Linear Algebra and Its Applications, Review Exercise 1.3

Review exercise 1.3. Find a 2 by 2 matrix A for which a_{12} = \frac{1}{2} and (a) A^2 = I, (b) A^{-1} = A^T, and (c) A^2 = A.

Answer: (a) We have a_{12} = \frac{1}{2} and A^2 = I. Since A^2 = I we see that A is its own inverse. Using the formula for the inverse of a 2 by 2 matrix we therefore have

A = \begin{bmatrix} a&\frac{1}{2} \\ c&d \end{bmatrix} = \frac{1}{ad - \frac{c}{2}} \begin{bmatrix} d&-\frac{1}{2} \\ -c&a \end{bmatrix} = A^{-1}

Since \frac{1}{2} = \frac{1}{ad - \frac{c}{2}} \cdot (-\frac{1}{2}), we have ad - \frac{c}{2} = -1. Then a = \frac{d}{ad - \frac{c}{2}} = -d or d = -a. Substituting for d we then have -a^2 - \frac{c}{2} = -1 or c = 2(1 - a^2). So we can freely choose a and then that value determines the values of c and d.

If we choose a = 1 then

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

so that

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

= \begin{bmatrix} 1 \cdot 1 + \frac{1}{2} \cdot 0&1 \cdot \frac{1}{2} + \frac{1}{2} \cdot (-1) \\ 0 \cdot 1 - 1 \cdot 0&0 \cdot \frac{1}{2} + (-1) \cdot (-1) \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

If we choose a = 0 then

A = \begin{bmatrix} 0&\frac{1}{2} \\ 2&0 \end{bmatrix}

and if we choose a = \frac{1}{2} then

A = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{3}{2}&-\frac{1}{2} \end{bmatrix}

There are an infinite number of other matrices for which a_{12} = \frac{1}{2} and A^2 = I, corresponding to other values of a = a_{11}.

(b) We have a_{12} = \frac{1}{2} and A^{-1} = A^T so that

AA^T = \begin{bmatrix} a&\frac{1}{2} \\ c&d \end{bmatrix} \begin{bmatrix} a&c \\ \frac{1}{2}&d \end{bmatrix}

= \begin{bmatrix} a^2 + \frac{1}{4}&ac + \frac{1}{2}d \\ ac + \frac{1}{2}d&c^2 + d^2 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}

Since a^2 + \frac{1}{4} = 1 we have a = \pm \frac{\sqrt{3}}{2}. We choose a = \frac{\sqrt{3}}{2}. Since ac + \frac{1}{2}d = \frac{\sqrt{3}}{2}c + \frac{1}{2}d = 0 we have d = -\sqrt{3}c. Finally, since c^2 + d^2 = c^2 + (-\sqrt{3}c)^2 = c^2 + 3c^2 = 1 we have c^2 = \frac{1}{4} or c = \pm \frac{1}{2}. We choose c = \frac{1}{2}.

Since a = \frac{\sqrt{3}}{2} and c = \frac{1}{2} we thus have

A = \begin{bmatrix} \frac{\sqrt{3}}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{\sqrt{3}}{2} \end{bmatrix}

and

AA^T = \begin{bmatrix} \frac{\sqrt{3}}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

Note that A = A^T in this case so that we also have A^2 = I. This value of A could have been derived from the formulas in (a) above by setting a = \frac{\sqrt{3}}{2}, so that d = -a = -\frac{\sqrt{3}}{2}, and c = 2(1 - a^2) = 2(1 - \frac{3}{4}) = \frac{1}{2}.

If we instead choose c = -\frac{1}{2} then

A = \begin{bmatrix} \frac{\sqrt{3}}{2}&\frac{1}{2} \\ -\frac{1}{2}&\frac{\sqrt{3}}{2} \end{bmatrix}

and

AA^T = \begin{bmatrix} \frac{\sqrt{3}}{2}&\frac{1}{2} \\ -\frac{1}{2}&\frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2}&-\frac{1}{2} \\ \frac{1}{2}&\frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

Finally, if we choose a = -\frac{\sqrt{3}}{2} then since ac + \frac{1}{2}d = -\frac{\sqrt{3}}{2}c + \frac{1}{2}d = 0 we have d = \sqrt{3}c. Since c^2 + d^2 = c^2 + (\sqrt{3}c)^2 = c^2 + 3c^2 = 1 we again have c^2 = \frac{1}{4} or c = \pm \frac{1}{2}.

Thus a = -\frac{\sqrt{3}}{2}, c = \pm \frac{1}{2}, and d = \sqrt{3}c produces two other matrices for which AA^T = I (and also A^2 = I for the first one):

A = \begin{bmatrix} -\frac{\sqrt{3}}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{\sqrt{3}}{2} \end{bmatrix} \qquad A = \begin{bmatrix} -\frac{\sqrt{3}}{2}&\frac{1}{2} \\ -\frac{1}{2}&-\frac{\sqrt{3}}{2} \end{bmatrix}

The above four matrices are the only matrices for which a_{12} = \frac{1}{2} and AA^T = I.

(c) We have a_{12} = \frac{1}{2} and A^2 = A, so that

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

= \begin{bmatrix} a^2 + \frac{1}{2}c&\frac{1}{2}a + \frac{1}{2}d \\ ca + dc&\frac{1}{2}c + d^2 \end{bmatrix} = \begin{bmatrix} a&\frac{1}{2} \\ c&d \end{bmatrix} = A

Since \frac{1}{2}a + \frac{1}{2}d = \frac{1}{2} we have a+d =1 or d = 1-a. Since a^2 + \frac{1}{2}c = a we have c = 2(a - a^2). Thus we can choose any value for a and then c and d will be determined accordingly. For example, if we choose a = 1 we have d = 1 - 1 = 0 and c = 2(1 - 1^2) = 0. We then have

A = \begin{bmatrix} 1&\frac{1}{2} \\ 0&0 \end{bmatrix}

and

A^2 = \begin{bmatrix} 1&\frac{1}{2} \\ 0&0 \end{bmatrix} \begin{bmatrix} 1&\frac{1}{2} \\ 0&0 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + \frac{1}{2} \cdot 0&1 \cdot \frac{1}{2} + \frac{1}{2} \cdot 0\\ 0 \cdot 1 + 0 \cdot \frac{1}{2}&0 \cdot \frac{1}{2} + 0 \cdot 0 \end{bmatrix} = \begin{bmatrix} 1&\frac{1}{2} \\ 0&0 \end{bmatrix} = A

Similarly if we choose a = 0 we have d = 1 - 0 = 1 and c = 2(0 - 0^2) = 0. We then have

A = \begin{bmatrix} 0&\frac{1}{2} \\ 0&1 \end{bmatrix}

and

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

Of particular interest is the matrix for which a = \frac{1}{2}. In that case we have d = 1 - \frac{1}{2} = \frac{1}{2} and c = 2(\frac{1}{2} - (\frac{1}{2})^2) = 2(\frac{1}{2} - \frac{1}{4}) = \frac{1}{2}. We then have

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

and

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}{4}+\frac{1}{4}&\frac{1}{4}+\frac{1}{4} \\ \frac{1}{4}+\frac{1}{4}&\frac{1}{4}+\frac{1}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&\frac{1}{2} \end{bmatrix} = A

There are an infinite number of other matrices for which a_{12} = \frac{1}{2} and A^2 = A, corresponding to other values of a = a_{11}.

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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