Question

For the matrix
A= [0. 1. 0. 0
0. 0. 1. 0
0. 0. 0. 1
1. 0. 0. 0]
What is the smallest positive integer k such that Ak = 1​

Answer 1

Correct Question:-

For the matrix $\smallA=\left[\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{array}\right],$ findthe smallest positive integer k such that $\smallA^k=I$ where I is 4×4 identity matrix.

Solution:-

We need to write A as $\smallA=PDP^{-1}$ where,

• P is matrix consisting of eigenvectors of A in each column.

• D is diagonal matrix with eigenvalues of A as entries.

The eigenvalues λ are given by,

$\small\longrightarrow |A-\lambda I|=0$

$\small\longrightarrow \left|\begin{array}{cccc}-\lambda&1&0&0\\0&-\lambda&1&0\\0&0&-\lambda&1\\1&0&0&-\lambda\end{array}\right|=0$

Expanding along R₁,

$\small\longrightarrow-\lambda\left|\begin{array}{ccc}-\lambda&1&1\\0&-\lambda&1\\0&0&-\lambda\end{array}\right|-\left|\begin{array}{ccc}0&1&0\\0&-\lambda&1\\1&0&-\lambda\end{array}\right|=0$

Expanding along C₁ in each determinant,

$\small\longrightarrow-\lambda\left[-\lambda\left|\begin{array}{cc}-\lambda&1\\0&-\lambda\end{array}\right|\right]-\left[\left|\begin{array}{cc}1&0\\-\lambda&1\end{array}\right|\right]=0$

$\small\longrightarrow\lambda^4-1=0$

$\small\text{ \Longrightarrow\lambda=e^{i\frac{2\pi}{4}r}=e^{i\frac{\pi}{2}r},\quad r\in\mathbb{N},\quad 0\leq r<4 }$

[Recall that the 'n' solutions of the equation xⁿ - a = 0 are given by $\small\text{ x=a^{\frac{1}{n}}e^{-\frac{2\pi}{n}r} }$ where r ∈ N and 0 ≤ r < n.]

$\small\text{ \Longrightarrow\lambda\in\left\{e^{i(0)},\ e^{i\frac{\pi}{2}},\ e^{i\pi},\ e^{i\frac{3\pi}{2}}\right\} }$

$\small\longrightarrow\lambda\in\left\{i,\ -1,\ -i,\ 1\right\}$

Here each λ has AM = 1. Then,

• $\smallD=\left[\begin{array}{cccc}i&0&0&0\\0&-1&0&0\\0&0&-i&0\\0&0&0&1\end{array}\right]$

Let us find eigenvector X for each λ. Since AM of each λ is 1, it is enough to solve and check in general,

$\small\longrightarrow(A-\lambda I)X=O$

$\small\longrightarrow\left[\begin{array}{cccc}-\lambda&1&0&0\\0&-\lambda&1&0\\0&0&-\lambda&1\\1&0&0&-\lambda\end{array}\right]\left[\begin{array}{c}a\\b\\c\\d\end{array}\right]=O$

$\small\longrightarrow\left[\begin{array}{c}-\lambda a+b\\-\lambda b+c\\-\lambda c+d\\-\lambda d+a\end{array}\right]=O$

Equating each entry we get,

$\small\longrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{1}{\lambda}$

$\small\Longrightarrow a:b:c:d=\lambda:\lambda^2:\lambda^3:1$

Thus,

$\small\longrightarrow X=\left[\begin{array}{c}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{c}\lambda\\\lambda^2\\\lambda^3\\1\end{array}\right]$

Then,

$\small\longrightarrow\big(X(i),\ X(-1),\ X(-i),\ X(1)\big)=\left(\left[\begin{array}{c}i\\-1\\-i\\1\end{array}\right],\ \left[\begin{array}{c}-1\\1\\-1\\1\end{array}\right],\ \left[\begin{array}{c}-i\\-1\\i\\1\end{array}\right],\ \left[\begin{array}{c}1\\1\\1\\1\end{array}\right]\right)$

We see each λ has GM = 1. Hence A is diagonalizable.

Now,

• $\smallP=\left[\begin{array}{cccc}i&-1&-i&1\\-1&1&-1&1\\-i&-1&i&1\\1&1&1&1\end{array}\right]$

P⁻¹ is found out in the figure.

• $\smallP^{-1}=\dfrac{1}{4}\left[\begin{array}{cccc}-i&-1&i&1\\-1&1&-1&1\\i&-1&-i&1\\1&1&1&1\end{array}\right]$

Now,

$\small\longrightarrow A=PDP^{-1}$

$\small\longrightarrow A^k=PD^kP^{-1}=I$

$\small\longrightarrow\dfrac{1}{4}\left[\begin{array}{cccc}i&-1&-i&1\\-1&1&-1&1\\-i&-1&i&1\\1&1&1&1\end{array}\right]\left[\begin{array}{cccc}i^k&0&0&0\\0&(-1)^k&0&0\\0&0&(-i)^k&0\\0&0&0&1\end{array}\right]\left[\begin{array}{cccc}-i&-1&i&1\\-1&1&-1&1\\i&-1&-i&1\\1&1&1&1\end{array}\right]=I$

$\small\longrightarrow\dfrac{1}{4}\left[\begin{array}{cccc}i^{k+1}&(-1)^{k+1}&(-i)^{k+1}&1\\-i^k&(-1)^k&-(-i)^k&1\\-i^{k+1}&(-1)^{k+1}&i(-i)^k&1\\i^k&(-1)^k&(-i)^k&1\end{array}\right]\left[\begin{array}{cccc}-i&-1&i&1\\-1&1&-1&1\\i&-1&-i&1\\1&1&1&1\end{array}\right]=I$

The product equals I whose element in 4th row and 4th column equals 1. On considering 4th row of first matrix and 4th column of second matrix in LHS, we get,

$\small\longrightarrow\dfrac{1}{4}[i^k+(-1)^k+(-i)^k+1]=1$

$\small\longrightarrow i^k+(-1)^k+(-i)^k=3$

If k is odd,

$\small\longrightarrow i^k-1-i^k=-1=3$

This is wrong so k must be even.

$\small\longrightarrow i^k+1+i^k=3,\quad 2i^k=2,\quad i^k=1$

$\small\Longrightarrow k=4m,\ m\in\mathbb{Z}$

If k is the smallest possible positive integer,

$\small\text{ \longrightarrow\underline{\underline{k=4}} }$