## 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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