Linear Algebra and Its Applications, exercise 1.4.23

Posted on August 16, 2010 by hecker

Exercise 1.4.23. Find the results of squaring, cubing, and in general raising to the power of n the following matrices:

A = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} \quad B = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} \quad C = AB = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2} \end{bmatrix}

Answer: For the matrix A we have

A^2 = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} = A

A^3 = A^2A = AA = A^2 = A

By induction we have A^n = A for all n.

For the matrix B we have

B^2 = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

B^3 = B^2B = IB = B

By induction we have B^n = I if n is even and B^n = B if n is odd.

For the matrix C we have

C^2 = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2} \end{bmatrix} \begin{bmatrix} \frac{1}{2}&-\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} = 0

C^3 = C^2C = 0 \cdot C = 0

By induction C^n = 0, n \ge 2.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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