pmatrix} 2 &0 &3 &4 \end{pmatrix},\begin{pmatrix} 0 &1 &1 &-1 \end{pmatrix},\begin{pmatrix} 0 &0 &-11/2 &-3 \end{pmatrix} \rangle .
Another, perhaps more convenient procedure, is to swap rows first,
\xrightarrow[]{\rho_1\leftrightarrow\rho_4} \;\xrightarrow[-2\rho_1+\rho_4]{-3\rho_1+\rho_3} \;\xrightarrow[]{-\rho_2+\rho_3} \;\xrightarrow[]{-\rho_3+\rho_4} \begin{pmatrix} 1 &0 &-4 &-1 \\ 0 &1 &1 &-1 \\ 0 &0 &11 &6 \\ 0 &0 &0 &0 \end{pmatrix}
leading to the basis  \langle \begin{pmatrix} 1 &0 &-4 &-1 \end{pmatrix},\begin{pmatrix} 0 &1 &1 &-1 \end{pmatrix}, \begin{pmatrix} 0 &0 &11 &6 \end{pmatrix} \rangle.

Problem 5
Find the rank of each matrix.
  1. \begin{pmatrix} 2 &1 &3 \\ 1 &-1 &2 \\ 1 &0 &3 \end{pmatrix}
  2. \begin{pmatrix} 1 &-1 &2 \\ 3 &-3 &6 \\ -2 &2 &-4 \end{pmatrix}
  3. \begin{pmatrix} 1 &3 &2 \\ 5 &1 &1 \\ 6 &4 &3 \end{pmatrix}
  4. \begin{pmatrix} 0 &0 &0 \\ 0 &0 &0 \\ 0 &0 &0 \end{pmatrix}
Answer
  1. This reduction
    \xrightarrow[-(1/2)\rho_1+\rho_3]{-(1/2)\rho_1+\rho_2} \;\xrightarrow[]{-(1/3)\rho_2+\rho_3} \begin{pmatrix} 2 &1 &3 \\ 0 &-3/2 &1/2 \\ 0 &0 &4/3 \end{pmatrix}
    shows that the row rank, and hence the rank, is three.
  2. Inspection of the columns shows that that the others are multiples of the first (inspection of the rows shows the same thing). Thus the rank is one. Alternatively, the reduction
    \begin{pmatrix} 1 &-1 &2 \\ 3 &-3 &6 \\ -2 &2 &-4 \end{pmatrix} \xrightarrow[2\rho_1+\rho_3]{-3\rho_1+\rho_2} \begin{pmatrix} 1 &-1 &2 \\ 0 &0 &0 \\ 0 &0 &0 \end{pmatrix}
    shows the same thing.
  3. This calculation
    \begin{pmatrix} 1 &3 &2 \\ 5 &1 &1 \\ 6 &4 &3 \end{pmatrix} \xrightarrow[-6\rho_1+\rho_3]{-5\rho_1+\rho_2} \;\xrightarrow[]{-\rho_2+\rho_3} \begin{pmatrix} 1 &3 &2 \\ 0 &-14 &-9 \\ 0 &0 &0 \end{pmatrix}
    shows that the rank is two.
  4. The rank is zero.

Problem 6
Find a basis for the span of each set.
  1. \{\begin{pmatrix} 1 &3 \end{pmatrix}, \begin{pmatrix} -1 &3 \end{pmatrix}, \begin{pmatrix} 1 &4 \end{pmatrix}, \begin{pmatrix} 2 &1 \end{pmatrix} \}\subseteq\mathcal{M}_{1 \! \times \! 2}
  2. \{\begin{pmatrix} 1 \\2 \\1 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ -3 \\ -3 \end{pmatrix} \}\subseteq\mathbb{R}^3
  3. \{1+x,1-x^2,3+2x-x^2\}\subseteq\mathcal{P}_3
  4. \{ \begin{pmatrix} 1 &0 &1 \\ 3 &1 &-1 \end{pmatrix}, \begin{pmatrix} 1 &0 &3 \\ 2 &1 &4 \end{pmatrix}, \begin{pmatrix} -1 &0 &-5 \\ -1 &-1 &-9 \end{pmatrix} \} \subseteq\mathcal{M}_{2 \! \times \! 3}
Answer
  1. This reduction
    \begin{pmatrix} 1 &3 \\ -1 &3 \\ 1 &4 \\ 2 &1 \end{pmatrix} \xrightarrow[\begin{array}{c}\\[-19pt]\scriptstyle -\rho_1+\rho_3 \\[-5pt] \scriptstyle -2\rho_1+\rho_4\end{array}]{\rho_1+\rho_2} \xrightarrow[(5/6)\rho_2+\rho_4]{-(1/6)\rho_2+\rho_3} \begin{pmatrix} 1 &3 \\ 0 &6 \\ 0 &0 \\ 0 &0 \end{pmatrix}
    gives  \langle \begin{pmatrix} 1 &3 \end{pmatrix},\begin{pmatrix} 0 &6 \end{pmatrix} \rangle .
  2. Transposing and reducing
    \begin{pmatrix} 1 &2 &1 \\ 3 &1 &-1 \\ 1 &-3 &-3 \end{pmatrix} \xrightarrow[-\rho_1+\rho_3]{-3\rho_1+\rho_2} \begin{pmatrix} 1 &2 &1 \\ 0 &-5 &-4 \\ 0 &-5 &-4 \end{pmatrix} \xrightarrow[]{-\rho_2+\rho_3} \begin{pmatrix} 1 &2 &1 \\ 0 &-5 &-4 \\ 0 &0 &0 \end{pmatrix}
    and then transposing back gives this basis.
    \langle \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ -5 \\ -4 \end{pmatrix} \rangle
  3. Notice first that the surrounding space is given as \mathcal{P}_3, not \mathcal{P}_2. Then, taking the first polynomial 1+1\cdot x+0\cdot x^2+0\cdot x^3 to be "the same" as the row vector \begin{pmatrix} 1 &1 &0 &0 \end{pmatrix}, etc., leads to
    \begin{pmatrix} 1 &1 &0 &0 \\ 1 &0 &-1 &0 \\ 3 &2 &-1 &0 \end{pmatrix} \xrightarrow[-3\rho_1+\rho_3]{-\rho_1+\rho_2} \;\xrightarrow[]{-\rho_2+\rho_3} \begin{pmatrix} 1 &1 &0 &0 \\ 0 &-1 &-1 &0 \\ 0 &0 &0 &0 \end{pmatrix}
    which yields the basis  \langle 1+x,-x-x^2 \rangle .
  4. Here "the same" gives
    \begin{pmatrix} 1 &0 &1 &3 &1 &-1 \\ 1 &0 &3 &2 &1 &4 \\ -1 &0 &-5 &-1 &-1 &-9 \end{pmatrix} \xrightarrow[\rho_1+\rho_3]{-\rho_1+\rho_2} \;\xrightarrow[]{2\rho_2+\rho_3} \begin{pmatrix} 1 &0 &1 &3 &1 &-1 \\ 0 &0 &2 &-1 &0 &5 \\ 0 &0 &0 &0 &0 &0 \end{pmatrix}
    leading to this basis.
    \langle \begin{pmatrix} 1 &0 &1 \\ 3 &1 &-1 \end{pmatrix}, \begin{pmatrix} 0 &0 &2 \\ -1 &0 &5 \end{pmatrix} \rangle
Problem 7
Which matrices have rank zero? Rank one?
Answer
Only the zero matrices have rank of zero. The only matrices of rank one have the form
\begin{pmatrix} k_1\cdot \rho \\ \vdots \\ k_m\cdot \rho \end{pmatrix}

where  \rho  is some nonzero row vector, and not all of the  k_i 's are zero. (Remark. We can't simply say that all of the rows are multiples of the first because the first row might be the zero row. Another Remark. The above also applies with "column" replacing "row".)

Problem 8
Given  a,b,c\in\mathbb{R} , what choice of  d  will cause this matrix to have the rank of one?

\begin{pmatrix} a &b \\ c &d \end{pmatrix}

Answer
If  a\neq 0  then a choice of  d=(c/a)b  will make the second row be a multiple of the first, specifically,  c/a times the first. If  a=0  and  b=0  then any non- 0  choice for  d  will ensure that the second row is nonzero. If  a=0 and  b\neq 0  and  c=0  then any choice for  d  will do, since the matrix will automatically have rank one (even with the choice of d=0). Finally, if  a=0  and  b\neq 0  and  c\neq 0 then no choice for  d  will suffice because the matrix is sure to have rank two.
Problem 9
Find the column rank of this matrix.
\begin{pmatrix} 1 &3 &-1 &5 &0 &4 \\ 2 &0 &1 &0 &4 &1 \end{pmatrix}
Answer
The column rank is two. One way to see this is by inspection— the column space consists of two-tall columns and so can have a dimension of at least two, and we can easily find two columns that together form a linearly independent set (the fourth and fifth columns, for instance). Another way to see this is to recall that the column rank equals the row rank, and to perform Gauss' method, which leaves two nonzero rows.
Problem 10
Show that a linear system with at least one solution has at most one solution if and only if the matrix of coefficients has rank equal to the number of its columns.
Answer
We apply Theorem 3.13. The number of columns of a matrix of coefficients A of a linear system equals the number n of unknowns. A linear system with at least one solution has at most one solution if and only if the space of solutions of the associated homogeneous system has dimension zero (recall: in the "\text{General}=\text{Particular}+\text{Homogeneous}" equation \vec{v}=\vec{p}+\vec{h}, provided that such a \vec{p} exists, the solution \vec{v}is unique if and only if the vector \vec{h} is unique, namely \vec{h}=\vec{0}). But that means, by the theorem, that n=r.

Problem 11
If a matrix is  5 \! \times \! 9 , which set must be dependent, its set of rows or its set of columns?
Answer
The set of columns must be dependent because the rank of the matrix is at most five while there are nine columns.
Problem 12
Give an example to show that, despite that they have the same dimension, the row space and column space of a matrix need not be equal. Are they ever equal?
Answer
There is little danger of their being equal since the row space is a set of row vectors while the column space is a set of columns (unless the matrix is  1 \! \times \! 1 , in which case the two spaces must be equal).
Remark. Consider
A=\begin{pmatrix} 1 &3 \\ 2 &6 \end{pmatrix}
and note that the row space is the set of all multiples of  \begin{pmatrix} 1 &3 \end{pmatrix}  while the column space consists of multiples of
\begin{pmatrix} 1 \\ 2 \end{pmatrix}
so we also cannot argue that the two spaces must be simply transposes of each other.
Problem 13
Show that the set  \{(1,-1,2,-3),(1,1,2,0),(3,-1,6,-6)\}  does not have the same span as  \{(1,0,1,0),(0,2,0,3)\} . What, by the way, is the vector space?
Answer
First, the vector space is the set of four-tuples of real numbers, under the natural operations. Although this is not the set of four-wide row vectors, the difference is slight— it is "the same" as that set. So we will treat the four-tuples like four-wide vectors.
With that, one way to see that  (1,0,1,0)  is not in the span of the first set is to note that this reduction
\begin{pmatrix} 1 &-1 &2 &-3 \\ 1 &1 &2 &0 \\ 3 &-1 &6 &-6 \end{pmatrix} \xrightarrow[-3\rho_1+\rho_3]{-\rho_1+\rho_2} \;\xrightarrow[]{-\rho_2+\rho_3} \begin{pmatrix} 1 &-1 &2 &-3 \\ 0 &2 &0 &3 \\ 0 &0 &0 &0 \end{pmatrix}
and this one
\begin{pmatrix} 1 &-1 &2 &-3 \\ 1 &1 &2 &0 \\ 3 &-1 &6 &-6 \\ 1 &0 &1 &0 \end{pmatrix} \xrightarrow[\begin{array}{c}\\[-19pt]\scriptstyle -3\rho_1+\rho_3 \\[-5pt] \scriptstyle -\rho_1+\rho_4\end{array}]{-\rho_1+\rho_2} \;\xrightarrow[-(1/2)\rho_2+\rho_4]{-\rho_2+\rho_3} \;\xrightarrow[]{\rho_3\leftrightarrow\rho_4} \begin{pmatrix} 1 &-1 &2 &-3 \\ 0 &2 &0 &3 \\ 0 &0 &-1 &3/2 \\ 0 &0 &0 &0 \end{pmatrix}
yield matrices differing in rank. This means that addition of(1,0,1,0) to the set of the first three four-tuples increases the rank, and hence the span, of that set. Therefore (1,0,1,0) is not already in the span.

Problem 14
Show that this set of column vectors
\left\{\begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix} \,\big|\, \text{there are }x, y, \text{ and } z \text{ such that } \begin{array}{*{3}{rc}r} 3x &+ &2y &+ &4z &= &d_1 \\ x & & &- &z &= &d_2 \\ 2x &+ &2y &+ &5z &= &d_3 \end{array} \right\}
is a subspace of  \mathbb{R}^3 . Find a basis.
Answer
It is a subspace because it is the column space of the matrix
\begin{pmatrix} 3 &2 &4 \\ 1 &0 &-1 \\ 2 &2 &5 \end{pmatrix}
of coefficients. To find a basis for the column space,
\{c_1\begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} +c_2\begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix} +c_3\begin{pmatrix} 4 \\ -1 \\ 5 \end{pmatrix} \,\big|\, c_1,c_2,c_3\in\mathbb{R}\}
we take the three vectors from the spanning set, transpose, reduce,
\begin{pmatrix} 3 &1 &2 \\ 2 &0 &2 \\ 4 &-1 &5 \end{pmatrix} \xrightarrow[-(4/3)\rho_1+\rho_3]{-(2/3)\rho_1+\rho_2} \;\xrightarrow[]{-(7/2)\rho_2+\rho_3} \begin{pmatrix} 3 &1 &2 \\ 0 &-2/3 &2/3 \\ 0 &0 &0 \end{pmatrix}
and transpose back to get this.
\langle \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ -2/3 \\ 2/3 \end{pmatrix} \rangle
Problem 15
Show that the transpose operation is linear:
{{(rA+sB)}^{\rm trans}} = r{{A}^{\rm trans}}+s{{B}^{\rm trans}}
for  r,s\in\mathbb{R}  and  A,B\in\mathcal{M}_{m \! \times \! n} .
Answer
This can be done as a straightforward calculation.

\begin{array}{rl} {{(rA+sB)}^{\rm trans}} &={{\begin{pmatrix} ra_{1,1}+sb_{1,1} &\ldots &ra_{1,n}+sb_{1,n} \\ \vdots & &\vdots \\ ra_{m,1}+sb_{m,1} &\ldots &ra_{m,n}+sb_{m,n} \end{pmatrix} }^{\rm trans}} \\ &=\begin{pmatrix} ra_{1,1}+sb_{1,1} &\ldots &ra_{m,1}+sb_{m,1} \\ \vdots \\ ra_{1,n}+sb_{1,n} &\ldots &ra_{m,n}+sb_{m,n} \end{pmatrix} \\ &=\begin{pmatrix} ra_{1,1} &\ldots &ra_{m,1} \\ \vdots \\ ra_{1,n} &\ldots &ra_{m,n} \end{pmatrix} +\begin{pmatrix} sb_{1,1} &\ldots &sb_{m,1} \\ \vdots \\ sb_{1,n} &\ldots &sb_{m,n} \end{pmatrix} \\ &=r{{A}^{\rm trans}}+s{{B}^{\rm trans}} \end{array}

Problem 16
In this subsection we have shown that Gaussian reduction finds a basis for the row space.
  1. Show that this basis is not unique— different reductions may yield different bases.
  2. Produce matrices with equal row spaces but unequal numbers of rows.
  3. Prove that two matrices have equal row spaces if and only if after Gauss-Jordan reduction they have the same nonzero rows.
Answer
  1. These reductions give different bases.
    \begin{pmatrix} 1 &2 &0 \\ 1 &2 &1 \end{pmatrix} \xrightarrow[]{-\rho_1+\rho_2} \begin{pmatrix} 1 &2 &0 \\ 0 &0 &1 \end{pmatrix} \qquad \begin{pmatrix} 1 &2 &0 \\ 1 &2 &1 \end{pmatrix} \xrightarrow[]{-\rho_1+\rho_2} \;\xrightarrow[]{2\rho_2} \begin{pmatrix} 1 &2 &0 \\ 0 &0 &2 \end{pmatrix}
  2. An easy example is this.
    \begin{pmatrix} 1 &2 &1 \\ 3 &1 &4 \end{pmatrix} \qquad \begin{pmatrix} 1 &2 &1 \\ 3 &1 &4 \\ 0 &0 &0 \end{pmatrix}
    This is a less simplistic example.
    \begin{pmatrix} 1 &2 &1 \\ 3 &1 &4 \end{pmatrix} \qquad \begin{pmatrix} 1 &2 &1 \\ 3 &1 &4 \\ 2 &4 &2 \\ 4 &3 &5 \end{pmatrix}
  3. Assume that  A  and  B  are matrices with equal row spaces. Construct a matrix  C  with the rows of  A  above the rows of  B , and another matrix  D  with the rows of  B  above the rows of  A .
    C=\begin{pmatrix} A \\ B \end{pmatrix} \qquad D=\begin{pmatrix} B \\ A \end{pmatrix}
    Observe that  C  and  D  are row-equivalent (via a sequence of row-swaps) and so Gauss-Jordan reduce to the same reduced echelon form matrix. Because the row spaces are equal, the rows of  B  are linear combinations of the rows of A  so Gauss-Jordan reduction on  C  simply turns the rows of  B  to zero rows and thus the nonzero rows of C are just the nonzero rows obtained by Gauss-Jordan reducing  A . The same can be said for the matrix  D — Gauss-Jordan reduction on  D  gives the same non-zero rows as are produced by reduction on  B  alone. Therefore,  A  yields the same nonzero rows as  C , which yields the same nonzero rows as  D , which yields the same nonzero rows as  B .
Problem 17
Why is there not a problem with Remark 3.14 in the case that  r  is bigger than  n ?
Answer
It cannot be bigger.
Problem 18
Show that the row rank of an  m \! \times \! n  matrix is at most  m . Is there a better bound?
Answer
The number of rows in a maximal linearly independent set cannot exceed the number of rows. A better bound (the bound that is, in general, the best possible) is the minimum of  m  and  n , because the row rank equals the column rank.

Problem 19
Show that the rank of a matrix equals the rank of its transpose.
Answer
Because the rows of a matrix  A  are turned into the columns of  {{A}^{\rm trans}}  the dimension of the row space of  A equals the dimension of the column space of  {{A}^{\rm trans}} . But the dimension of the row space of  A  is the rank of  A  and the dimension of the column space of  {{A}^{\rm trans}}  is the rank of  {{A}^{\rm trans}} . Thus the two ranks are equal.
Problem 20
True or false: the column space of a matrix equals the row space of its transpose.
Answer
False. The first is a set of columns while the second is a set of rows.
This example, however,
A=\begin{pmatrix} 1 &2 &3 \\ 4 &5 &6 \end{pmatrix}, \qquad {{A}^{\rm trans}}=\begin{pmatrix} 1 &4 \\ 2 &5 \\ 3 &6 \end{pmatrix}
indicates that as soon as we have a formal meaning for "the same", we can apply it here:
\mathop{{\text{Columnspace}}}(A)=[\{ \begin{pmatrix} 1 \\ 4 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 6 \end{pmatrix} \} ]
while
\mathop{{\mbox{Rowspace}}}({{A}^{\rm trans}})=[\{ \begin{pmatrix} 1 &4 \end{pmatrix}, \begin{pmatrix} 2 &5 \end{pmatrix}, \begin{pmatrix} 3 &6 \end{pmatrix} \} ]
are "the same" as each other.
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