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This is an exercise from linear algebra and optimization by Gill, I do exercises to be prepared for my final exam and these are not homework questions!

Exercise $\mathbf{6.1.}\,$ Consider the following matrix $A$ and vector $b$: $$ A=\begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \\ \end{pmatrix},\quad b=\begin{pmatrix} 3 \\ 2 \\ 1 \\ \end{pmatrix}. $$ $\textbf{(a)}$ What is the rank of $A$? Give a general form for any vector in the range of $A$.

$\textbf{(b)}$ Show that the dimension of the null space of $A^T$ is two, and display two linearly independent vectors $z_1$ and $z_2$ in $\operatorname{null}(A^T)$. Give a general form for every vector in $\operatorname{null}(A^T)$.

$\textbf{(c)}$ Find the vectors $b_R\in\operatorname{range}(A)$ and $b_N\in\operatorname{range}(A^T)$ such that $b=b_R+b_N$.

$\textbf{(d)}$ Give the general form of $b_A$ such that $b_R=Ab_A$. (Hint: consider all vectors $q$ such that $Aq=0$.)

For part (a), I think $rank(A)=1$ since all the columns are linear combination of each other. As for a general form for any vector in the range of $A$, when I write $Ax=b$ I get an over-determined system:

$$2x_1+4x_2=3$$ $$x_1+2x_2=2$$ $$x_1+2x_2=1$$

So I'm not sure about the general form.

As for part (b), since the rank is two, dimension of $N(A^T)$ must be $3-1=1$, right?

And no idea about the rest. Any help would be greatly appreciated.

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  • $\begingroup$ If you know row reduction and properties of row reduced matrix then apply. It will be easier to get all answer together. $\endgroup$ – Dutta Jan 10 '14 at 10:37
  • $\begingroup$ Your solution to (a) is correct. For (b) note that if the dimension of the null space is two and you already found two linearly independent vectors in the null space, those form a base. Now that directly gives you a general form of all null vectors. $\endgroup$ – Listing Jan 10 '14 at 10:38
  • $\begingroup$ For (c) note that $\text{null}(A^T)$ are exactly the vectors that are not in the image of $A$ (aka $\text{range}(A)$). The image of $A$ gets spanned by $(2,1,1)$ thus you can write $b$ as $\alpha z_1+\beta z_2 + \gamma (2,1,1)$. where $z_1,z_2$ are from (b). $\endgroup$ – Listing Jan 10 '14 at 10:41
  • $\begingroup$ @Listing: Thank you, that helped to understand the problem. $\endgroup$ – Gigili Jan 10 '14 at 12:07
  • $\begingroup$ Thank you for the edit @Aðøbe. $\endgroup$ – Gigili Jan 11 '14 at 2:14
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If you do row reduction, you find $$ \begin{pmatrix} 2 & 4 \\ 1 & 2 \\ 1 & 2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 \\ 0 & 0 \\ 0 & 0 \end{pmatrix} $$ which means that the first column is a basis for the column space of $A$ (which is better terminology than “range of $A$”, in my opinion). So the general form of the vectors in the column space of $A$ is $$ \begin{pmatrix} 2a \\ a \\ a \end{pmatrix},\quad \text{$a$ any scalar} $$

By the rank nullity theorem, the null space of $A$ has dimension $1$; the equation defining it is $$ x_1+2x_2=0 $$ so a basis for it is the single vector $$ \begin{pmatrix} -2 \\ 1 \end{pmatrix} $$

The null space of $A^T$ has indeed dimension $2$; the row reduction on $A^T$ is $$ \begin{pmatrix} 2 & 1 & 1 \\ 4 & 2 & 2 \end{pmatrix} \to \begin{pmatrix} 2 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix} $$ so the equation defining the null space is $2x_1+x_2+x_3=0$ and a basis for it is $$ \left\{ \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \,, \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \right\} $$ Writing $b=b_R+b_N$ should now be easy: the system to solve is $$ \left(\begin{array}{ccc|c} 2 & -1 & -1 & 3 \\ 1 & 2 & 0 & 2 \\ 1 & 0 & 2 & 1 \end{array}\right) $$ but you can as well find the orthogonal projection of $b$ on the column space of $A$: $$ b_R= \frac{(2\ 1\ 1)\begin{pmatrix}3\\2\\1\end{pmatrix}} {(2\ 1\ 1)\begin{pmatrix}2\\1\\1\end{pmatrix}} \begin{pmatrix}2\\1\\1\end{pmatrix} =\frac{9}{6}\begin{pmatrix}2\\1\\1\end{pmatrix} =\begin{pmatrix}3\\3/2\\3/2\end{pmatrix} $$ and $b_N=b-b_R$.

With this last idea it's easy to solve the last point.

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  • $\begingroup$ Thank you for your detailed answer. Where did you use the fact that the first column is a basis for the column space (because it has leading $1$, right?)? And how did you come up with that general form, $\begin{pmatrix} 2a \\ a \\ a \end{pmatrix}$? $\endgroup$ – Gigili Jan 10 '14 at 11:56
  • $\begingroup$ @Gigili I used it for the last part, when computing $b_R$ with the orthogonal projection. Since the first column is a basis for the column space, every vector in the column space is that vector times a scalar. $\endgroup$ – egreg Jan 10 '14 at 11:59
  • $\begingroup$ Ah, right! I was thinking of $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ as the first column but it's the first column in REF. Thanks, it's clear now. $\endgroup$ – Gigili Jan 10 '14 at 12:04
  • $\begingroup$ Sorry, but I still don't get the last part about $b$. Where did that system come from? Did you multiply the basis by $-1$? Then what's the first row? How did you calculate $b_R$? $\endgroup$ – Gigili Jan 10 '14 at 12:07
  • $\begingroup$ @Gigili It's a common error, I find it all the time in students' papers. ;-) The first non zero column in REF is always that one! It can't be in the column space of every matrix, can it? $\endgroup$ – egreg Jan 10 '14 at 12:07

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