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If $$ A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$$

Then find $A^n$

I have tried solving it using diagonalization $PDP^{-1}$ but I am getting only one independent eigenvector i.e $$ \begin{bmatrix} 2\\ 1 \end{bmatrix}$$ Please tell me the correct method to solve it.

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You can split the matrix $A$ into $B+ I$ where $B= \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$ Now it is interesting to observe that $B^2=O$, the null matrix So using binomial $A^n=(I+B)^n= C^n_0 I^n + C^n_1 I^{n-1} B + C^n_2 I^{n-2} B^2 +\cdots+ C^n_n B^n$ Which simplifies to: $A^n= I + nB$ :)

Note: we are able to use binomial this way because, one of the matrix is identity matrix and the product becomes commutative.

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Indeed, $A$ has one and only one eigenvalue: $1$. And $(2,1)$ is an eigenvector corresponding to that eigevalue. Now, consider the vector $(1,0)$. Then$$A.(1,0)=(3,1)=(2,1)+(1,0).$$So, if $M=\left[\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right]$, then$$M^{-1}.A.M=\begin{bmatrix}1&1\\0&1\end{bmatrix},$$or$$A=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}.M^{-1}.$$So,\begin{align}A^n&=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}^n.M^{-1}\\&=\begin{bmatrix}2&1\\1&0\end{bmatrix}.\begin{bmatrix}1&n\\0&1\end{bmatrix}.\begin{bmatrix}0&1\\1&-2\end{bmatrix}\\&=\begin{bmatrix}2n+1&-4n\\n&1-2n\end{bmatrix}.\end{align}

Note that the equality $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]^n=\left[\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right]$ has to do with the fact that$$(\forall x,y\in\mathbb R):\begin{bmatrix}1&x\\0&1\end{bmatrix}.\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}1&x+y\\0&1\end{bmatrix}.$$

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  • $\begingroup$ Can be good to mention this $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ is a very famous and important matrix. It turns matrix multiplication into addition for the off-diagonal element: $\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&b\\0&1\end{bmatrix} = \begin{bmatrix}1&a+b\\0&1\end{bmatrix}$. $\endgroup$ – mathreadler Apr 25 at 13:25
  • $\begingroup$ @lhf I'e added it back. Thank you. $\endgroup$ – José Carlos Santos Apr 25 at 13:26
  • $\begingroup$ You hit 21 reviewed edits again... Awesome! $\endgroup$ – YuiTo Cheng Apr 25 at 14:21
  • $\begingroup$ @mathreadler Nice suggestion. I've added that to my answer. $\endgroup$ – José Carlos Santos Apr 25 at 16:03
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This matrix is not diagonalizable. Start multiplying and look for patterns. E.g.,

\begin{align*} \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix}^2 &= \begin{bmatrix} 5 & -8 \\ 2 & -3\end{bmatrix} \\ \begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix}^3 &= \begin{bmatrix} 7 & -12 \\ 3 & -5\end{bmatrix} \\ &\vdots \end{align*} Does this give you any ideas?

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  • $\begingroup$ Yes, I have tried squaring it, but I wanted to know is there any another approach. $\endgroup$ – sawan kumawat Apr 25 at 13:17
  • $\begingroup$ My answer gives the derivation of exact pattern $\endgroup$ – Tojrah Apr 25 at 13:37
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$$|A-\lambda I| = (3-\lambda)(-1-\lambda) - 1\times(-4)\\= 4-3-2\lambda+\lambda^2 \\= 1-2\lambda+\lambda^2=\\(1-\lambda)^2$$

so $\lambda =1$ double root.

Which eigenspace?

$$\text{null}(A-\lambda I)\\\begin{bmatrix}2&-4\\1&-2\end{bmatrix}$$

We see $[2,1]^T$ is eigenvector but also there can be only one because of degeneracy.

Therefore $A$ is not diagonalizable. You have to try find some other type of matrix than diagonal matrix for $D$ to be if you want to put $A$ on form $PDP^{-1} =A$. For example you can try

  1. Jordan Canonical Form or
  2. Smith's form
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Since you have a defective eigenvalue, to proceed with your method you’ll need to find a generalized eigenvector and put the matrix into Jordan normal form. However, there’s no need to go to all that trouble since powers of the matrix can be computed without working out the full decomposition.

Using the same method as the one in my answer to another of your questions (in fact, the two problems are congruent) we write $A^n=aI+bA$ and solve the system of equations generated by substituting $A$’s eigenvalues for $A$: $$a+b=1 \\ b=n,$$ (the second equation is obtained by differentiating $a+b\lambda=\lambda^n$) therefore, as in your other question, $A^n=nA-(n-1)I$.

More generally, if the repeated eigenvalue is $\lambda$ and $A$ is not a multiple of the identity, we get the system of equations $$a+b=\lambda^n \\ b=n\lambda^{n-1}$$ so $$A^n=(1-n)\lambda^n I+n\lambda^{n-1} A = \lambda^n I+n\lambda^{n-1}(A-\lambda I).$$ From Cayley-Hamilton, we know that $(A-\lambda I)^2=0$, so this result can also be obtained by writing $A$ as the sum of the identity and the nilpotent matrix $N=A-I$ and then applying the Binomial Theorem, as has been suggested in other answers. This is much less work that computing generalized eigenvectors and then performing tedious matrix multiplications to find $A^n$.

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