## Linear Algebra and Its Applications, exercise 1.4.23

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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