Section6.4Constructing Matrices With Given Row, Column, or Nullspace Properties
We offer now an opportunity to use our right-brains for a change. In this section we will learn to solve the problem “given a set of properties about \(\A\in\R^{m\times n},\)” is such an \(\A\) possible and if so how do we characterize such an \(\A\text{?}\) This will help deepen our understanding of the relationships between row spaces, column spaces, and nullspaces and in particular their dmensions. The first step in any such investigation should always be a check of consistency of the given properties against dimensional considerations such as Theorem 6.1.40. We call such a check the plausibility/size check. FFFFIIIIXXXX to include how to generate zero divisors.
Example6.4.1.
If possible construct a matrix \(\A\) whose column space and nullspace both contain \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: We perform the plausibility/size check first. For a \(3\)-vector to be in col\(\left(\A\right)\) we need \(\A\in\R^{3\times n}\) and for a \(3\)-vector to be in the nullspace we must have \(\A\in\R^{m\times 3}.\) Since both of these must be true we must have \(\A\in\R^{3\times 3}.\) We need dim(col\((\A))\ge1,\) and dim\((N(\A))\ge1,\) which is not precluded since by Theorem 6.1.40 we must have only that dim(col\((\A))\)+ dim\((N(\A))=n=3.\) So such an \(\A\) is possible.
At this point it’s really a question of how efficiently we can construct such a matrix. We put \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\) as the first column of \(\A,\) and since \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\in N(\A)\) we have the following relation between the columns \(\A_{\ast1},\A_{\ast2},\) and \(\A_{\ast3}\text{:}\)\(-3\A_{\ast1}+4\A_{\ast2}-\A_{\ast3}=\left(\begin{array}{r}0\\0\\0 \end{array} \right).\) Fixing \(\A_{\ast1}=\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\) yields
and we see that we can choose any \(\A_{\ast2},\A_{\ast3}\) for which this relation holds. My strategy is to choose \(\A_{\ast2}=\left(\begin{array}{r}0\\0\\0 \end{array} \right)\) (think simple!) and figure out what that demands of \(\A_{\ast3}.\) We then have
If possible construct a matrix \(\A\) whose column space and nullspace are both spanned by \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: This is certainly not possible. We must have \(\A\in\R^{3\times 3}\) which fixes \(n=3.\) For \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\) to span \(N(\A)\) we must have dim\((N(\A))\le1,\) and for \(\left(\begin{array}{r}-3\\4\\-1 \end{array} \right)\) to span col\((\A)\) we must have dim(col\((\A))\le1.\) Thus dim(col\((\A))+\) dim\((N(\A))\le2\ne n=3,\) so Theorem 6.1.40 is violated, and no such matrix exists.
Example6.4.3.
If possible, construct a matrix \(\A\) whose column space and nullspace both contain \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}4\\-4\\8 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: Such a matrix \(\A\) is not possible. We must have \(\A\in\R^{3\times 3}\) so that \(n=3.\) The vectors \(\left(\begin{array}{r}2\\1\\-3 \end{array} \right)\) and \(\left(\begin{array}{r}4\\-4\\8 \end{array} \right)\) are linearly independent (linear independence for two vectors is only obtained when neither vector is a scalar multiple of the other) so any basis for either \(N(\A)\) or col\((\A))\) must have at least two elements. So dim\((N(\A))\ge2\) and dim(col\((\A))\ge2.\) Such a matrix would satisfy dim(col\((\A))+\) dim\((N(\A)\ge4>n=3,\) violating Theorem 6.1.40. So no such matrix exists.
Example6.4.4.
If possible, construct a matrix \(\A\) whose column space contains \(\left(\begin{array}{r}-2\\3\\-7 \end{array} \right)\) and whose nullspace consists of all multiples of (i.e. is spanned by) \(\left(\begin{array}{r}1\\-3\\3\\6 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: Here such an \(\A\) must be in \(\R^{3\times 4}\) with one free variable and three pivot variables. We can let the first column \(\A_{\ast1}\) of \(\A\) be \(\A_{\ast1}=\left(\begin{array}{r}-2\\3\\-7 \end{array} \right),\) and we can let \(\A_{\ast2}\) and \(\A_{\ast3}\) be whatever we want as long as \(\{\A_{\ast1},\A_{\ast2},\A_{\ast3}\}\) is linearly independent (remember we can only have one free variable). Further, we can use the fourth column \(\A_{\ast4}\) to ensure that the nullspace relation \(\A_{\ast1}-3\A_{\ast2}+3\A_{\ast3}+6\A_{\ast4}=\vec{0}\) holds. So, if we set
If possible construct a matrix \(\A\) whose column space contains \(\left(\begin{array}{r}-2\\3\\7 \end{array} \right)\) and \(\left(\begin{array}{r}-4\\1\\6 \end{array} \right)\) and whose nullspace is spanned by \(\left(\begin{array}{r}1\\-2\\6 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: Here we must have \(\A\in\R^{3\times 3}\) so \(n=3,\) which fits with the two given linearly independent columns and a one-dimensional nullspace. We start with
We finish by establishing some convenient machinery for the case in which the only given condition regards the nullspace.
Definition6.4.6.
Let \(S\) be a subspace of \(\R^n.\) Then the orthogonal complement \(S^\perp\) of \(S\) is defined to the set of all vectors in \(\R^n\) which are orthogonal to every vector in \(S.\) That is
\begin{equation}
S^\perp=\left\{\vec{x}\in\R^n\,|\,\fa\vec{y}\in S,\,\vec{x}\perp\vec{y}\right (\hbox{that is, }\vec{x}\cdot\vec{y}=0)\}\tag{6.4.2}
\end{equation}
Example6.4.7.
Let \(S\) equal the \(xy\)-plane, as a subspace of \(\R^3.\) Then \(S^\perp\) equals the \(z\)-axis:
Suppose we wish to construct a matrix, \(\A,\) whose nullspace is spanned by some set of vectors, \(S=\{\vec{v_1},\vec{v_2},...,\vec{v_k} \}\) (that is, \(N(\A)=\text{ span } (S)).\) We can proceed by first inserting those vectors in a matrix \(\B\) as rows so that \(\text{ row } (\B)=\text{ span } (S).\) Then we find the nullspace \(N(\B)\) of \(\B.\) Next, we create the matrix \(\A\) whose rows are the basis vectors of \(N(\B).\) That is, \(N(\B)=\text{ row } (\A).\) It follows that \((N(\B))^\perp = (\text{ row } (\A))^\perp.\) By Theorem 6.4.9 we see \((N(\B))^\perp = \text{ row } (\B)\) while by Corollary 6.4.10\((\text{ row } (\A))^\perp = N(\A).\) Thus \(N(\A)=\text{ row } (\B)=\text{ span } (S)\) as desired. We illustrate this process in the next example.
Example6.4.13.
If possible, construct a matrix \(\A\) whose nullspace is spanned by \(\left(\begin{array}{r}1\\2\\4\\8 \end{array} \right).\) If this is not possible, explain carefully why it is not possible.
Solution: For such a constraint to be satisfied, we must have one free variable (dim \((N(\A))=1\)) and four columns, so there will be three pivot variables. Thus we can take \(\A\in\R^{3\times 4}.\) (Note: We could also take \(\A\) to have more than \(3\) rows, but in total there will be \(3\) linearly independent rows so we just illustrate with an example with \(\A \in \R^{3 \times 4}.\)) By Remark 6.4.12 we first seek \(N\left(\left(\begin{array}{rrrr}1\amp 2\amp 4\amp 8 \end{array} \right)\right).\) We have the pivot variable \(x_1\) and free variables \(x_2,x_3,\) and \(x_4.\) Setting \(x_2=1,\,x_3=x_4=0,\) then \(x_3=1,\,x_2=0=x_4,\) and finally \(x_4=1,\,x_2=x_3=0,\) we obtain, respectively, basis vectors for \(N(\B)\) to be:
Find a matrix \(\A\) whose nullspace consists of all linear combinations of \(\left(\begin{array}{r}1\\2\\3\\4 \end{array} \right)\) and \(\left(\begin{array}{r}5\\6\\7\\8 \end{array} \right),\) or determine that such a matrix does not exist.
Solution: Such a matrix does indeed exist. First, the plausibility/size check: For \(N(\A)\) to contain, for example, \(\left(\begin{array}{r}1\\2\\3\\4 \end{array} \right)\)\(n\) must be equal to \(4.\) We have dim\(N(\A))=2\) since there are two basis vectors for \(N(\A),\) so by Theorem 6.1.40 we must have two pivot variables so dim(row\((\A))=n=2.\) Thus we can find a matrix \(\A\in\R^{2\times 4}.\) In light of Remark 6.4.12 we form
\begin{align*}
x_1+2x_2+3x_3+4x_4\amp =0\qquad\text{ and }\\
-4x_2-8x_3-12x_4\amp =0
\end{align*}
which shows that we have pivot variables \(x_1,x_2\) and free variables \(x_3,x_4.\) Setting \(x_3=1,\,x_4=0\) and subsequently \(x_3=0,\,x_4=1\) we obtain, respectively, basis vectors for the nullspace of \(\B\) to be
Note that dim(row\((\A))=2=\) dim(col\((\A)),\) and dim(col\((\A))+\) dim\((N(\A))=n=4,\) as is necessary.
Remark6.4.15.
Note that dim\((N(\A))\) can equal \(0\) only under the circumstances that there is no basis for \(N(\A)\) which only occurs when \(N(\A)=\left\{\vec{0}\right\}.\)
Remark6.4.16.
Note that the conditions “\(\begin{pmatrix}a\\b\\c \end{pmatrix} \in N(\A)\)” and “\(\left\{\begin{pmatrix}a\\b\\c \end{pmatrix} \right\}\) spans \(N(\A)\)” mean very different things! The former means only that \(N(\A)\) has dimension at least \(1\) and the latter means that \(N(\A)\) has dimension at most \(1.\) Thus by Remark 6.4.15, if any of \(a,b,c\ne0\) in “\(\left\{\begin{pmatrix}a\\b\\c \end{pmatrix} \right\}\) spans \(N(\A)\)” we can conclude that dim\(N(\A))=1.\)
Remark6.4.17.
If we know that, for instance, \(\begin{pmatrix}1\\2\\3 \end{pmatrix} \in\text{ col } (\A)\) and we find that, also, \(\A\in\R^{3\times4},\) we are left with the task of filling out three additional columns of \(\A.\) This task is often set against some nullspace requirement, such as “\(\left\{\begin{pmatrix}2\\1\\4\\9 \end{pmatrix} ,\begin{pmatrix}{1}\\0\\5\\7 \end{pmatrix} \right\}\) spans \(N(\A)\)” which, since \(\left\{\begin{pmatrix}2\\1\\4\\9 \end{pmatrix} ,\begin{pmatrix}{1}\\0\\5\\7 \end{pmatrix} \right\}\) is linearly independent, means that by Definition 5.4.44 dim\(N(\A)=2\) so by Theorem 6.1.40, dim(col\((\A)\))\(=4-2=2.\) This fact allows us to choose another column \(\A_{\ast2}\) of \(\A\) as long as \(\left\{\begin{pmatrix}1\\2\\3 \end{pmatrix} ,\A_{\ast2}\right\}\) is linearly independent.
