In this section we analyze under what conditions we can turn the 3rd and the 4th diagonal elements of the last three slices of đ_{j} into zeros.

Our main idea is as follows. For each combination in *T*_{1}, combining with expression (9), we can obtain the relationship between *e*_{p}, *g*_{p} and *h*_{p}, *i*_{p} (*p* = 2, 3, 4). Through the relationship between them, we give conditions that can turn the 3rd and 4th diagonal elements of the last three slices of đ_{j} into zeros. Under these conditions, we can turn đ_{j} into canonical form (6).

#### Theorem 4.1

*If one of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*)* *is not equal to zero*, *the remaining eight of them are equal to zeros*, *there are 6 combinations in* *T*_{1}, *i.e*., *e*_{p}, *g*_{p} *(**p* = 2, 3, 4*)*. *If* *e*_{x} â 0, *the other of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*)* *are equal to zeros*, *under conditions* *i*_{x} = *i*_{y} = *i*_{z} = 0,*h*_{y}*j*_{z} â *j*_{y} *h*_{z}, *or if g*_{x} â 0, *the other of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*)* *are equal to zeros*, *under conditions* *h*_{x} = *h*_{y} = *h*_{z} = 0, *j*_{y}*i*_{z} â *i*_{y} *j*_{z}, *where* *x*, *y*, *z* â {2, 3, 4} *and* *x* â *y* â *z*, *we can turn* đ_{j} *into canonical form (6)*.

#### Proof

Here, we only consider the combination that *e*_{2} â 0, the remaining eight of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combinations that *e*_{3} â 0 (or *e*_{4} â 0 or *g*_{2} â 0 or *g*_{3} â 0 or *g*_{4} â 0), the remaining eight of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros can be proved in a similar way as the combination that *e*_{2} â 0, the remaining eight of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros.

If *e*_{2} â 0, *e*_{3} = *e*_{4} = *f*_{2} = *f*_{3} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we have *i*_{3} = *i*_{4} = 0, then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {h}_{3}& {j}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& {i}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

Because the mode-3 rank of đ_{j} is 4, then *h*_{3}*j*_{4} â *j*_{3}*h*_{4}. Next, we show why the mode-3 rank of đ_{j} is not equal to 4 if *h*_{3}*j*_{4} = *j*_{3}*h*_{4}. Suppose *h*_{3} = 0, then *j*_{3}*h*_{4} = 0. This yields three possibilities. The first case is *j*_{3} = 0,*h*_{4} â 0, then all the entries of the third slice are equal to zeros. The second case is *j*_{3} â 0,*h*_{4} = 0, if *j*_{4} = 0, then all the entries of the fourth slice are equal to zeros; if *j*_{4} â 0, by subtracting *j*_{4}/*j*_{3} times the third slice from the fourth slice (or *j*_{3}/*j*_{4} times the fourth slice from the third slice), we can turn *j*_{4} (or *j*_{3}) into zero, then all the entries of the fourth (or third) slice are equal to zeros. The third case is *j*_{3} = 0,*h*_{4} = 0, then all the entries of the third slice are equal to zeros. The situations where we suppose that *j*_{4} = 0 or *j*_{3} = 0 or *h*_{4} = 0 can be dealt with analogously. Suppose *h*_{3},*j*_{4},*j*_{3},*h*_{4} are nonzero, by subtracting *h*_{4}/*h*_{3} times the third slice from the fourth slice (or *h*_{3}/*h*_{4} times the fourth slice from the third slice), we can turn *h*_{4} and *j*_{4} (or *h*_{3} and *j*_{3}) into zeros. Then all the entries of the third slice (or the fourth slice) are equal to zeros. From the above discussion we can see that, if *h*_{3}*j*_{4} = *j*_{3}*h*_{4}, there always exists a slice whose entries are all equal to zeros. This implies that the mode-3 rank of đ_{j} is 3. Thus we draw the conclusion that *h*_{3}*j*_{4} â *j*_{3}*h*_{4}.

Now we prove that *i*_{2} must be equal to zero, because *i*_{2} can be turned into zero only through (2, 3) or (3, 4) entry of the second slice or through (2, 4) entry of the third or (2, 4) entry of the fourth slice. However, (2, 3) and (3, 4) entries of the second slice, (2, 4) entry of the third and fourth slices are all equal to zeros. This means in the process of transforming đ_{j} into canonical form, *i*_{2} canât be turned into zero. Therefore, we must have *i*_{2} equal to zero.

Now we prove how we turn đ_{j} into canonical form if *e*_{2} â 0,*e*_{3} = *e*_{4} = *f*_{2} = *f*_{3} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, under conditions that *i*_{2} = *i*_{3} = *i*_{4} = 0 and *h*_{3}*j*_{4} â *j*_{3}*h*_{4}. In fact, under these conditions, đ_{j} have the following form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {h}_{3}& {j}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

Firstly, we discuss how to standardize the last two slices. Because *h*_{3}*j*_{4} â *j*_{3}*h*_{4}, suppose *h*_{3} = 0, then *j*_{3}*h*_{4} â 0. If *j*_{4} = 0, then the third slice only has its (1, 4) entry *j*_{3} nonzero, the fourth slice only has its (2, 3) entry *h*_{4} nonzero. We normalize them to one. By exchanging the third slice and the fourth slice, then we obtain the following form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& 1& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$(10)

If *j*_{4} â 0, by subtracting *j*_{4}/*j*_{3} times the third slice from the fourth slice, we can turn *j*_{4} into zero. After this, the third slice only has *j*_{3} is nonzero, the fourth slice only has *h*_{4} is nonzero. Similarly, we normalize them to one. By exchanging the third slice and the fourth slice, then we obtain (10). The situations where we suppose that *j*_{4} = 0 or *j*_{3} = 0 or *h*_{4} = 0 can be dealt with analogously.

Next, we discuss how to standardize the second slice. If *h*_{2} = *j*_{2} = 0, then we have transformed đ_{j} into canonical form. If *h*_{2},*j*_{2} are nonzero, we can turn them into zeros. The specific method is: *h*_{2} can be turned into zero by subtracting *h*_{2} times the third slice from the second slice, *j*_{2} can be turned into zero by subtracting *j*_{2} times the fourth slice from the second slice.

Consequently, if *e*_{x} â 0, the remaining eight of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{x} = *i*_{y} = *i*_{z} = 0 and *h*_{y}*j*_{z} â *j*_{y}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 2, *y* = 4, *z* = 3, we can turn đ_{j} into canonical form.ââĄ

#### Theorem 4.3

*If two of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*) are not equal to zeros*, *the remaining seven of them are equal to zeros*, *there are 15 combinations in* *T*_{1}, *i.e*. *e*_{2}*e*_{3}, *e*_{2}*e*_{4}, *e*_{3}*e*_{4}, *g*_{2}*g*_{3}, *g*_{2}*g*_{4}, *g*_{3}*g*_{4}, *e*_{2}*f*_{2}, *e*_{3}*f*_{3}, *e*_{4}*f*_{4}, *f*_{2}*g*_{2}, *f*_{3}*g*_{3}, *f*_{4}*g*_{4}, *e*_{2}*g*_{2}, *e*_{3}*g*_{3}, *e*_{4}*g*_{4}. *For each combinations*, *we can turn* đ_{j} *into canonical form under the following conditions*:

$$\begin{array}{}{\displaystyle {e}_{x}{e}_{y}\xe2\x890,{i}_{x}={i}_{y}={i}_{z}=0,{\stackrel{~}{h}}_{y}{j}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{h}_{z};}\\ {g}_{x}{g}_{y}\xe2\x890,{h}_{x}={h}_{y}={h}_{z}=0,{\stackrel{~}{i}}_{y}{j}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{i}_{z};\\ {e}_{x}{f}_{x}\xe2\x890,{i}_{y}={i}_{z}=0,{h}_{y}{j}_{z}\xe2\x89{j}_{y}{h}_{z};\\ {f}_{x}{g}_{x}\xe2\x890,{h}_{y}={h}_{z}=0,{i}_{y}{j}_{z}\xe2\x89{j}_{y}{i}_{z};\\ {e}_{x}{g}_{x}\xe2\x890,{i}_{y}={h}_{y}=0,{i}_{z}{h}_{z}\xe2\x890,{j}_{y}\xe2\x890,{h}_{x}{i}_{z}={i}_{x}{h}_{z};\\ ({e}_{x}{g}_{x}\xe2\x890,{i}_{y}{h}_{y}{i}_{z}{h}_{z}\xe2\x890,{h}_{x}{i}_{y}={i}_{x}{h}_{y},{h}_{y}{j}_{z}\xe2\x89{j}_{y}{h}_{z});\\ where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x,y,z\xe2\x88\x88\{2,3,4\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\xe2\x89y\xe2\x89z.\end{array}$$

#### Proof

Firstly, we discuss the combination that *e*_{2}*e*_{3} â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combinations *e*_{2}*e*_{4}, *e*_{3}*e*_{4}, *g*_{2}*g*_{3}, *g*_{2}*g*_{4}, *g*_{3}*g*_{4} can be similar as in the discussion with *e*_{2}*e*_{3}.

If *e*_{2}*e*_{3} â 0, *e*_{4} = *f*_{2} = *f*_{3} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we can obtain *i*_{4} = 0 and *e*_{2}*i*_{3} = *e*_{3}*i*_{2}. Because *e*_{2}*e*_{3} â 0, then *i*_{2} and *i*_{3} are zero or nonzero at the same time. This yields two possibilities.

The first case is *i*_{2} = *i*_{3} = 0. Then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& {e}_{3}& {h}_{3}& {j}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

By subtracting *e*_{3}/*e*_{2} times the second slice from the third slice, we can turn *e*_{3} into zero. Then we obtain the following form:

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {\stackrel{~}{h}}_{3}& {\stackrel{~}{j}}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right],}\end{array}$$

where *hÌ*_{3} = *h*_{3} â *e*_{3}*h*_{2}/*e*_{2},*jÌ*_{3} = *j*_{3} â *e*_{3}*j*_{2}/*e*_{2}. According to Theorem 4.1, under condition *hÌ*_{3}*j*_{4} â *jÌ*_{3}*h*_{4}, we can turn đ_{j} into canonical form. Analogously, by subtracting *e*_{2}/*e*_{3} times the third slice from the second slice, we can turn *e*_{2} into zero, under condition *hÌ*_{2}*j*_{4} â *jÌ*_{2}*h*_{4}, where *hÌ*_{2} = *h*_{2} â *e*_{2}*h*_{3}/*e*_{3}, *jÌ*_{2} = *j*_{2} â *e*_{2}*j*_{3}/*e*_{3}, we can also turn đ_{j} into canonical form.

The second case is *i*_{2}*i*_{3} â 0. By subtracting *e*_{3}/*e*_{2} times the second slice from the third slice, we can turn *e*_{3}, *i*_{3} into zeros according to *e*_{2}*i*_{3} = *e*_{3}*i*_{2}. Then we obtain the following form:

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {\stackrel{~}{h}}_{3}& {\stackrel{~}{j}}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& {i}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right],}\end{array}$$

where *hÌ*_{3} = *h*_{3} â *e*_{3}*h*_{2}/*e*_{2}, *jÌ*_{3} = *j*_{3} â *e*_{3}*j*_{2}/*e*_{2}. According to Theorem 4.1, only under conditions *i*_{2} = 0 and *hÌ*_{3}*j*_{4} â *jÌ*_{3} *h*_{4}, we can turn đ_{j} into canonical form. This contradicts the fact that *i*_{2}*i*_{3} â 0.

Thus we conclude that if *e*_{x}*e*_{y} â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{x} = *i*_{y} = *i*_{z} = 0, *hÌ*_{y}*j*_{z} â *jÌ*_{y} *h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 3, *y* = 2, *z* = 4, we can turn đ_{j} into canonical form (6).

Next, we discuss the combination that *e*_{2}*f*_{2} â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combinations *f*_{2}*g*_{2}, *e*_{3}*f*_{3}, *f*_{3}*g*_{3}, *e*_{4}*f*_{4}, *f*_{4}*g*_{4} can be similar as in the discussion with *e*_{2}*f*_{2}.

If *e*_{2}*f*_{2}â 0, *e*_{3} = *e*_{4} = *f*_{3} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we can obtain *i*_{3} = *i*_{4} = 0. Because the mode-3 rank of đ_{j} is 4, then *h*_{3}*j*_{4}â *j*_{3}*h*_{4}. The way to standardize the last two slices is the same as in Theorem 4.1, so we donât repeat it here. Now, we show how to standardize the second slice. The way to turn *h*_{2} and *j*_{2} into zeros is the same as in Theorem 4.1. It remains to consider how to turn *i*_{2} into zero. In fact, if *i*_{2} = 0, then we have transformed đ_{j} into canonical form. If *i*_{2} â 0, for every slice of đ_{j}, by subtracting *i*_{2}/*f*_{2} times column 3 from column 4 and adding *i*_{2}/*f*_{2} times row 4 to row 3 , we can turn *i*_{2} into zero. Then the (1, 4) entry of slice three is turned into â *i*_{2}/*f*_{2}. Next, by subtracting â *i*_{2}/*f*_{2} times the fourth slice form the third slice, we can turn (1, 4) entry of slice three into zero.

Consequently, if *e*_{x}*f*_{x}â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p}(*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{y} = *i*_{z} = 0 and *h*_{y}*j*_{z}â *j*_{x}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 2, *y* = 4, *z* = 3, we can turn đ_{j} into canonical form.

Next, we discuss the combination that *e*_{2}*g*_{2}â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combinations *e*_{3}*g*_{3}, *e*_{4}*g*_{4} can be similar as in the discussion with *e*_{2}*g*_{2}.

If *e*_{2}*g*_{2}â 0, *e*_{3} = *e*_{4} = *f*_{2} = *f*_{3} = *f*_{4} = *g*_{3} = *g*_{4} = 0, according to (9), we have *e*_{2}*i*_{3} = *h*_{3}*g*_{2}, *e*_{2}*i*_{4} = *h*_{4}*g*_{2}. Because *e*_{2}*g*_{2}â 0, so we must have *i*_{3} and *h*_{3} being zero or nonzero at the same time, *i*_{4} and *h*_{4} being zero or nonzero at the same time. This yields four possibilities.

The first case is *i*_{3} = *h*_{3} = 0, *i*_{4}*h*_{4}â 0. Then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& 0& {j}_{3}& 0& 0& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& {i}_{2}& 0& 0& 0& 0& 0& 0& 0& {i}_{4}\\ 0& 0& 1& 0& 0& 0& 0& {g}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

Because the mode-3 rank of đ_{j} is 4, then *j*_{3} â 0. Next, we normalize *j*_{3} to one, and it can be used to turn *j*_{2} and *j*_{4} into zeros if *j*_{2} and *j*_{4} are nonzero. After this, by exchanging the third slice and the fourth slice, then we can obtain the following form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& 0& 0& 0& {h}_{4}& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& 0& 0& {i}_{2}& 0& 0& 0& {i}_{4}& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& {g}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

If *h*_{2} and *i*_{2} are equal to zeros, then we have turned đ_{j} into canonical form. If one of *h*_{2}, *i*_{2} is equal to zero, we canât turn đ_{j} into canonical form. In fact, if *h*_{2} = 0,*i*_{2} â 0, for every slice, by subtracting *i*_{2}/*g*_{2} times row 3 from row 2 and adding *i*_{2}/*g*_{2} times column 2 to column 3, we can turn *i*_{2} into zero. But meanwhile, *h*_{2} is turned into nonzero. Similarly, if *h*_{2} â 0,*i*_{2} = 0, while turning *h*_{2} into zero, *i*_{2} is turned into nonzero. Similarly to the discussion of the above, if *h*_{2} and *i*_{2} are nonzero, after turn *h*_{2} into zero, *i*_{2} is turned into *Ä©*_{2} = *i*_{2}+*h*_{2}*g*_{2}/*e*_{2}. While turning *Ä©*_{2} into zero, *h*_{2} is turned into nonzero. If we add a restriction condition *h*_{2}*i*_{4} = *i*_{2}*h*_{4}, by subtracting *h*_{4}/*h*_{2} times the third slice from the second slice, we can turn *h*_{2} and *i*_{2} into zeros. Hence, we have turned đ_{j} into canonical form. From the above discussion we can see that, if *i*_{3} = *h*_{3} = 0,*i*_{4}*h*_{4}â 0, only under condition *h*_{2}*i*_{4} = *i*_{2}*h*_{4} we can turn đ_{j} into canonical form.

The second case is *i*_{3}*h*_{3} â 0,*i*_{4} = *h*_{4} = 0. This situation can be similar as in the discussion with *i*_{3} = *h*_{3} = 0,*i*_{4}*h*_{4}â 0.

The third case is *i*_{3} = *h*_{3} = *i*_{4} = *h*_{4} = 0. Then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& 0& {j}_{3}& 0& 0& 0& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& {i}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& {g}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

If one of *j*_{3}, *j*_{4} is equal to zero, then the mode-3 rank of đ_{j} is equal to 3. If *j*_{3} and *j*_{4} are all equal to zeros, then the mode-3 rank of đ_{j} is equal to 2. If *j*_{3} and *j*_{4} are nonzero, by subtracting *j*_{4}/*j*_{3} times the third slice from the fourth slice, we can turn *j*_{4} into zero. Then the mode-3 rank of đ_{j} is equal to 3. Consequently, if *i*_{3} = *h*_{3} = *i*_{4} = *h*_{4} = 0, we canât turn đ_{j} into canonical form.

The fourth case is *i*_{3}*h*_{3}*i*_{4}*h*_{4}â 0. It follows from *e*_{2}*i*_{3} = *h*_{3}*g*_{2}, *e*_{2}*i*_{4} = *h*_{4}*g*_{2} that *i*_{3}*h*_{4} = *i*_{4}*h*_{3}. This means the vectors (*h*_{3}, *i*_{3}) and (*h*_{4}, *i*_{4}) are proportional. If the vectors (*h*_{3}, *j*_{3}) and (*h*_{4}, *j*_{4}) are also proportional, then the mode-3 rank of đ_{j} is not equal to 4. Therefore, we conclude that *h*_{3}*j*_{4} â *h*_{4}*j*_{3}. Similarly to the discussion of the case that *i*_{3} = *h*_{3} = 0,*i*_{4}*h*_{4}â 0. If we add a restriction condition *h*_{2}*i*_{4} = *i*_{2}*h*_{4} or *h*_{2}*i*_{3} = *i*_{2}*h*_{3}, we can turn *h*_{2} and *i*_{2} into zero. Hence, if *i*_{3}*h*_{3}*i*_{4}*h*_{4}â 0, under conditions *h*_{3}*j*_{4} â *h*_{4}*j*_{3}, *h*_{2}*i*_{4} = *i*_{2}*h*_{4} or *h*_{3}*j*_{4} â *h*_{4}*j*_{3}, *h*_{2}*i*_{3} = *i*_{2}*h*_{3}, we can turn đ_{j} into canonical form.

Based on the above argument, we draw the conclusion that if *e*_{x}*g*_{x} â 0, the remaining seven of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{y} = *h*_{y} = 0, *i*_{z}*h*_{z} â 0, *j*_{y} â 0, *h*_{x}*i*_{z} = *i*_{x}*h*_{z} or *i*_{y}*h*_{y}*i*_{z}*h*_{z} â 0, *h*_{x}*i*_{y} = *i*_{x}*h*_{y}, *h*_{y}*j*_{z} â *j*_{y}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 2, *y* = 4, *z* = 3, we can turn đ_{j} into canonical form.ââĄ

#### Theorem 4.4

*If three of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*) are not equal to zeros*, *the remaining six of them are equal to zeros*, *there have 5 combinations in* *T*_{1}, *i.e*. *e*_{2}*e*_{3}*e*_{4}, *g*_{2}*g*_{3}*g*_{4}, *e*_{2}*f*_{2}*g*_{2}, *e*_{3}*f*_{3}*g*_{3}, *e*_{4}*f*_{4}*g*_{4}. *For each combinations*, đ_{j} *can be turned into canonical form under the following conditions*:

$$\begin{array}{}{\displaystyle {e}_{x}{e}_{y}{e}_{z}\xe2\x890,{i}_{x}={i}_{y}={i}_{z}=0,{\stackrel{~}{h}}_{y}{\stackrel{~}{j}}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{\stackrel{~}{h}}_{z};}\\ {g}_{x}{g}_{y}{g}_{z}\xe2\x890,{h}_{x}={h}_{y}={h}_{z}=0,{\stackrel{~}{i}}_{y}{\stackrel{~}{j}}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{\stackrel{~}{i}}_{z};\\ {e}_{x}{f}_{x}{g}_{x}\xe2\x890,{i}_{y}={h}_{y}=0,{i}_{z}{h}_{z}\xe2\x890,{j}_{y}\xe2\x890;\\ ({e}_{x}{f}_{x}{g}_{x}\xe2\x890,{i}_{y}{h}_{y}{i}_{z}{h}_{z}\xe2\x890,{j}_{y}{h}_{z}\xe2\x89{h}_{y}{j}_{z}\phantom{\rule{thinmathspace}{0ex}});\\ where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x,y,z\xe2\x88\x88\{2,3,4\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\xe2\x89y\xe2\x89z.\end{array}$$

#### Proof

Firstly, we discuss the combination that *e*_{2}*e*_{3}*e*_{4} â 0, the remaining six of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combination *g*_{2}*g*_{3}*g*_{4} can be similar as in the discussion with *e*_{2}*e*_{3}*e*_{4}.

If *e*_{2}*e*_{3}*e*_{4}â 0,*f*_{2} = *f*_{3} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we have *e*_{3}*i*_{2} = *i*_{3}*e*_{2}, *e*_{4}*i*_{2} = *e*_{2}*i*_{4}, *e*_{4}*i*_{3} = *e*_{3}*i*_{4}. Because *e*_{2}*e*_{3}*e*_{4} â 0, so we must have *i*_{2}, *i*_{3}, *i*_{4} being zero or nonzero at the same time. This yields two possibilities. The first case is *i*_{2} = *i*_{3} = *i*_{4} = 0. The second case is *i*_{2}*i*_{3}*i*_{4} â 0. Similarly to the discussion of the situation *e*_{2}*e*_{3} â 0 in Theorem 4.3, only for the case that *i*_{2} = *i*_{3} = *i*_{4} = 0 we can turn đ_{j} into canonical form. Now, we show if *i*_{2} = *i*_{3} = *i*_{4} = 0, under what conditions we can turn đ_{j} into canonical form. In fact, if *i*_{2} = *i*_{3} = *i*_{4} = 0, then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& {e}_{3}& {h}_{3}& {j}_{3}& 0& {e}_{4}& {h}_{4}& {j}_{4}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right].}\end{array}$$

Through subtracting *e*_{3}/*e*_{2} times the second slice from the third slice and *e*_{4}/*e*_{2} times the second slice from the fourth slice, we can turn *e*_{3} and *e*_{4} into zeros. Then đ_{j} is of the form

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccccccc}1& 0& 0& 0& 0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {\stackrel{~}{h}}_{3}& {\stackrel{~}{j}}_{3}& 0& 0& {\stackrel{~}{h}}_{4}& {\stackrel{~}{j}}_{4}\\ 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right],}\end{array}$$

where *hÌ*_{3} = *h*_{3} â *e*_{3}*h*_{2}/*e*_{2}, *jÌ*_{3} = *j*_{3} â *e*_{3}*j*_{2}/*e*_{2}, *hÌ*_{4} = *h*_{4} â *e*_{4}*h*_{2}/*e*_{2}, *jÌ*_{4} = *j*_{4} â *e*_{4}*j*_{2}/*e*_{2}. According to the mode-3 rank of đ_{j} is 4, then *hÌ*_{3}*jÌ*_{4} â *jÌ*_{3}*hÌ*_{4}. Thus we draw the conclusion that if *e*_{x}*e*_{y}*e*_{z} â 0, the remaining six of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under the conditions *i*_{x} = *i*_{y} = *i*_{z} = 0, *hÌ*_{y}*jÌ*_{z} â *jÌ*_{y} *hÌ*_{z}, where {*x*, *y*, *z*} â {2, 3, 4} and *x* â *y* â *z*, we can turn đ_{j} into canonical form.

Next, we discuss the combination that *e*_{2}*f*_{2}*g*_{2} â 0, the remaining six of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combinations *e*_{3}*f*_{3}*g*_{3}, *e*_{4}*f*_{4}*g*_{4} can be similar as in the discussion with *e*_{2}*f*_{2}*g*_{2}.

If *e*_{2}*f*_{2}*g*_{2}â 0, *e*_{3} = *e*_{4} = *f*_{3} = *f*_{4} = *g*_{3} = *g*_{4} = 0, according to (9), we have *e*_{2}*i*_{3} = *h*_{3}*g*_{2}, *e*_{2}*i*_{4} = *h*_{4}*g*_{2}. Because *e*_{2}*g*_{2} â 0, then *i*_{3} and *h*_{3} are zero or nonzero at the same time, *i*_{4} and *h*_{4} are zero or nonzero at the same time. This yields four possibilities. The proof of these four cases is almost identical as the case that *e*_{2}*g*_{2} â 0 in Theorem 4.3, the major change is that *h*_{2} can be turned into zero by *e*_{2}, *i*_{2} can be turned into zero by *f*_{2}. Consequently, if *e*_{x}*f*_{x}*g*_{x}â 0, the remaining six of *e*_{p}, *f*_{p}, *g*_{p}(*p* = 2, 3, 4) are zeros, under conditions *i*_{y} = *h*_{y} = 0, *i*_{z}*h*_{z} â 0, *j*_{y} â 0 or *i*_{y}*h*_{y}*i*_{z}*h*_{z} â 0, *h*_{y}*j*_{z} â *j*_{y}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 2, *y* = 4, *z* = 3, we can turn đ_{j} into canonical form.ââĄ

#### Theorem 4.5

*If four of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*) are not equal to zero*, *the remaining five of them are equal to zeros*, *there have 6 combinations in* *T*_{1}, *i.e*. *e*_{2}*e*_{3}*f*_{2}*f*_{3}, *e*_{2}*e*_{4}*f*_{2}*f*_{4}, *e*_{3}*e*_{4}*f*_{3}*f*_{4}, *f*_{2}*f*_{3}*g*_{2}*g*_{3}, *f*_{2}*f*_{4}*g*_{2}*g*_{4}, *f*_{3}*f*_{4}*g*_{3}*g*_{4}. *For each combination*, đ_{j} *can be turned into canonical form under the following conditions*:

$$\begin{array}{}{\displaystyle {e}_{x}{e}_{y}{f}_{x}{f}_{y}\xe2\x890,{i}_{z}=0,{i}_{x},{i}_{y}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}being\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}zero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}or\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}nonzero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}at\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}same\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}time,{\stackrel{~}{h}}_{y}{j}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{h}_{z};}\\ {f}_{x}{f}_{y}{g}_{x}{g}_{y}\xe2\x890,{h}_{z}=0,{h}_{x},{h}_{y}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}being\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}zero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}or\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}nonzero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}at\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}same\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}time,{\stackrel{~}{j}}_{y}{i}_{z}\xe2\x89{\stackrel{~}{i}}_{y}{j}_{z};\\ where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x,y,z\xe2\x88\x88\{2,3,4\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\xe2\x89y\xe2\x89z.\end{array}$$

#### Proof

Here, we only discuss the combination that *e*_{2}*e*_{3}*f*_{2}*f*_{3} â 0, the remaining five of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4)are equal to zeros. The combinations *e*_{2}*e*_{4}*f*_{2}*f*_{4}, *e*_{3}*e*_{4}*f*_{3}*f*_{4}, *f*_{2}*f*_{3}*g*_{2}*g*_{3}, *f*_{2}*f*_{4}*g*_{2}*g*_{4}, *f*_{3}*f*_{4}*g*_{3}*g*_{4} can be similar as in the discussion with *e*_{2}*e*_{3}*f*_{2}*f*_{3}.

If *e*_{2}*e*_{3}*f*_{2}*f*_{3}â 0, *e*_{4} = *f*_{4} = *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we have *i*_{4} = 0 and *e*_{2}*i*_{3} = *e*_{3}*i*_{2}. The proof of this case is almost identical as the combination that *e*_{2}*e*_{3} â 0 in Theorem 4.3, the major change is that *i*_{2} (or *i*_{3}) can be turned into zero by *f*_{2} (or *f*_{3}) if *i*_{2} (or *i*_{3}) is nonzero.

Thus we conclude that if *e*_{2}*e*_{3}*f*_{2}*f*_{3}â 0, the remaining five of *e*_{p}, *f*_{p}, *g*_{p}(*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{z} = 0, *i*_{x}, *i*_{y} being zero or not at the same time, *hÌ*_{y}*j*_{z} â *jÌ*_{y}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4 or *x* = 3, *y* = 2, *z* = 4, we can turn đ_{j} into canonical form.ââĄ

#### Theorem 4.6

*If six of* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*) are not equal to zero*, *the remaining three of them are equal to zeros*, *there we have 5 combinations in* *T*_{1}, *i.e*. *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4}, *f*_{2}*f*_{3}*f*_{4}*g*_{2}*g*_{3}*g*_{4}, *e*_{2}*f*_{2}*g*_{2}*e*_{3}*f*_{3}*g*_{3}, *e*_{2}*f*_{2}*g*_{2}*e*_{4}*f*_{4}*g*_{4}, *e*_{3}*f*_{3}*g*_{3}*e*_{4}*f*_{4}*g*_{4}. *For each combination*, đ_{j} *can be turned into canonical form under the following conditions*:

$$\begin{array}{}{\displaystyle {e}_{x}{e}_{y}{e}_{z}{f}_{x}{f}_{y}{f}_{z}\xe2\x890,{i}_{x},{i}_{y},{i}_{z}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}being\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}zero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}or\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}nonzero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}at\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}same\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}time,{\stackrel{~}{h}}_{y}{\stackrel{~}{j}}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{\stackrel{~}{h}}_{z};}\\ {f}_{x}{f}_{y}{f}_{z}{g}_{x}{g}_{y}{g}_{z}\xe2\x890,{h}_{x},{h}_{y},{h}_{z}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}being\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}zero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}or\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}nonzero\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}at\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}same\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}time,{\stackrel{~}{j}}_{y}{\stackrel{~}{i}}_{z}\xe2\x89{\stackrel{~}{i}}_{y}{\stackrel{~}{j}}_{z};\\ {e}_{x}{f}_{x}{g}_{x}{e}_{y}{f}_{y}{g}_{y}\xe2\x890,{\stackrel{~}{h}}_{y}={\stackrel{~}{i}}_{y}=0,{h}_{z}{i}_{z}\xe2\x890,{\stackrel{~}{j}}_{y}\xe2\x890;\\ ({e}_{x}{f}_{x}{g}_{x}{e}_{y}{f}_{y}{g}_{y}\xe2\x890,{\stackrel{~}{h}}_{y}{\stackrel{~}{i}}_{y}\xe2\x890,{h}_{z}={i}_{z}=0,{j}_{z}\xe2\x890);\\ ({e}_{x}{f}_{x}{g}_{x}{e}_{y}{f}_{y}{g}_{y}\xe2\x890,{\stackrel{~}{h}}_{y}{\stackrel{~}{i}}_{y}{h}_{z}{i}_{z}\xe2\x890,{\stackrel{~}{h}}_{y}{j}_{z}\xe2\x89{\stackrel{~}{j}}_{y}{h}_{z});\\ where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x,y,z\xe2\x88\x88\{2,3,4\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\xe2\x89y\xe2\x89z.\end{array}$$

#### Proof

Firstly, we discuss the combination that *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4} â 0, the remaining three of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros. The combination *f*_{2}*f*_{3}*f*_{4}*g*_{2}*g*_{3}*g*_{4} can be similar as in the discussion with *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4}. If *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4}â 0, *g*_{2} = *g*_{3} = *g*_{4} = 0, according to (9), we have *e*_{2}*i*_{3} = *e*_{3}*i*_{2}, *e*_{4}*i*_{2} = *e*_{2}*i*_{4}, *e*_{4}*i*_{3} = *e*_{3}*i*_{4}. The proof of this combination is almost identical as the combination that *e*_{2}*e*_{3}*e*_{4} â 0 in Theorem 4.4, the major change is that *i*_{2} (or *i*_{3} or *i*_{4}) can be turned into zero by *f*_{2} (or *f*_{3} or *f*_{4}) if *i*_{2} (or *i*_{3} or *i*_{4}) is nonzero. Thus we conclude that if *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4}â 0, the remaining three of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under conditions *i*_{x}, *i*_{y}, *i*_{z} being zero or not at the same time, *hÌ*_{y}*jÌ*_{z} â *jÌ*_{y}*hÌ*_{z}, where *x*, *y*, *y* â {2, 3, 4} and *x* â *y* â *z*, we can turn đ_{j} into canonical form.

Next, we discuss the combination that *e*_{2}*f*_{2}*g*_{2}*e*_{3}*f*_{3}*g*_{3} â 0, the remaining three of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4)are equal to zeros. The combinations *e*_{2}*f*_{2}*g*_{2}*e*_{3}*f*_{3}*g*_{3}, *e*_{2}*f*_{2}*g*_{2}*e*_{4}*f*_{4}*g*_{4}, *e*_{3}*f*_{3}*g*_{3}*e*_{4}*f*_{4}*g*_{4} can be similar as in the discussion with *e*_{2}*f*_{2}*g*_{2}*e*_{3}*f*_{3}*g*_{3}. If *e*_{2}*f*_{2}*g*_{2}*e*_{3}*f*_{3}*g*_{3}â 0, *e*_{4} = *f*_{4} = *g*_{4} = 0, according to (9), we have *e*_{3}*i*_{2} â *e*_{2}*i*_{3}+*h*_{3}*g*_{2} â *g*_{3}*h*_{2} = 0, *e*_{2}*i*_{4} = *h*_{4}*g*_{2} and *e*_{3}*i*_{4} = *h*_{4}*g*_{3}. Because *e*_{2}*e*_{3}*g*_{2}*g*_{3} â 0 and *e*_{3}/*e*_{2} = *g*_{3}/*g*_{2} = *Î±*, then we obtain *e*_{2}*Ä©*_{3} = *g*_{2}*hÌ*_{3}, where *Ä©*_{3} = *i*_{3} â *Î±* *i*_{2}, *hÌ*_{3} = *h*_{3} â *Î±* *h*_{2}. On the other hand, because *e*_{2}*e*_{3}*g*_{2}*g*_{3} â 0, so *hÌ*_{3} and *Ä©*_{3} are zero or nonzero at the same time, *h*_{4} and *i*_{4} are zero or nonzero at the same time. This yields four possibilities. These four situations can be discussed as the situation that *e*_{2}*f*_{2}*g*_{2} â 0, the remaining six of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros.

Consequently, if *e*_{x}*f*_{x}*g*_{x}*e*_{y}*f*_{y}*g*_{y}â 0, the remaining three of *e*_{p}, *f*_{p}, *g*_{p} (*p* = 2, 3, 4) are equal to zeros, under conditions *hÌ*_{y} = *Ä©*_{y} = 0, *h*_{z}*i*_{z} â 0, *jÌ*_{y} â 0, or *hÌ*_{y}*Ä©*_{y} â 0, *h*_{z} = *i*_{z} = 0, *j*_{z} â 0, or *hÌ*_{y}*Ä©*_{y}*h*_{z}*i*_{z} â 0, *hÌ*_{y}*j*_{z} â *jÌ*_{y}*h*_{z}, where *x* = 2, *y* = 3, *z* = 4, or *x* = 2, *y* = 4, *z* = 3, we can turn đ_{j} into canonical form.ââĄ

#### Theorem 4.7

*If* *e*_{p}, *f*_{p}, *g*_{p} *(**p* = 2, 3, 4*) are all nonzero*, đ_{j} *can be turned into canonical form under the following conditions*:

$$\begin{array}{}{\displaystyle {e}_{x}{e}_{y}{e}_{z}{f}_{x}{f}_{y}{f}_{z}{g}_{x}{g}_{y}{g}_{z}\xe2\x890,{\stackrel{~}{h}}_{y}={\stackrel{~}{i}}_{y}=0,{\stackrel{~}{h}}_{z}={\stackrel{~}{i}}_{z}\xe2\x890,{\stackrel{~}{j}}_{y}\xe2\x890;}\\ ({e}_{x}{e}_{y}{e}_{z}{f}_{x}{f}_{y}{f}_{z}{g}_{x}{g}_{y}{g}_{z}\xe2\x890,{\stackrel{~}{h}}_{y}{\stackrel{~}{i}}_{y}{\stackrel{~}{h}}_{z}{\stackrel{~}{i}}_{z}\xe2\x890,{\stackrel{~}{j}}_{y}{\stackrel{~}{h}}_{z}\xe2\x89{\stackrel{~}{h}}_{y}{\stackrel{~}{j}}_{z});\\ where\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x,y,z\xe2\x88\x88\{2,3,4\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\xe2\x89y\xe2\x89z.\end{array}$$

#### Proof

According to the second conclusion of Property 2.1, if *e*_{p}, *f*_{p}, *g*_{p} ( *p* = 2, 3, 4 ) are all nonzero, the vectors (*e*_{p}, *f*_{p}, *g*_{p}) (*p* = 2, 3, 4) being proportional. Then we can turn *e*_{3}, *e*_{4}, *f*_{3}, *f*_{4}, *g*_{3}, *g*_{4} into zeros due to the vectors (*e*_{p}, *f*_{p}, *g*_{p}) (*p* = 2, 3, 4) are proportional. Then we obtain the following for the last three slices of đ_{j}:

$$\begin{array}{}{\displaystyle \left[\begin{array}{cccccccccccc}0& {e}_{2}& {h}_{2}& {j}_{2}& 0& 0& {\stackrel{~}{h}}_{3}& \stackrel{~}{{j}_{3}}& 0& 0& {\stackrel{~}{h}}_{4}& {\stackrel{~}{j}}_{4}\\ 0& 0& {f}_{2}& {i}_{2}& 0& 0& 0& {\stackrel{~}{i}}_{3}& 0& 0& 0& {\stackrel{~}{i}}_{4}\\ 0& 0& 0& {g}_{2}& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]}\end{array}$$

where *hÌ*_{3} = *h*_{3} â *Î±* *h*_{2}, *jÌ*_{3} = *j*_{3} â *Î±* *j*_{2}, *hÌ*_{4} = *h*_{4} â *ÎČ* *h*_{2}, *jÌ*_{4} = *j*_{4} â *ÎČ* *j*_{2} and *Î±* = *e*_{3}/*e*_{2}, *ÎČ* = *e*_{4}/*e*_{2}.

Now we show that equality *e*_{2}*Ä©*_{3} = *g*_{2}*hÌ*_{3} and *e*_{2}*Ä©*_{4} = *g*_{2}*hÌ*_{4} holds. Firstly, we write *e*_{3} = *Î±* *e*_{2}, *g*_{3} = *Î±* *g*_{2}, *e*_{4} = *ÎČ* *e*_{2}, *g*_{4} = *ÎČ* *g*_{2}. Next, by substituting them into the first two equations of (9), we obtain *e*_{2}*Ä©*_{3} = *g*_{2}*hÌ*_{3} and *e*_{2}*Ä©*_{4} = *g*_{2}*hÌ*_{4}. Because *e*_{2}*g*_{2} â 0, then *Ä©*_{3} and *hÌ*_{3} are zero or nonzero at the same time, *Ä©*_{4} and *hÌ*_{4} are zero or nonzero at the same time. This yields four possibilities. These four situation can be discussed just as the situation that *e*_{2}*e*_{3}*e*_{4}*f*_{2}*f*_{3}*f*_{4} â 0, the remaining three of *e*_{p}, *f*_{p}, *g*_{p}(*p* = 2, 3, 4) are equal to zeros.

Consequently, if *e*_{x}*f*_{x}*g*_{x}*e*_{y}*f*_{y}, *g*_{y}*e*_{z}*f*_{z}*g*_{z} are nonzero, under conditions *hÌ*_{y} = *Ä©*_{y} = 0,*hÌ*_{z} = *Ä©*_{z} â 0,*jÌ*_{y} â 0 or *hÌ*_{y}*Ä©*_{y}*hÌ*_{z}*Ä©*_{z} â 0, *jÌ*_{y}*hÌ*_{z} â *hÌ*_{y}*jÌ*_{z}, we can turn đ_{j} into canonical form.ââĄ

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