Find values of $a$ and $b$ that make matrix orthogonal

Question

Given the matrix

$$A=\begin{bmatrix}1/2&a\\b&1/2\\ \end{bmatrix}$$

find the values of $a$ and $b$ that make it orthogonal.

So far I have tried using dot product $$(1/2)a+(1/2)b=0$$ and we can conclude that $a=-b$ and $b=-a$. I also tried the following theorem

$$A^T=A^{-1}$$

so

$$\begin{bmatrix}1/2&b\\a&1/2\\ \end{bmatrix}= \begin{bmatrix} (2+\frac{4ab}{1-4ab})&\frac{-4a}{1-4ab}\\ \frac{-4b}{1-4ab}&\frac{2}{1-4ab}\\ \end{bmatrix}$$

Can someone tell if I'am on the right track and point me in the right direction? Thanks!

Answer 1

Orthogonal matrix means $$AA^T=I$$ Hence, $$\begin{bmatrix} \frac{1}{2} & a\\ b& \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{2} & b\\ a& \frac{1}{2} \end{bmatrix}=\begin{bmatrix} \frac{1}{4}+a^2 & \frac{1}{2}(a+b)\\ \frac{1}{2}(a+b)& \frac{1}{4}+b^2 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0& 1\end{bmatrix}$$ which implies $$a=-b=\pm\frac{\sqrt{3}}{2}$$

Answer 2

A square matrix $A$ is orthogonal iff $A^TA=AA^T=I$. Computing $A^TA$, we have $$ A^TA = \begin{bmatrix} \frac12& b\\ a &\frac12 \end{bmatrix}\begin{bmatrix} \frac12& a\\ b &\frac12 \end{bmatrix} = \begin{bmatrix} \frac14 + b^2& \frac12(a+b)\\ \frac12(a+b) &\frac14 + a^2. \end{bmatrix} $$ From $A^TA$ we obtain the system of equations \begin{align} \frac14 + b^2 &= 1\\ \frac14 + a^2 &= 1\\ \frac12 (a+b) &= 0, \end{align} which has solution set $$(a,b) = \pm\left(\frac{\sqrt3}2, -\frac{\sqrt3}2 \right).$$

Answer 3

We know $A^TA=I$ where $I$ is the identity matrix. So applying this, we have $a^2+0.25=1$ giving us $a= \pm \frac{\sqrt 3}{2}$. Next, $\frac{b} {2}+\frac{a}{2}=0$ giving us $b=-a$.

Answer 4

The following answer is very close to the previous ones but

• it uses your first attempt;
• it exempts us to calculate the $2\times2$ matrix $A^TA$ or $AA^T.$

$A$ is orthogonal iff:

• its two columns are orthogonal i.e., as you correctly calculated: $\frac12a+\frac12b=0,$
• and each column has norm $1,$ i.e. $$\frac14+b^2=a^2+\frac14=1.$$ The solutions are therefore $$(a,b)=\pm\left(\frac{\sqrt3}2,-\frac{\sqrt3}2\right).$$

Answer 5

Another way, since all second order orthogonal (rotation) matrix are in the form

$$A_\theta=\begin{bmatrix}\cos \theta& -\sin \theta\\\sin \theta&\cos\theta\\ \end{bmatrix}$$

and since in this case

$$\cos\theta =\frac12 \implies \theta=\pm\frac\pi 3 +2k\pi \implies \sin\theta =\pm\frac{\sqrt 3}2$$

therefore

$$A=\begin{bmatrix}\frac12& \pm\frac{\sqrt 3}2\\\mp\frac{\sqrt 3}2&\frac12\\ \end{bmatrix}$$

Answer 6

I'll answer in two parts.

1. Since your title is ambiguous, "Find values of a and b that make matrix orthogonal", it seems appropriate to account for the situation where the matrix is real and to underline that it is not valid for a complex matrix.
2. Secondly, you ask the question: "Can someone tell if I'm on the right track and point me in the right direction?"

A real square matrix is said to be orthogonal if its transpose is equal to its inverse. In other words, a real matrix A is orthogonal if A^T = A^-1, where A^T denotes the transpose of A and A^-1 denotes the inverse of A.

This condition guarantees that the dot product of different rows (or columns, in the case of a square matrix) will be zero, which means that they are orthogonal to each other. Additionally, the dot product of a row with itself (or a column with itself) will be one, which means that each row (or column) is of unit length, or normalized.

For complex matrices, the concept of orthogonality is generalized to unitarity. A complex square matrix U is said to be unitary if its conjugate transpose U^H is also its inverse. In other words, U*U^H = I, where U^H is the conjugate transpose (also called Hermitian transpose) of U, * represents matrix multiplication, and I is the identity matrix.

Example: $$ U = \begin{bmatrix} 2+3i & 5-2i \\ 4-i & 1+i \end{bmatrix} $$

$$ U^{T} = \begin{bmatrix} 2+3i & 4-i \\ 5-2i & 1+i \end{bmatrix} $$

$$ U^{H} = \begin{bmatrix} 2-3i & 4+i \\ 5+2i & 1-i \end{bmatrix} $$

For your matrix, it seems to be implied that you are asked to find real values for a and b, such that A is orthogonal.

We have the matrix A, defined as:

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

Given A^T = A^-1, we can multiply with A on each side, such that:

$$AA^{T} = AA^{-1}$$

A matrix multiplied by its inverse gives, by definition, the identity matrix:

$$AA^{-1} = I$$

$$\therefore AA^{T}=I$$

The transpose of a matrix is obtained by interchanging its rows and columns. That is, the element in the i-th row and j-th column of the original matrix becomes the element in the j-th row and i-th column of the transposed matrix. For some generalised 2 x 2 matrix A:

$$A = \begin{bmatrix} a & b \\ c & d \ \end{bmatrix}$$

We find the transpose:

$$A^{T} = \begin{bmatrix} a & c \\ b & d \ \end{bmatrix}$$

Applying this to your matrix A:

$$A^{T} =\begin{bmatrix} \frac{1}{2} & b \\ a & \frac{1}{2} \\ \end{bmatrix}$$

A matrix product for some arbitrary 2 x 2 matrix is given by:

$$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}\cdot \begin{bmatrix} w & y \\ z & x \\ \end{bmatrix} = \begin{bmatrix} aw + by & ax + bz \\ cw + dy & cx + dz \end{bmatrix} $$

The matrix product for your particular matrix and its transpose are given by:

$$\begin{bmatrix} \frac{1}{2} & a \\ b & \frac{1}{2} \\ \end{bmatrix}\cdot \begin{bmatrix} \frac{1}{2} & b \\ a & \frac{1}{2} \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{4} + a^2 & \frac{1}{2}(a + b) \\ \frac{1}{2}(a + b) & \frac{1}{4} + b^2 \end{bmatrix} $$

From here, you can take each equation in the matrix product and set them equal to the corresponding value in the corresponding matrix index in the identity matrix.

For the equations on the diagonal, we have:

$$ \frac{1}{4} + a^2 = 1\\ \frac{1}{4} + b^2 = 1 $$ For the non-diagonals, we get:

$$\frac{1}{2}(a + b) = \frac{1}{2}(a + b) = 0$$

Since the first two equations each consist of only one unknown per equation, they can be solved by simple rearrangement. The non-diagonal equation can be solved in terms of either a or b. It also shows the relation, which could be interested depending on the situation. In terms of a:

$$\frac{1}{2}(a+b) = 0 \rightarrow a+b = 0 \rightarrow a = -b $$

In an exam setting, where computers are allowed (nearly all of my exams prior to graduating have permitted the usage of computers), it would be convenient to have a solver prepared, instead of having to calculate this by hand. Therefore, I have included some Mathematica code, which performs all of the actions above. This would also help future questioners to quickly verify whether they have done it correctly or not, for their own matrix.

n = 2;
(*Size of the matrix,replace with the desired value*)
A = Transpose[{{1/2, a}, {b, 1/2}}].{{1/2, a}, {b, 1/2}};
(*Example nxn matrix,replace with your own matrix*)
MatrixForm[A] (*Add this line to display the matrix A*)
lowerHalfEquations = Table[A[[i, j]], {i, 2, n}, {j, 1, i - 1}];
upperHalfEquations = Table[A[[i, j]], {i, 2, n}, {j, 1, i - 1}];
lowerHalfEquations;
individualEquations = # == 0 & /@ Flatten[lowerHalfEquations];
individualEquations
n = Length[A];(*Size of the matrix*)equations = 
 Join[Table[A[[i, i]] == 1, {i, n}], individualEquations];
solutions = Solve[equations, Variables[A]];
If[Length[solutions] > 0, 
 MatrixForm /@ solutions, "No solutions found."]

Also works with, say, a 3 x 3 matrix:

    n = 3;
(*Size of the matrix,replace with the desired value*)
A = Transpose[{{a, -a, 0}, {b, b, -b}, {x, y, z}}].{{a, -a, 0}, {b, 
     b, -b}, {x, y, z}};
(*Example nxn matrix,replace with your own matrix*)
MatrixForm[A] (*Add this line to display the matrix A*)
lowerHalfEquations = Table[A[[i, j]], {i, 2, n}, {j, 1, i - 1}];
upperHalfEquations = Table[A[[i, j]], {i, 2, n}, {j, 1, i - 1}];
lowerHalfEquations;
individualEquations = # == 0 & /@ Flatten[lowerHalfEquations];
individualEquations
n = Length[A];(*Size of the matrix*)equations = 
 Join[Table[A[[i, i]] == 1, {i, n}], individualEquations];
solutions = Solve[equations, Variables[A]];
If[Length[solutions] > 0, 
 MatrixForm /@ solutions, "No solutions found."]

1. Output for your 2 x 2 Matrix
2. Output for my 3 x 3 example