Find {eq}\displaystyle A^n {/eq} for {eq}\displaystyle n = 1,\ 2,\ 3,\ \cdots {/eq} for {eq}\displaystyle A = \begin {bmatrix}0& -1\\ 1& 0 \end{bmatrix} {/eq}.

Question:

Find {eq}\displaystyle A^n {/eq} for {eq}\displaystyle n = 1,\ 2,\ 3,\ \cdots {/eq} for {eq}\displaystyle A = \begin {bmatrix}0& -1\\ 1& 0 \end{bmatrix} {/eq}.

Powers of Matrix:

Let {eq}A {/eq} be any square matrix and {eq}P {/eq} be a non-singular matrix. Then, using diagonalization, we can find out the diagonal matrix {eq}D {/eq}, which is given by the formula {eq}D={{P}^{-1}}AP {/eq}. Using the below-mentioned steps, we can find out the power of a given matrix {eq}A {/eq}:

(1): First, find out the eigenvalues of the given matrix.

(2): Determine the matrix {eq}P {/eq} by determining the eigenvectors of the corresponding eigenvalues of the given matrix.

(3): Use diagonalization {eq}D={{P}^{-1}}AP {/eq} to obtain the diagonal matrix {eq}D {/eq}.

(4): The power matrix will be determined by using the formula {eq}{{A}^{n}}=P{{D}^{n}}{{P}^{-1}} {/eq}.

Answer and Explanation: 1

Find out the matrix {eq}A-\lambda I {/eq} by substituting the values of {eq}A {/eq} and {eq}I {/eq} into the equation {eq}A-\lambda I {/eq}.

{eq}\begin{aligned} A-\lambda I&=\left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]-\lambda \left[ \begin{array}{ccc}1&0\\0&1\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]-\left[ \begin{array}{ccc}\lambda&0\\0&\lambda\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}0-\lambda&-1-0\\1-0&0-\lambda\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}-\lambda&-1\\1&-\lambda\\\end{array} \right] \\ \end{aligned} {/eq}

Calculate the determinant of {eq}A-\lambda I {/eq} by using the formula {eq}\det \left[ \begin{array}{ccc}a\; & \;b\;\\c\; & \;d\;\\\end{array} \right]\;=\;ad-bc {/eq}.

{eq}\begin{aligned} \left| A-\lambda I \right|&=\left( -\lambda \right)\left( -\lambda \right)-\left( -1 \right)\left(1 \right) \\ &={{\lambda }^{2}}+1 \\ \end{aligned} {/eq}

Find out the roots of the equation {eq}\left| A-\lambda I \right|=0 {/eq} by substituting {eq}{{\lambda }^{2}}+1 {/eq} for {eq}\left| A-\lambda I \right| {/eq}.

{eq}\begin{aligned} {{\lambda }^{2}}+1&=0 \\ {{\lambda }^{2}}&=-1 \\ \lambda &=\sqrt{-1} \\&=\pm i \\ \end{aligned} {/eq}

Thus, the eigenvalues of {eq}A {/eq} are {eq}-i {/eq} and {eq}i {/eq}.

Let {eq}{{X}_{1}}=\left[ \begin{array}{ccc}{{x}_{1}}\\{{y}_{1}}\\\end{array} \right] {/eq} be an eigenvector with respect to the eigenvalue {eq}\lambda=-i {/eq}, so {eq}\left( A+iI \right){{X}_{1}}=\mathbf{0} {/eq}.

Substitute the values of {eq}A {/eq}, {eq}I {/eq}, and {eq}{{X}_{1}} {/eq} into the equation {eq}\left( A+iI \right){{X}_{1}}=\mathbf{0} {/eq} and simplify.

{eq}\begin{aligned} \left( \left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]+i\left[ \begin{array}{ccc}1&0\\0&1\\\end{array} \right] \right)\left[ \begin{array}{ccc}{{x}_{1}}\\{{y}_{1}}\\\end{array} \right]&=\mathbf{0} \\ \left( \left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]+\left[ \begin{array}{ccc}i&0\\0&i\\\end{array} \right] \right)\left[ \begin{array}{ccc}{{x}_{1}}\\{{y}_{1}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}0+i&-1-0\\1-0&0+i\\\end{array} \right]\left[ \begin{array}{ccc}{{x}_{1}}\\{{y}_{1}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}i&-1\\1&i\\\end{array} \right]\left[ \begin{array}{ccc}{{x}_{1}}\\{{y}_{1}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}i{{x}_{1}}-{{y}_{1}}\\{{x}_{1}}+i{{y}_{1}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \end{aligned} {/eq}

From the obtained matrix, it can be observed that {eq}i{{x}_{1}}-{{y}_{1}}=0 {/eq} and {eq}{{x}_{1}}+i{{y}_{1}}=0 {/eq}. Thus, {eq}{{x}_{1}}=-i{{y}_{1}} {/eq}.

Let {eq}{{y}_{1}}=1 {/eq}, so {eq}{{x}_{1}}=-i {/eq}. Thus, the eigenvector is {eq}{{X}_{1}}=\left[ \begin{array}{ccc}-i\\1\\\end{array} \right] {/eq} for the eigenvalue {eq}-i {/eq}.

Let {eq}{{X}_{2}}=\left[ \begin{array}{ccc}{{x}_{2}}\\{{y}_{2}}\\\end{array} \right] {/eq} be an eigenvector with respect to the eigenvalue {eq}\lambda=i {/eq}, so {eq}\left( A-iI \right){{X}_{2}}=\mathbf{0} {/eq}.

Substitute the values of {eq}A {/eq}, {eq}I {/eq}, and {eq}{{X}_{2}} {/eq} into the equation {eq}\left( A-iI \right){{X}_{2}}=\mathbf{0} {/eq} and simplify.

{eq}\begin{aligned} \left( \left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]-i\left[ \begin{array}{ccc}1&0\\0&1\\\end{array} \right] \right)\left[ \begin{array}{ccc}{{x}_{2}}\\{{y}_{2}}\\\end{array} \right]&=\mathbf{0} \\ \left( \left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]-\left[ \begin{array}{ccc}i&0\\0&i\\\end{array} \right] \right)\left[ \begin{array}{ccc}{{x}_{2}}\\{{y}_{2}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}0-i&-1-0\\1-0&0-i\\\end{array} \right]\left[ \begin{array}{ccc}{{x}_{2}}\\{{y}_{2}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}-i&-1\\1&-i\\\end{array} \right]\left[ \begin{array}{ccc}{{x}_{2}}\\{{y}_{2}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \left[ \begin{array}{ccc}-i{{x}_{2}}-{{y}_{2}}\\{{x}_{2}}-i{{y}_{2}}\\\end{array} \right]&=\left[ \begin{array}{ccc}0\\0\\\end{array} \right] \\ \end{aligned} {/eq}

From the obtained matrix, it can be observed that {eq}-i{{x}_{2}}-{{y}_{2}}=0 {/eq} and {eq}{{x}_{2}}-i{{y}_{2}}=0 {/eq}. Thus, {eq}{{x}_{2}}=i{{y}_{2}} {/eq}.

Let {eq}{{y}_{2}}=1 {/eq}, so {eq}{{x}_{2}}=i {/eq}. Thus, the eigenvector is {eq}{{X}_{2}}=\left[ \begin{array}{ccc}i\\1\\\end{array} \right] {/eq} for the eigenvalue {eq}i {/eq}.

From the obtained eigenvector, it can be observed that {eq}{{X}_{1}}=\left[ \begin{array}{ccc}-i\\1\\\end{array} \right] {/eq} and {eq}{{X}_{2}}=\left[ \begin{array}{ccc}i\\1\\\end{array} \right] {/eq}. Thus, {eq}P=\left[ \begin{array}{ccc}-i & i\\1 & 1\\\end{array} \right] {/eq}.

Find out the inverse of the matrix {eq}P {/eq} by using the formula {eq}{{\left[ \begin{array}{ccc}a\; & \;b\;\\c\; & \;d\;\\\end{array} \right]}^{-1}}=\dfrac{1}{\det \left[ \begin{array}{ccc}a\; & \;b\;\\c\; & \;d\;\\\end{array} \right]}\left[ \begin{array}{ccc}d\; & \;-b\;\\-c\; & \;a\;\\\end{array} \right] {/eq}.

{eq}\begin{aligned} {{P}^{-1}}&=\dfrac{1}{\det \left[ \begin{array}{ccc}-i&i\\1&1\\\end{array} \right]}\left[ \begin{array}{ccc}1&-i\\-1&-i\\\end{array} \right] \\ &=\dfrac{1}{\left( -i \right)\left( 1 \right)-\left( i \right)\left( 1 \right)}\left[ \begin{array}{ccc}1&-i\\-1&-i\\\end{array} \right] \\ &=\dfrac{1}{-2i}\left[ \begin{array}{ccc}1&-i\\-1&-i\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}\dfrac{i}{2}&\dfrac{1}{2}\\-\dfrac{i}{2}&\dfrac{1}{2}\\\end{array} \right] \\ \end{aligned} {/eq}

Substitute the value of {eq}A {/eq} and the obtained values of {eq}P {/eq} and {eq}{{P}^{-1}} {/eq} into the equation {eq}D={{P}^{-1}}AP {/eq}, and simplify.

{eq}\begin{aligned} D&=\left[ \begin{array}{ccc}\dfrac{i}{2}&\dfrac{1}{2}\\-\dfrac{i}{2}&\dfrac{1}{2}\\\end{array} \right]\left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right]\left[ \begin{array}{ccc}-i&i\\1&1\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}\dfrac{1}{2}&-\dfrac{i}{2}\\\dfrac{1}{2}&\dfrac{i}{2}\\\end{array} \right]\left[ \begin{array}{ccc}-i&i\\1&1\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}-i&0\\0&i\\\end{array} \right] \\ \end{aligned} {/eq}

Since {eq}D {/eq} is a diagonal matrix, {eq}{{D}^{n}} {/eq} is also a diagonal matrix. Thus, {eq}{{D}^{n}}=\left[ \begin{array}{ccc}{{\left( {{a}_{1}} \right)}^{n}} & 0\\0 & {{\left( {{a}_{2}} \right)}^{n}}\\\end{array} \right] {/eq}.

Hence, {eq}{{D}^{n}}=\left[ \begin{array}{ccc} {{\left( -i \right)}^{n}} & 0\\0 & {{ i }^{n}} \\\end{array} \right] {/eq}.

By using the formula of the power matrix, {eq}{{A}^{n}}=P{{D}^{n}}{{P}^{-1}} {/eq}, find out the value of {eq}{{A}^{n}} {/eq} by substituting the obtained values of {eq}P {/eq}, {eq}{{P}^{-1}} {/eq}, and {eq}{{D}^{n}} {/eq} into the equation {eq}{{A}^{n}}=P{{D}^{n}}{{P}^{-1}} {/eq}.

{eq}\begin{aligned} {{A}^{n}}&=\left[ \begin{array}{ccc}-i&i\\1&1\\\end{array} \right]\left[ \begin{array}{ccc}{{\left( -i \right)}^{n}}&0\\0&{{i}^{n}}\\\end{array} \right]\left[ \begin{array}{ccc}\dfrac{i}{2}&\dfrac{1}{2}\\-\dfrac{i}{2}&\dfrac{1}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}{{\left( -i \right)}^{n+1}}&{{i}^{n+1}}\\{{\left( -i \right)}^{n}}&{{i}^{n}}\\\end{array} \right]\left[ \begin{array}{ccc}\dfrac{i}{2}&\dfrac{1}{2}\\-\dfrac{i}{2}&\dfrac{1}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}\dfrac{{{\left( -i \right)}^{n}}-{{i}^{n+2}}}{2}&\dfrac{{{\left( -i \right)}^{n+1}}+{{i}^{n+1}}}{2}\\\dfrac{i{{\left( -i \right)}^{n}}-{{i}^{n+1}}}{2}&\dfrac{{{\left( -i \right)}^{n}}+{{i}^{n}}}{2}\\\end{array} \right] \\ \end{aligned} {/eq}

Thus, {eq}{{\mathbf{A}}^{\mathbf{n}}}\mathbf{=}\left[ \begin{array}{ccc}\dfrac{{{\left( \mathbf{-i} \right)}^{\mathbf{n}}}\mathbf{-}{{\mathbf{i}}^{\mathbf{n+2}}}}{\mathbf{2}} & \dfrac{{{\left( \mathbf{-i} \right)}^{\mathbf{n+1}}}\mathbf{+}{{\mathbf{i}}^{\mathbf{n+1}}}}{\mathbf{2}}\\\dfrac{\mathbf{i}{{\left( \mathbf{-i} \right)}^{\mathbf{n}}}\mathbf{-}{{\mathbf{i}}^{\mathbf{n+1}}}}{\mathbf{2}} & \dfrac{{{\left( \mathbf{-i} \right)}^{\mathbf{n}}}\mathbf{+}{{\mathbf{i}}^{\mathbf{n}}}}{\mathbf{2}}\\\end{array} \right] {/eq}.

It is given that {eq}A=\left[ \begin{array}{ccc}0&-1\\1&0\\\end{array} \right] {/eq}. Thus, {eq}{{\mathbf{A}}^{\mathbf{n}}}\mathbf{=}\left[ \begin{array}{ccc}\mathbf{0} & \mathbf{-1}\\\mathbf{1} & \mathbf{0}\\\end{array} \right] {/eq} for {eq}\mathbf{n=1} {/eq}.

Substitute {eq}2 {/eq} for {eq}n {/eq} into the equation {eq}{{A}^{n}}=\left[ \begin{array}{ccc}\dfrac{{{\left( -i \right)}^{n}}-{{i}^{n+2}}}{2} & \dfrac{{{\left( -i \right)}^{n+1}}+{{i}^{n+1}}}{2}\\\dfrac{i{{\left( -i \right)}^{n}}-{{i}^{n+1}}}{2} & \dfrac{{{\left( -i \right)}^{n}}+{{i}^{n}}}{2}\\\end{array} \right] {/eq} and simplify.

{eq}\begin{aligned} {{A}^{2}}&=\left[ \begin{array}{ccc}\dfrac{{{\left( -i \right)}^{2}}-{{i}^{2+2}}}{2}&\dfrac{{{\left( -i \right)}^{2+1}}+{{i}^{2+1}}}{2}\\\dfrac{i{{\left( -i \right)}^{2}}-{{i}^{2+1}}}{2}&\dfrac{{{\left( -i \right)}^{2}}+{{i}^{2}}}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}\dfrac{\left( -1 \right)-\left( 1 \right)}{2}&\dfrac{\left( i \right)+\left( -i \right)}{2}\\\dfrac{i\left( -1 \right)-\left( -i \right)}{2}&\dfrac{\left( -1 \right)+\left( -1 \right)}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}-1&0\\0&-1\\\end{array} \right] \\ \end{aligned} {/eq}

Thus, {eq}{{\mathbf{A}}^{\mathbf{n}}}\mathbf{=}\left[ \begin{array}{ccc}\mathbf{-1} & \mathbf{0}\\\mathbf{0} & \mathbf{-1}\\\end{array} \right] {/eq} for {eq}\mathbf{n=2} {/eq}.

Substitute {eq}3 {/eq} for {eq}n {/eq} into the equation {eq}{{A}^{n}}=\left[ \begin{array}{ccc}\dfrac{{{\left( -i \right)}^{n}}-{{i}^{n+2}}}{2} & \dfrac{{{\left( -i \right)}^{n+1}}+{{i}^{n+1}}}{2}\\\dfrac{i{{\left( -i \right)}^{n}}-{{i}^{n+1}}}{2} & \dfrac{{{\left( -i \right)}^{n}}+{{i}^{n}}}{2}\\\end{array} \right] {/eq} and simplify.

{eq}\begin{aligned} {{A}^{3}}&=\left[ \begin{array}{ccc}\dfrac{{{\left( -i \right)}^{3}}-{{i}^{3+2}}}{2}&\dfrac{{{\left( -i \right)}^{3+1}}+{{i}^{3+1}}}{2}\\\dfrac{i{{\left( -i \right)}^{3}}-{{i}^{3+1}}}{2}&\dfrac{{{\left( -i \right)}^{3}}+{{i}^{3}}}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}\dfrac{\left( i \right)-\left( i \right)}{2}&\dfrac{\left( 1 \right)+\left( 1 \right)}{2}\\\dfrac{i\left( i \right)-\left( 1 \right)}{2}&\dfrac{\left( i \right)+\left( -i \right)}{2}\\\end{array} \right] \\ &=\left[ \begin{array}{ccc}0&1\\-1&0\\\end{array} \right] \\ \end{aligned} {/eq}

Thus, {eq}{{\mathbf{A}}^{\mathbf{n}}}\mathbf{=}\left[ \begin{array}{ccc}\mathbf{0} & \mathbf{1}\\\mathbf{-1} & \mathbf{0}\\\end{array} \right] {/eq} for {eq}\mathbf{n=3} {/eq}.


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Diagonalization: Definition & Example

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Chapter 6 / Lesson 4
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Diagonalization is the process of transforming a matrix into diagonal form. Diagonal matrices represent the eigenvalues of a matrix in a clear manner. This lesson will focus on finding the diagonalized form of a simple matrix.


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