Prove that $A^n = nA - (n-1)I$

Question

Let $$ A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ Prove that $$A^n = nA - (n-1)I$$ where $I$ is identity matrix.

I have tried solving it using the Cayley-Hamilton theorem, getting $$A^{2n} = {(2A- I)}^n$$ However, I am unable to convert it to the desired form.

Answer 1

A consequence of the Cayley-Hamilton theorem is that any analytic function $f$ of $A$ can be written in the form $aI+bA$ for some scalars $a$ and $b$. It’s also the case that if $\lambda$ is an eigenvalue of $A$, then $f(\lambda)=a+b\lambda$. This gives you a system of linear equations for $a$ and $b$. If there are repeated eigenvalues, differentiate to generate enough independent equations.

By inspection, the only eigenvalue of $A$ is $1$, so we have the system $$a+b=1 \\ b=n,$$ from which $A^n = (1-n)I+nA = nA-(n-1)I$.

If you’re not familiar with eigenvalues, then split $A$ into $I+N$, where $N^2=0$, and apply the Binomial Theorem, as suggested in other answers.

Answer 2

Since $A^2=2A-I$, we get

$$A^3 = A \cdot A^2 = A \cdot (2A - I) = 2A^2 - A = 2(2A - I) - A = 3A - 2I$$ $$A^4 = A \cdot A^3 = A \cdot (3A - 2I) = 3A^2 - 2A = 3(2A - I) - 2A = 4A - 3I$$

Can you get it from here?

Answer 3

Hint: Use induction and Cayley–Hamilton with $A^2=2A-I$: $$ A^{n+1} = A^n A = (nA - (n-1)I) A = n A^2 -(n-1)A = \cdots $$

Answer 4

Write $A=I+B$. Then $B^2=0$ and so the binomial theorem gives $$ A^n = (I+B)^n = I + nB = I + n(A-I) = nA - (n-1)I $$

Answer 5

I am sure this question has a duplicate, however, since I am not able to find the duplicate I give you here an answer.

My suggestion is to use induction. For $n=1$, you have that $A^1=A=A-(1-1)I=A$, hence the base case works.

Suppose it holds for $n$, we are going to prove for $n+1$.

$$A^{n+1}=AA^n=A\big(nA-(n-1)I\big)=nA^2-(n-1)A=nA^2-nA+A$$

since $A^2=2A-I$, you get $$n(2A-I)-nA+A=nA-nI+A=(n+1)A-\big((n+1)-1\big)I=(n+1)A-nI$$

Answer 6

For $n\ge 3$, using $$ (x+1)^n=x^2P(x)+nx+1 $$ where $P(x)$ is a polynomial with degree $n-2$, one has $$ x^n=(x-1)^2P(x-1)+n(x-1)+1. $$ Since $A^2=2A-I$ or $(A-I)^2=0$, one has $$ A^n=(A-1)^2P(A-1)+n(A-I)+I=nA-(n-1)I.$$

Answer 7

The case $n=1$ is obvious and Cayley-Hamilton gives the case $n=2$: Since $A^2-tr(A)A+det(A)I=0$ for a $2\times 2$ matrix, we have $A^2-2A+I=0$, that is $A^2=2A-I$.

For the other exponents, you can use induction: if $A^n=nA-(n-1)I$, then

$$ A^{n+1} = AA^n = A(nA-(n-1)I)=nA^2-(n-1)A=n(2A-I)-(n-1)A $$

so $A^{n+1} =(n+1)A-nI$.

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Here is another method: we have $A=I+N$, where $n= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$. Since $I$ and $N$ commute, we can use the Binomial formula for $A^n=(I+N)^2$, but $N^2=0$ so it simplifies to

$$ A^n = I^n + nI^{n-1}N = I + nN = I + n(A-I) = nA - (n-1)I$$

Answer 8

$A= B+I $where$ B= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$ Notice that here $B^2=O $the null matrix. Using binomial on$ A^n= (B+I)^n=C_0^n I^n + C_1^n I^{n-1} B + C_2^n I^{n-2} B^2 + .......+ C_n^n B^n$ Which simplifies to $A^n=I+ n B = I+ n(A-I) = nA-(n-1)I$

Answer 9

$$A=\begin{bmatrix} 1 & 0 \\1 & 1\\\end{bmatrix}$$ $$A^2=\begin{bmatrix} 1 & 0 \\1 & 1\\\end{bmatrix}\begin{bmatrix} 1 & 0 \\1 & 1\\\end{bmatrix}=\begin{bmatrix} 1 & 0 \\2 & 1\\\end{bmatrix}$$ $$A^3=A^2A=\begin{bmatrix} 1 & 0 \\2 & 1\\\end{bmatrix}\begin{bmatrix} 1 & 0 \\1 & 1\\\end{bmatrix}=\begin{bmatrix} 1 & 0 \\3 & 1\\\end{bmatrix}$$

$n^th$ term

$A^n$=A^(n-1)A$=\begin{bmatrix} 1 & 0 \\n-1 & 1\\\end{bmatrix}\begin{bmatrix} 1 & 0 \\1 & 1\\\end{bmatrix}=\begin{bmatrix} 1 & 0 \\n & 1\\\end{bmatrix}$

and $$nA=\begin{bmatrix} n & 0 \\n & n\\\end{bmatrix}$$

Hence

$$A^n-nA=\begin{bmatrix} 1 & 0 \\n & 1\\\end{bmatrix}-\begin{bmatrix} n & 0 \\n & n\\\end{bmatrix}$$

$$A^n-nA=\begin{bmatrix} 1-n & 0 \\0 & 1-n\\\end{bmatrix}=(1-n)\begin{bmatrix} 1 & 0 \\0 & 1\\\end{bmatrix}$$

$$A^n-nA=(1-n)I$$ $$A^n=nA+(1-n)I$$ $$A^n=nA-(n-1)I$$

$$Hence Proved$$

Answer 10

With so many answers I'm surprised none started by simplifying the right-hand side viz. $A^{n}=\left(\begin{array}{cc} 1 & 0\\ n & 1 \end{array}\right)$, at which point the inductive step is trivial viz. $\left(\begin{array}{cc} 1 & 0\\ k & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 1 & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ k+1 & 1 \end{array}\right)$. In fact, $\left(\begin{array}{cc} 1 & 0\\ a & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0\\ b & 1 \end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ a+b & 1 \end{array}\right)$ gives a well-known isomorphism to the Abelian group $\Bbb R$ (or for that matter anything whose elements you'd use for matrix entries) forms under addition.