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Sometimes, if you want students to manipulate some sort of special matrices (like, asking if they are invertible or ask for a LU decomposition, etc.), you have to possibility to state all properties of that special matrix in a rigorous way. Moreover, since it is a matrix, you can additionally show a picture of that matrix with a lot of $\ldots$ in it.

The problem, however, is that for some students the picture is helpful; for other students, it is distracting or misleading.

An example question:

Let $A$ be a tridiagonal matrix (i.e., $a_{ij}=0$ for $j>i+1$ and for $i>j+1$) with an non zero element $a_{n1}\neq 0$, i.e., the matrix looks like $$A = \begin{pmatrix} * & * & 0 & \ldots & \ldots & 0 \\ * & \ddots & \ddots & \ddots & & \vdots \\ 0 & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & & \ddots & \ddots & \ddots & * \\ * & 0 & \ldots & 0 & * & * \end{pmatrix}$$ where $*$ means that on this location, there is an arbitrary real number which can be different from zero. Let $LU=A$ be a LU-decomposition of $A$. Perform an LU decomposition of $A$ in order to find the locations in $L$ and $U$ which are zero for sure.

Several students manipulated the matrix and arrived at some point to calculate things like $$\frac{*}{*}=1, \quad *-*=0$$ etc. which is not true since the interpret $*$ as a fixed number.

How can one avoid such misinterpretations of the picture? What went wrong when students perform calculations like above? Should one avoid pictures like the above? Should one write even more text to make sure the meaning of $*$?

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  • $\begingroup$ You should explicitly tell the students that $\ast$ just means that the entry is an arbitrary element of the ring (may be, field) over which this matrix is defined and that each $\ast$ is possibly a different element! Then, the computations arising from this mistaken understanding will be minimised. $\endgroup$ – kan Apr 7 '14 at 12:56
  • $\begingroup$ @kan I wrote that explicitly in the question, but that was within a lot of other text clarifying other things as well - Students tend to look only at the picture then. $\endgroup$ – Markus Klein Apr 7 '14 at 13:04
  • $\begingroup$ @MarkusKlein It might help if you could give the full text of the question. $\endgroup$ – Jim Belk Apr 7 '14 at 14:26
  • $\begingroup$ @JimBelk I've added (the translation into English of) the whole question. $\endgroup$ – Markus Klein Apr 7 '14 at 15:34
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Well, here's some possibly relevant experience. I gave the following question on a recent (in-class) exam in a sopohmore linear algebra course.

Suppose that $$ \newcommand{\?}{\scriptsize\text{?}\,} A \;=\; \left[\begin{array}{@{\;\;}r@{\;\;}r@{\;\;}r@{\;\;}} 2 & \? & \phantom{+}x \\[6pt] \? & 4 & \? \\[6pt] \? & -1 & \? \end{array}\right] \qquad\text{and}\qquad A^{-1} \;=\; \left[\begin{array}{@{\;\;}r@{\;\;}r@{\;\;}r@{\;\;}} 7 & \? & 3 \\[6pt] \? & \phantom{+}2 & \phantom{+}0 \\[6pt] 1 & \? & 1 \end{array}\right], $$ where each ? indicates an unknown entry. Use this information to find the value of $x$.

Between the use of question marks and the explanatory text, I didn't seem to have any trouble with students misinterpreting the question.

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    $\begingroup$ There was really no confusion over whether all the ?'s were the same number? I would think you would have to write something like "where each ? indicates a (possibly different) unknown entry." $\endgroup$ – Chris Cunningham♦ Apr 7 '14 at 15:22
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    $\begingroup$ @ChrisCunningham I didn't experience any. I've only used the question once. $\endgroup$ – Jim Belk Apr 7 '14 at 16:49
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The pitfall of this notation is: In mathematics, if something has the same (variable) name, it is the same thing. Because of that, this notation needs to be read in a relaxed way, which probably only comes when you have some kind of rough understanding of the topic already.

This is why I'd say it depends on the context if such a notation is a good choice. I think it's a nice choice to illustrate an example or something similar in a lecture or tutorium. In this context it can be made clear that this is an illustration, and if there is something unclear, you can talk about it.

For an exam or exercise sheet, there is no easy way to communicate that this matrix is just a relaxed illustration and not an actual written-down version of the matrix. I think for situations like this, it's better to be on the pedantic side and write

$$A = \begin{pmatrix} a_{11} & a_{12} & 0 & \ldots & \ldots & 0 \\ a_{21} & \ddots & \ddots & \ddots & & \vdots \\ 0 & \ddots & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & & \ddots & \ddots & \ddots & a_{n-1 \ n-1} \\ a_{n\ 1} & 0 & \ldots & 0 & a_{n \ n-1} & a_{n n} \end{pmatrix}.$$

The extra effort in typing is not too much, but the benefit is a huge one.

That having said, it surprises me as well that your students regarded $*$ as an actual name for a variable.

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My approach: Give them a matrix with random entries, and ask for the zeroes in the decomposition of that particular matrix.

You should be able to generate a suitably generic matrix after a few tries. If you use Mathematica, you can test this as:

$$\mathtt{ B = Table[Random[Integer, \{1, 9\}], \{6\}, \{6\}] }$$ $$\mathtt{ C = \{\{1, 1, 0, 0, 0, 0\}, \{1, 1, 0, 0, 0, 0\}, \{0, 0, 0, 0, 0, 0\},\\ \{0, 0, 0, 0, 0, 0\}, \{0, 0, 0, 0, 1, 1\}, \{1, 0, 0, 0, 1, 1\}\} }$$ $$\mathtt{A = B\ C}$$ $$\mathtt{ LUDecomposition[A] }$$

Find an $A$ whose decomposition has zeroes only in the right places, and then just give them that $A$, e.g. $$ 2\ 5\ 0\ 0\ 0\ 0 \\ 6\ 3\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 0\ 0 \\ 0\ 0\ 0\ 0\ 1\ 3 \\ 6\ 0\ 0\ 0\ 2\ 5 $$

Note 1: The definition of $A$ here used the Mathematica code for element-wise product, where $\mathtt{B.C}$ would have given the matrix product instead.

Note 2: I think the example shows that I understood the pattern incorrectly, but I hope the approach is clear anyway!

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  • $\begingroup$ But this then becomes an entirely different, and much easier, exercise. $\endgroup$ – Tobias Kildetoft Apr 7 '14 at 13:37
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    $\begingroup$ @TobiasKildetoft Or even more complicated since people tend to calculate numbers instead of calculating if it is either zero or not. $\endgroup$ – Markus Klein Apr 7 '14 at 13:43
  • $\begingroup$ Also, you have to calculate the entries in case some of the presumed non-zero entries are zero by coincidence. (You might avoid this by chosing “relatively trancendent“ entries such as $\pi, e, 2^\pi$, but that might not be a good idea either.) $\endgroup$ – Wrzlprmft Apr 7 '14 at 14:48
  • $\begingroup$ @Wrzlprmft, specifying lots of algebraically independent entries is distracting and inelegant for a test. That's why I recommend generating random matrices with simple entries, and checking to see that what results is sufficiently generic -- if not, roll the dice again. $\endgroup$ – user173 Apr 7 '14 at 15:00
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    $\begingroup$ @MattF.: 1) As I stated, I did not consider this a good idea. 2) The problem is not to select matrices such that there are no coincidental zeros but that your students cannot know this and thus have to calculate the entries. $\endgroup$ – Wrzlprmft Apr 7 '14 at 15:07

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